toby
/
zotero-db
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity
zotero-db/storage/EVM8PYZR/.zotero-ft-cache

13017 lines
368 KiB
Plaintext
Raw Blame History

This file contains invisible Unicode characters

This file contains invisible Unicode characters that are indistinguishable to humans but may be processed differently by a computer. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

This file contains Unicode characters that might be confused with other characters. If you think that this is intentional, you can safely ignore this warning. Use the Escape button to reveal them.

5th EDITION
PHYSICS
for SCIENTISTS and ENGINEERS
DOUGLAS GIANCOLI

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 1
03/07/20 16:04
Editorial Director: Jeanne Zalesky Content Development: Margy Kuntz, Andrea Giancoli Project Managers: Cynthia Rae Abbott, Elisa Mandelbaum, Francesca Monaco, Karen Misler, Rebecca Dunn Production Vendor: CodeMantra Interior Composition: Preparé Italia, Battipaglia (SA), Italy Copyeditor: Joanna Dinesmore Proofreaders: Andrea Giancoli, Carol Reitz, and Clare Romeo Art House: Lachina Creative Design Managers: Mark Ong, Derek Bacchus, Emily Friel, SPi Global Rights & Permissions Manager: Ben Ferrini SPi Global Photo Researcher: Eric Schrader and Mary Teresa Giancoli Manufacturing Buyer: Stacey Wienberger
Copyright © 2020, 2008, 2000, 1989, 1984 by Douglas C. Giancoli. Published by Pearson Education, Inc. All Rights Reserved. Printed in the United States of America. This publication is protected by copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise. For information regarding permissions, request forms and the appropriate contacts within the Pearson Education Global Rights & Permissions department, please visit www.pearsoned.com/permissions/.
Photo credits appear on page A – 77, which constitutes a continuation of this copyright page.
PEARSON, ALWAYS LEARNING and Mastering™ Physics are exclusive trademarks in the U.S. and/or other countries owned by Pearson Education, Inc. or its affiliates.
Unless otherwise indicated herein, any third-party trademarks that may appear in this work are the property of their respective owners and any references to third-party trademarks, logos or other trade dress are for demonstrative or descriptive purposes only. Such references are not intended to imply any sponsorship, endorsement, authorization, or promotion of Pearson’s products by the owners of such marks, or any relationship between the owner and Pearson Education, Inc. or its affiliates, authors, licensees or distributors.
Library of Congress Cataloging-in-Publication Data
Giancoli, Douglas C., author. Title: Physics for scientists & engineers with modern physics/Douglas C. Giancoli. Other titles: Physics for scientists and engineers with modern physics Description: Fifth edition. | Upper Saddle River, N.J. : Pearson Education, Inc., [2019] | Includes bibliographical references and index.
Contents: Introduction, measurement, estimating — Describing motion: kinematics in one dimension — Kinematics in two or three dimensions; vectors — Dynamics: Newton’s laws of motion — Using Newton’s laws: friction, circular motion, drag forces — Gravitation and Newton’s synthesis — Work and energy — Conservation of energy — Linear momentum — Rotational motion — Angular momentum; general rotation — Static equilibrium; elasticity and fracture — Fluids — Oscillations — Wave motion — Sound — Temperature, thermal expansion, and the ideal gas law — Kinetic theory of gases — Heat and the first law of thermodynamics — Second law of thermodynamics — Electric charge and electric field — Gauss’s law — Electric potential — Capacitance, dielectrics, electric energy storage — Electric currents and resistance — DC circuits — Magnetism — Sources of magnetic field — Electromagnetic induction and Faraday’s law — Inductance, electromagnetic oscillations, and AC circuits — Maxwell’s equations and electromagnetic waves — Light: reflection and refraction — Lenses and optical instruments — The wave nature of light: interference and polarization — Diffraction — The special theory of relativity — Early quantum theory and models of the atom — Quantum mechanics — Quantum mechanics of atoms — Molecules and solids — Nuclear physics and radioactivity — Nuclear energy; effects and uses of radiation — Elementary particles — Astrophysics and cosmology. Identifiers: LCCN 2019015435 | ISBN 9780134378053 (v.1) | ISBN 0134378059 (v.1) Subjects: LCSH: Physics--Textbooks. Classification: LCC QC21.3 .G539 2019 | DDC 530--dc23 LC record available at https://lccn.loc.gov/2019015435
ISBN 10:  0-321-99227-X; ISBN 13:  978-0-32-199227-7 (Student Edition) ISBN 10:  0-134-37806-7; ISBN 13:  978-13-437806-0 (Classic Student Edition) ISBN 10:  0-134-37808-3; ISBN 13:  978-0-13-437808-4 (Looseleaf Edition)

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 2
www.pearson.com
03/07/20 16:04
Contents
Applications List
xii
Preface
xvi
To Students
xx
Use of Color
xxi
  Estimate how much this lake weighs.
1 Introduction, Measurement,
Estimating
1
1 – 1 How Science Works
2
1 – 2 Models, Theories, and Laws
3
1 – 3 Measurement and Uncertainty;
Significant Figures
3
1 –   4  Units, Standards, and the SI System
6
1 – 5 Converting Units
9
1 – 6 Order of Magnitude: Rapid Estimating 11
*1 – 7 Dimensions and Dimensional Analysis 14
Questions, MisConceptions, Problems   15 – 19
2 Describing Motion: Kinematics
in One Dimension
20
2 – 1 Reference Frames and Displacement
21
2 – 2 Average Velocity
22
2 – 3 Instantaneous Velocity
24
2 – 4 Acceleration
27
2 – 5 Motion at Constant Acceleration
30
2 – 6 Solving Problems
33
2 – 7 Freely Falling Objects
37
*2 – 8 Variable Acceleration; Integral Calculus 43
Questions, MisConceptions, Problems   45 – 53
3 Kinematics in Two or Three Dimensions; Vectors 54
3 – 1 Vectors and Scalars
55
3 – 2 Addition of Vectors—Graphical Methods 55
3 – 3 Subtraction of Vectors, and
Multiplication of a Vector by a Scalar
57
3 – 4 Adding Vectors by Components
58
3 – 5 Unit Vectors
62
3 – 6 Vector Kinematics
62
3 – 7 Projectile Motion
65
3 – 8 Solving Problems Involving Projectile
Motion
67
3 – 9 Relative Velocity
73
Questions, MisConceptions, Problems   76 – 84
4 Dynamics: Newton’s Laws of Motion
85
4 – 1 Force
86
4 – 2 Newton’s First Law of Motion
86
4 – 3 Mass
88
4 – 4 Newton’s Second Law of Motion
88
4 – 5 Newton’s Third Law of Motion
91
4 – 6 Weight—the Force of Gravity;
and the Normal Force
94
4 – 7 Solving Problems with Newton’s Laws:
Free-Body Diagrams
97
4 – 8 Problem Solving—A General Approach 104
Questions, MisConceptions, Problems   105 – 115
5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces 116
5 – 1 5 – 2 5 – 3 5 – 4 5 – 5 *5 – 6
Using Newton’s Laws with Friction
117
Uniform Circular Motion—Kinematics 123
Dynamics of Uniform Circular Motion 126
Highway Curves: Banked and Unbanked 130
Nonuniform Circular Motion
133
Velocity-Dependent Forces:
Drag and Terminal Velocity
134
Questions, MisConceptions, Problems   136 – 144
  iii

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 3
03/07/20 16:04
Gravitation and
Newton’s Synthesis
145
6 – 1 6 –   2 
6 – 3 6 – 4 6 – 5
6 –   6  6 – 7 *6 – 8 *6 – 9
Newton’s Law of Universal ­Gravitation 146
Vector Form of Newton’s Law of
Universal Gravitation
149
Gravity Near the Earth’s Surface
149
Satellites and “Weightlessness”
152
Planets, Kepler’s Laws, and
­Newton’s Synthesis
155
Moon Rises an Hour Later Each Day 161
Types of Forces in Nature
161
Gravitational Field
162
Principle of Equivalence;
Curvature of Space; Black Holes
163
Questions, MisConceptions, Problems   165 – 171
7 Work and Energy
172
7 –   1  Work Done by a Constant Force
173
7 – 2 Scalar Product of Two Vectors
176
7 – 3 Work Done by a Varying Force
177
7 – 4 Kinetic Energy and the
Work-Energy Principle
181
Questions, MisConceptions, Problems   186 – 193
8 Conservation of Energy 194
8 – 1 Conservative and ­Nonconservative Forces 195
8 – 2 Potential Energy
197
8 – 3 Mechanical Energy and Its Conservation 200
8 – 4 Problem Solving Using
Conservation of Mechanical Energy
201
8 –   5  The Law of Conservation of Energy
207
8 – 6 Energy Conservation with Dissipative Forces: Solving Problems 208
8 – 7 Gravitational Potential Energy and
Escape Velocity
210
8 – 8 Power
213
8 – 9 Potential Energy Diagrams;
Stable and Unstable ­Equilibrium
215
*8 – 10 Gravitational Assist (Slingshot)
216
Questions, MisConceptions, Problems   218 – 226
Linear Momentum
227
9 –   1  Momentum and Its Relation to Force 228
9 – 2 Conservation of Momentum
230
9 – 3 Collisions and Impulse
234
9 – 4 Conservation of Energy and
Momentum in Collisions
235
9 – 5 Elastic Collisions in One Dimension
236
9 – 6 Inelastic Collisions
239
9 –   7  Collisions in 2 or 3 Dimensions
241
9 – 8 Center of Mass (CM)
244
9 – 9 Center of Mass and Translational Motion 248
*9 – 10 Systems of Variable Mass; Rocket Propulsion 251
Questions, MisConceptions, Problems   254 – 263
10 Rotational Motion
264
10 – 1 Angular Quantities
265
10 – 2 Vector Nature of Angular ­Quantities 270
10 – 3 Constant Angular Acceleration
270
10 – 4 Torque
271
10 – 5 Rotational Dynamics; Torque and
Rotational Inertia
274
10 – 6 Solving Problems in Rotational
Dynamics
276
10 – 7 Determining Moments of Inertia
279
10 – 8 Rotational Kinetic Energy
281
10 – 9 Rotation plus Translational Motion; Rolling 283
*10 –   1  0 Why Does a Rolling Sphere Slow Down? 289
Questions, MisConceptions, Problems   291 – 301
11 Angular Momentum; General Rotation
302
11 – 1 Angular Momentum : Objects
Rotating About a Fixed Axis
303
11 – 2 Vector Cross Product; Torque as a Vector 307
11 – 3 Angular Momentum of a Particle
309
11 – 4 Angular Momentum and Torque for a System of Particles; General Motion 310
11 – 5 Angular Momentum and
Torque for a Rigid Object
312
11 – 6 Conservation of Angular Momentum 315
*11 – 7 The Spinning Top and Gyroscope
317
11 –   8  Rotating Frames of Reference; Inertial Forces 318
*11 – 9 The Coriolis Effect
319
Questions, MisConceptions, Problems   322 – 330
iv  CONTENTS

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 4
03/07/20 16:04
12 Static Equilibrium; Elasticity and Fracture
331
12 – 1 12 – 2 *12 – 3 12 – 4 12 – 5 12 – 6 *12 – 7 *12 – 8
The Conditions for Equilibrium
332
Solving Statics Problems
334
Applications to Muscles and Joints
339
Stability and Balance
341
Elasticity; Stress and Strain
342
Fracture
345
Trusses and Bridges
347
Arches and Domes
350
Questions, MisConceptions, Problems   353 – 364
5FB2 5FB1
m2 5g
m1 5g
13 Fluids
365
13 – 1 Phases of Matter
366
13 – 2 Density and Specif ic Gravity
366
13 – 3 Pressure in Fluids
367
13 – 4 Atmospheric Pressure and
Gauge Pressure
371
13 – 5 Pascal’s Principle
371
13 – 6 Measurement of Pressure;
Gauges and the Barometer
372
13 – 7 Buoyancy and Archimedes’ Principle 374
13 –   8  Fluids in Motion; Flow Rate and
the Equation of Continuity
378
13 – 9 Bernoulli’s Equation
380
13 – 10 Applications of ­Bernoulli’s Principle:
Torricelli, ­Airplanes, Baseballs,
Blood Flow
382
13 – 11 Viscosity
385
*13 – 12 Flow in Tubes: Poiseuille’s Equation,
Blood Flow
385
*13 – 13 Surface Tension and Capillarity
386
*13 – 14 Pumps, and the Heart
388
Questions, MisConceptions, Problems   390 – 398
14 Oscillations
399
14 – 1 14 – 2 14 – 3
14 – 4
14 – 5 *14 – 6
14 – 7 14 – 8
Oscillations of a Spring
400
Simple Harmonic Motion
402
Energy in the Simple
Harmonic Oscillator
408
Simple Harmonic Motion Related
to Uniform Circular Motion
410
The Simple Pendulum
411
The Physical Pendulum and
the Torsion Pendulum
412
Damped Harmonic Motion
414
Forced Oscillations; Resonance
417
Questions, MisConceptions, Problems   420 – 427
15 Wave Motion
428
15 – 1 Characteristics of Wave Motion
429
15 – 2 Types of Waves:
Transverse and Longitudinal
431
15 – 3 Energy Transported by Waves
435
15 – 4 Mathematical Representation of a
Traveling Wave
437
*15 – 5 The Wave Equation
440
15 – 6 The Principle of Superposition
441
15 – 7 Reflection and Transmission
443
15 – 8 Interference
444
15 – 9 Standing Waves; Resonance
446
15 – 10 Refraction
449
15 – 11 Diffraction
450
Questions, MisConceptions, Problems   452 – 459
1 Sound
460
16 – 1 16 – 2
16 – 3 16 – 4
*16 – 5
16 – 6 16 – 7 *16 – 8 *16 – 9
Characteristics of Sound
461
Mathematical Representation
of Longitudinal Waves
462
Intensity of Sound: Decibels
464
Sources of Sound:
Vibrating Strings and Air Columns
467
Quality of Sound, and Noise;
Superposition
472
Interference of Sound Waves; Beats
473
Doppler Effect
476
Shock Waves and the Sonic Boom
480
Applications: Sonar, ­Ultrasound,
and Medical Imaging
481
Questions, MisConceptions, Problems   484 – 491
CONTENTS  v 
GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 5
03/07/20 16:04
Temperature,
17 Thermal Expansion, and the Ideal Gas Law
492
17 – 1 Atomic Theory of Matter
493
17 – 2 Temperature and Thermometers
495
17 – 3 Thermal Equilibrium and the
Zeroth Law of Thermodynamics
497
17 – 4 Thermal Expansion
497
*17 – 5 Thermal Stresses
501
17 – 6 The Gas Laws and
Absolute Temperature
502
17 – 7 The Ideal Gas Law
503
17 – 8 Problem Solving with the
Ideal Gas Law
504
17 –   9  Ideal Gas Law in Terms of ­Molecules:
Avogadro’s Number
506
*17 – 10 Ideal Gas Temperature Scale—
a Standard
507
Questions, MisConceptions, Problems   509 – 515
18 Kinetic Theory of Gases 516
18 –   1 
18 – 2 18 –   3  18 – 4 18 – 5
18 – 6 18 – 7 18 – 8
The Ideal Gas Law and the Molecular
Interpretation of Temperature
516
Distribution of Molecular Speeds
520
Real Gases and Changes of Phase
522
Vapor Pressure and Humidity
524
Temperature Decrease of Boiling Water
with Altitude
526
Van der Waals Equation of State
527
Mean Free Path
528
Diffusion
530
Questions, MisConceptions, Problems   532 – 537
1 Heat and the First Law of Thermodynamics
538
19 – 1 Heat as Energy Transfer
539
19 – 2 Internal Energy
540
19 – 3 Specific Heat
541
19 – 4 Calorimetry—Solving ­Problems
542
19 – 5 Latent Heat
545
19 – 6 The First Law of ­Thermodynamics
549
19 – 7 Thermodynamic Processes and
the First Law
551
19 – 8 Molar Specific Heats for Gases,
and the Equipartition of Energy
556
19 – 9 Adiabatic Expansion of a Gas
559
19 – 10 Heat Transfer: Conduction,
Convection, Radiation
560
Questions, MisConceptions, Problems   568 – 575
20 Second Law of Thermodynamics
576
20 – 1 The Second Law of
Thermodynamics—Introduction
577
20 – 2 Heat Engines
578
20 – 3 The Carnot Engine; Reversible and Irreversible Processes 580
20 – 4 Refrigerators, Air Conditioners, and
Heat Pumps
584
20 – 5 Entropy
587
20 –   6  Entropy and the Second Law of
Thermodynamics
590
20 – 7 Order to Disorder
593
20 – 8 Unavailability of Energy; Heat Death 594
20 – 9 Statistical Interpretation of Entropy
and the Second Law
595
*20 – 10 Thermodynamic Temperature;
Third Law of Thermodynamics
597
20 – 11 Thermal Pollution, Global Warming,
and Energy Resources
598
Questions, MisConceptions, Problems   601 – 608
vi  CONTENTS 
GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 6
03/07/20 16:04
21 Electric Charge and Electric Field
609
21 – 1 Static Electricity; Electric Charge and
Its Conservation
610
21 – 2 Electric Charge in the Atom
611
21 – 3 Insulators and Conductors
611
21 – 4 Induced Charge; the Electroscope
612
21 – 5 Coulomb’s Law
613
21 – 6 The Electric Field
618
21 – 7 Electric Field Calculations for
Continuous Charge Distributions
622
21 – 8 Field Lines
626
21 – 9 Electric Fields and Conductors
627
21 –   1  0 Motion of a Charged Particle in
an Electric Field
628
21 – 11 Electric Dipoles
629
*21 – 12 Electric Forces in Molecular Biology:
DNA Structure and Replication
631
Questions, MisConceptions, Problems   634 – 642
22 Gauss’s Law
643
22 – 1 Electric Flux
644
22 – 2 Gauss’s Law
645
22 – 3 Applications of Gauss’s Law
647
*22 – 4 Experimental Basis of Gauss’s and
Coulomb’s Laws
652
Questions, MisConceptions, Problems   653 – 659
23 Electric Potential
660
23 – 1 Electric Potential Energy and
Potential Difference
661
23 – 2 Relation between Electric Potential
and Electric Field
664
23 – 3 Electric Potential Due to Point Charges 666
23 – 4 Potential Due to Any Charge Distribution 669
23 – 5 Equipotential Lines and Surfaces
670
23 – 6 Potential Due to Electric Dipole;
Dipole Moment
671
23 – 7 E5 Determined from V
672
23 – 8 Electrostatic Potential Energy; the
Electron Volt
674
23 – 9 Digital; Binary Numbers; Signal Voltage 676
*23 – 10 TV and Computer Monitors
679
*23 – 11 Electrocardiogram (ECG or EKG)
682
Questions, MisConceptions, Problems   684 – 691
24 Capacitance, Dielectrics, Electric Energy Storage 692
24 – 1 Capacitors
692
24 – 2 Determination of Capacitance
694
24 – 3 Capacitors in Series and Parallel
698
24 – 4 Storage of Electric Energy
700
24 – 5 Dielectrics
703
*24 – 6 Molecular Description of Dielectrics 706
Questions, MisConceptions, Problems   708 – 716
25 Electric Current and Resistance
717
25 – 1 The Electric Battery
718
25 – 2 Electric Current
720
25 – 3 Ohm’s Law: Resistance and Resistors 722
25 – 4 Resistivity
724
25 – 5 Electric Power
726
25 – 6 Power in Household Circuits
729
25 – 7 Alternating Current
730
25 – 8 Microscopic View of Electric Current 732
*25 – 9 Superconductivity
735
*25 – 10 Electrical Conduction in the Human
Nervous System
736
Questions, MisConceptions, Problems   739 – 746
2 DC Circuits
747
26 – 1 EMF and Terminal Voltage
748
26 –   2  Resistors in Series and in Parallel
749
26 – 3 Kirchhoff’s Rules
754
26 –   4  EMFs in Series and in Parallel;
Charging a Battery
757
26 – 5 RC Circuits—Resistor and Capacitor
in Series
759
26 – 6 Electric Hazards and Safety
764
26 – 7 Ammeters and Voltmeters—Measurement
Affects Quantity Measured
767
Questions, MisConceptions, Problems   771 – 781
CONTENTS  vii

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 7
03/07/20 16:04
27 Magnetism
782
27 – 1 Magnets and Magnetic Fields
782
27 – 2 Electric Currents Produce ­Magnetic
Fields
785
27 –   3  Force on an Electric Current in a
Magnetic Field; Definition of B5
786
27 – 4 Force on an Electric Charge
Moving in a Magnetic Field
788
27 – 5 Torque on a Current Loop;
Magnetic Dipole Moment
793
27 – 6 Applications: Motors, Loudspeakers,
Galvanometers
795
27 – 7 Discovery and Properties of the
Electron
797
27 – 8 The Hall Effect
799
27 – 9 Mass Spectrometer
800
Questions, MisConceptions, Problems   802 – 810
28 Sources of Magnetic Field 811
28 –   1  Magnetic Field Due to a Straight Wire 812
28 – 2 Force between Two Parallel Wires
813
28 – 3 Definitions of the Ampere and the
Coulomb
814
28 – 4 Ampère’s Law
815
28 –   5  Magnetic Field of a Solenoid and
a Toroid
819
28 – 6 Biot-Savart Law
821
28 –   7  Magnetic Field Due to a Single
Moving Charge
824
28 – 8 Magnetic Materials—Ferromagnetism 824
28 – 9 Electromagnets and
Solenoids—Applications
826
28 – 10 Magnetic Fields in Magnetic
Materials; Hysteresis
827
*28 – 11 Paramagnetism and Diamagnetism
828
Questions, MisConceptions, Problems   830 – 837
2 Electromagnetic Induction and
Faraday’s Law
838
29 – 1 Induced EMF
839
29 –   2  Faraday’s Law of Induction; Lenz’s Law 840
29 –   3  EMF Induced in a Moving Conductor 845
29 – 4 Electric Generators
846
29 – 5 Back EMF and Counter Torque;
Eddy Currents
848
29 – 6 Transformers and Transmission of Power 851
29 –   7  A Changing Magnetic Flux Produces an
Electric Field
854
*29 – 8 Information Storage: Magnetic and
Semiconductor
856
*29 – 9 Applications of Induction:
Microphone, Seismograph, GFCI
858
Questions, MisConceptions, Problems   860 – 868
30 Inductance, Electromagnetic Oscillations, and AC Circuits 869
30 – 1 Mutual Inductance
870
30 – 2 Self-Inductance; Inductors
872
30 –   3  Energy Stored in a Magnetic Field
874
30 – 4 LR Circuits
875
30 – 5 LC Circuits and Electromagnetic
Oscillations
877
30 – 6 LC Oscillations with Resistance
(LRC Circuit)
880
30 – 7 AC Circuits and Reactance
881
30 – 8 LRC Series AC Circuit; Phasor Diagrams 885
30 – 9 Resonance in AC Circuits
887
30 – 10 Impedance Matching
888
*30 – 11 Three-Phase AC
889
Questions, MisConceptions, Problems   890 – 897
31 Maxwell’s Equations and Electromagnetic Waves 898
31 – 1 Changing Electric Fields Produce
Magnetic Fields; Displacement
Current
899
31 – 2 Gauss’s Law for Magnetism
902
31 – 3 Maxwell’s Equations
903
31 – 4 Production of Electromagnetic Waves 903
31 – 5 Electromagnetic Waves, and Their Speed,
Derived from Maxwell’s Equations
905
31 – 6 Light as an Electromagnetic Wave
and the Electromagnetic Spectrum
909
31 – 7 Measuring the Speed of Light
912
31 – 8 Energy in EM Waves; the Poynting Vector 913
31 – 9 Radiation Pressure
915
31 – 10 Radio and Television;
Wireless Communication
917
Questions, MisConceptions, Problems   921 – 925
viii  CONTENTS

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 8
03/07/20 16:04
32 Light: Reflection and Refraction
926
32 – 1 The Ray Model of Light
927
32 – 2 Reflection; Image Formation by a
Plane Mirror
927
32 – 3 Formation of Images by
Spherical Mirrors
931
32 – 4 Seeing Yourself in a
Magnifying Mirror (Concave)
936
32 – 5 Convex (Rearview) Mirrors
938
32 – 6 Index of Refraction
939
32 – 7 Refraction: Snell’s Law
939
32 – 8 The Visible Spectrum and Dispersion 941
32 – 9 Total Internal Reflection; Fiber Optics 943
*32 – 10 Refraction at a Spherical Surface
946
Questions, MisConceptions, Problems   949 – 957
33 Lenses and Optical Instruments
958
33 – 1 Thin Lenses; Ray Tracing and
Focal Length
959
33 – 2 The Thin Lens Equation
962
33 – 3 Combinations of Lenses
966
33 – 4 Lensmaker’s Equation
968
33 – 5 Cameras: Film and Digital
970
33 – 6 The Human Eye; Corrective Lenses
975
33 – 7 Magnifying Glass
979
33 – 8 Telescopes
980
33 – 9 Compound Microscope
983
33 – 10 Aberrations of Lenses and Mirrors
984
Questions, MisConceptions, Problems   986 – 994
The Wave Nature of Light:
34 Interference and Polarization
995
34 – 1 Waves vs. Particles; Huygens’
Principle and Diffraction
996
34 –   2  Huygens’ Principle and the Law of
Refraction; Mirages
997
34 – 3 Interference – Young’s Double-Slit
Experiment
998
*34 – 4 Intensity in the Double-Slit
Interference Pattern
1002
34 – 5 Interference in Thin Films
1004
34 – 6 Michelson Interferometer
1010
34 – 7 Polarization
1010
*34 – 8 Liquid Crystal Displays (LCD)
1014
*34 – 9 Scattering of Light by the Atmosphere 1015
34 – 10 Brightness: Lumens and Luminous Intensity 1016
*34 – 11 Efficiency of Lightbulbs
1016
Questions, MisConceptions, Problems   1018 – 1024
35 Diffraction
1025
35 –   1  Diffraction by a Single Slit or Disk
1026
*35 – 2 Intensity in Single-Slit Diffraction
Pattern
1028
*35 – 3 Diffraction in the Double-Slit Experiment 1031
35 – 4 Interference vs. Diffraction
1033
35 – 5 Limits of Resolution; Circular Apertures 1033
35 – 6 Resolution of Telescopes and
Microscopes; the l Limit
1035
35 –   7  Resolution of the Human Eye and
Useful Magnification
1037
35 – 8 Diffraction Grating
1037
35 – 9 The Spectrometer and Spectroscopy 1040
*35 –   1  0 Peak Widths and Resolving Power for a
Diffraction Grating
1041
35 – 11 X-Rays and X-Ray Diffraction
1043
*35 – 12 X-Ray Imaging and Computed
Tomography (CT Scan)
1045
*35 – 13 Specialty Microscopes and Contrast 1048
Questions, MisConceptions, Problems   1049 – 1054
CONTENTS  ix

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 9
03/07/20 16:04
3 The Special Theory of Relativity
1055
36 – 1 Galilean – Newtonian Relativity
1056
36 – 2 The Michelson – Morley Experiment 1058
36 – 3 Postulates of the Special Theory
of Relativity
1061
36 – 4 Simultaneity
1062
36 – 5 Time Dilation and the Twin Paradox 1064
36 – 6 Length Contraction
1070
36 – 7 Four-Dimensional Space – Time
1072
36 – 8 Galilean and Lorentz Transformations 1072
36 – 9 Relativistic Momentum
1077
36 – 10 The Ultimate Speed 36 – 11 E = mc2; Mass and Energy
1079 1080
36 – 12 Doppler Shift for Light
1085
36 – 13 The Impact of Special Relativity
1086
Questions, MisConceptions, Problems   1088 – 1094
37 Early Quantum Theory and
Models of the Atom
1095
37 – 1 Blackbody Radiation;
Planck’s Quantum Hypothesis
1096
37 – 2 Photon Theory of Light and the
Photoelectric Effect
1098
37 –   3  Energy, Mass, and Momentum of a
Photon
1101
37 – 4 Compton Effect
1102
37 – 5 Photon Interactions; Pair Production 1104
37 – 6 Wave – Particle Duality; the Principle of
Complementarity
1105
37 – 7 Wave Nature of Matter
1106
37 – 8 Electron Microscopes
1108
37 – 9 Early Models of the Atom
1110
37 –   1  0 Atomic Spectra: Key to the Structure
of the Atom
1111
37 – 11 The Bohr Model
1113
37 – 12 de Broglie’s Hypothesis Applied to Atoms 1120
Questions, MisConceptions, Problems   1121 – 1127
Appendices
A Mathematical Formulas B Derivatives and Integrals C Numerical Integration D More on Dimensional Analysis E Gravitational Force Due to a
Spherical Mass Distribution F Differential Form of Maxwell’s
Equations G Selected Isotopes
Answers to Odd-Numbered Problems Index Photo Credits
A–1 A–6 A–8 A–12
A–13
A–16 A–18 A–23 A–47 A–77
38 Quantum Mechanics
1128
38 – 1 Quantum Mechanics—A New Theory 1129
38 – 2 The Wave Function and Its Interpretation;
the Double-Slit Experiment
1129
38 – 3 The Heisenberg Uncertainty Principle 1131
38 – 4 Philosophic Implications; Probability Versus Determinism
1135
38 – 5 The Schrödinger Equation in One Dimension—Time-Independent Form 1136
*38 – 6 Time-Dependent Schrödinger Equation 1138
38 – 7 Free Particles; Plane Waves and
Wave Packets
1140
38 –   8  Particle in an Infinitely Deep Square
Well Potential (a Rigid Box)
1142
38 – 9 Finite Potential Well
1147
38 – 10 Tunneling through a Barrier
1149
Questions, MisConceptions, Problems   1152 – 1157
39 Quantum Mechanics of
Atoms
1158
39 – 1 Quantum-Mechanical View of Atoms 1159
39 – 2 Hydrogen Atom: Schrödinger
Equation and Quantum Numbers
1159
39 – 3 Hydrogen Atom Wave Functions
1163
39 – 4 Multielectron Atoms; the Exclusion Principle
1166
39 – 5 Periodic Table of Elements
1167
39 – 6 X-Ray Spectra and Atomic Number 1169
*39 – 7 Magnetic Dipole Moment;
Total Angular Momentum
1171
39 – 8 Fluorescence and Phosphorescence
1174
39 – 9 Lasers
1175
*39 – 10 Holography
1178
Questions, MisConceptions, Problems   1180 – 1185
40 Molecules and Solids
1186
40 – 1 Bonding in Molecules
1187
40 – 2 Potential-Energy Diagrams
for Molecules
1189
40 – 3 Weak (van der Waals) Bonds
1192
40 – 4 Molecular Spectra
1196
40 – 5 Bonding in Solids
1202
40 – 6 Free-Electron Theory of Metals; Fermi Energy
1203
40 – 7 Band Theory of Solids
1208
40 – 8 Semiconductors and Doping
1210
40 – 9 Semiconductor Diodes, LEDs, OLEDs 1212
40 – 10 Transistors: Bipolar and MOSFETs
1218
40 – 11 Integrated Circuits, 10-nm Technology 1219
Questions, MisConceptions, Problems   1220 – 1225
x  CONTENTS

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 10
03/07/20 16:04
0
1
2
3
4
5
6N
41 Nuclear Physics and Radioactivity
1226
41 –   1  Structure and Properties of the Nucleus 1227
41 – 2 Binding Energy and Nuclear Forces 1230
41 – 3 Radioactivity
1233
41 – 4 Alpha Decay
1234
41 – 5 Beta Decay
1237
41 – 6 Gamma Decay
1239
41 – 7 Conservation of Nucleon Number and
Other Conservation Laws
1240
41 – 8 Half-Life and Rate of Decay
1240
41 – 9 Decay Series
1245
41 – 10 Radioactive Dating
1246
41 – 11 Detection of Particles
1248
Questions, MisConceptions, Problems   1250 – 1255
42 Nuclear Energy; Effects and Uses of Radiation 1256
42 – 1 Nuclear Reactions and the
Transmutation of Elements
1257
42 – 2 Cross Section
1260
42 – 3 Nuclear Fission; Nuclear Reactors
1261
42 – 4 Nuclear Fusion
1266
42 – 5 Passage of Radiation Through Matter;
Biological Damage
1271
42 – 6 Measurement of Radiation—Dosimetry 1272
*42 – 7 Radiation Therapy
1276
*42 – 8 Tracers in Research and Medicine
1277
*42 – 9 Emission Tomography: PET and SPECT 1278
*42 – 10 Nuclear Magnetic Resonance (NMR);
Magnetic Resonance Imaging (MRI) 1279
Questions, MisConceptions, Problems   1283 – 1288
43 Elementary Particles
1289
43 – 1 High-Energy Particles and Accelerators 1290
43 – 2 Beginnings of Elementary Particle
Physics—Particle Exchange
1296
43 – 3 Particles and Antiparticles
1299
43 – 4 Particle Interactions and
Conservation Laws
1300
43 – 5 Neutrinos
1302
43 – 6 Particle Classification
1304
43 – 7 Particle Stability and Resonances
1306
43 – 8 Strangeness? Charm?
Towards a New Model
1307
43 – 9 Quarks
1308
43 – 10 The Standard Model: QCD and
Electroweak Theory
1311
43 – 11 Grand Unified Theories
1314
43 – 12 Strings and Supersymmetry
1317
Questions, MisConceptions, Problems   1318 – 1231
44 Astrophysics and Cosmology
1322
44 – 1 Stars and Galaxies
1323
44 – 2 Stellar Evolution: Birth and Death
of Stars, Nucleosynthesis
1326
44 – 3 Distance Measurements
1332
44 – 4 General Relativity: Gravity and the
Curvature of Space
1334
44 – 5 The Expanding Universe: Redshift
and Hubble’s Law
1338
44 –   6  The Big Bang and the Cosmic
Microwave Background
1342
44 – 7 The Standard Cosmological Model:
Early History of the Universe
1345
44 – 8 Inflation: Explaining Flatness,
Uniformity, and Structure
1348
44 – 9 Dark Matter and Dark Energy
1350
44 – 10 Large-Scale Structure of the Universe 1353
44 – 11 Gravitational Waves : LIGO
1354
44 – 12 Finally . . .
1354
Questions, MisConceptions, Problems   1356 – 1360
Appendices
A Mathematical Formulas B Derivatives and Integrals C Numerical Integration D More on Dimensional Analysis E Gravitational Force Due to a
Spherical Mass Distribution F Differential Form of Maxwell’s
Equations G Selected Isotopes
Answers to Odd-Numbered Problems Index Photo Credits
A–1 A–6 A–8 A–12
A–13
A–16 A–18 A–23
CONTENTS  xi

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 11
03/07/20 16:04
Applications (Selected) to Medicine and Biology and to Engineering, Environment, Everyday Life, Etc.
(Entries with a star * include material new to this edition)
Chapter 1
Viruses attack cell
7
Heartbeats in a lifetime
12
Number of nucleons in human body 17
Lung capacity
19
Building collapse
2, 332, 346–7
The 8000-m peaks
9
Making estimates: volume of a lake 11
Page thickness
12
Building height by triangulation
12
Earth radius estimate
13, 18
Fermi estimates
13
Particulate pollution
18
Global positioning satellite
18
Computer chips
18
Chapter 2
Airport runway design
32
*Car air bag inflation time
32, 254
Car braking distance
35, 183
CD bit size, bit rate, playing time 48, 53
*Baseball
49, 82, 83, 84, 172
Basketball
50, 83, 109
Golf putt, uphill or down
52
Rapid transit system
53
Chapter 3
Helicopter supply drop
54, 72, 83
*Sports
54, 65, 69, 71, 76, 77, 79, 80, 81, 82, 83, 84
Kicked football
69, 71
Truck escape lane
79, 110
Golf on the Moon
82
Extreme sports
83
Chapter 4
How we can walk
92
Whiplash
106
*Force heart exerts
107
Rocket
85, 92, 108, 233, 252, 395
Skater pushoff
91
What force accelerates a car
92
You weigh less in a falling elevator 96
Hockey
98
Elevator, discomfort
101, 108
Mechanical advantage, pulley 102, 188
Accelerometer
102
*Sports *Bear sling
Tug of war
106, 107, 108, 109, 110, 112 106, 355 106, 255
Car accident “g’s”
107
Optical tweezers
108, 916, 923
*Tightrope walker
109
Basketball shot
109
Mountain climbers 110, 114, 115, 193, 370
City planning, cars on hill
112
Bicycling
112, 114
Supermarket ramp design
113
Doomsday asteroid
114, 262
*Car stuck in mud
115
Chapter 5
Centrifugation
126
Skiing
116, 121, 136
Push or pull a sled?
120
Skier speed in air vs. on snow
121
*Simulating gravity 126, 136, 141, 168, 171
*Uranium enrichment, reactor, bomb 126
Ferris wheel
129
Avoid skidding on a curve
130–2
Banked highway curves
132
Cross-country skiing friction
136
Rotating space station 136, 141, 168
*Rotor ride Airplane bank /turn
137, 143 137, 144
Roller coaster upside down
141
Car flying up off road
141
Rock climbing friction
143
Chapter 6
Weightlessness
154–5
*Astronauts in orbit Gravity on tall peaks
145, 155, 165 150
Oil and mineral exploration 150, 165, 167
Satellites, spacecraft 145, 152–5, 168, 169
Geostationary satellites
153
Free fall, for athletes
155
Planets
155–8, 167
Determining the Sun’s mass
158
Planets around other stars 158, 250, 262
*Ocean tides Lagrange point
159, 165, 170 160
*Moon’s orbit, periods, phases,
diagram
161, 169
*Eclipses Curved space
161 163–4
Black holes
164, 167
White dwarfs
167
Comets, asteroids, moons 168, 169, 171
GPS
169
Milky Way Galaxy
171
Chapter 7 Baseball pitch Car stopping distance r v2 Lever
*Pulley Jet catapults
Bicycle, sprockets (teeth)
Climbing rope stretch
172 183 187, 334 188 189 192, 299 193
Chapter 8
Stair-climbing power
213
*ATP
216
Hike over logs
218
Pole vault
194, 203–4, (189)
Downhill ski runs
194
Roller coaster
198, 202, 209
Escape velocity from Earth
or Moon
212
Power needs of car
214
Efficiency of engine
215
*Gravitational assist High jump
216–7, 224, 263 220
Bungee jump
221
Lunar module landing
222
Escape velocity from solar system 223
Ski jump
225
Long jump
225
Chapter 9
Impulse in fall: break a leg?
257
Billiard balls
227, 230, 237, 242
Tennis serve
229, 234
Rocket propulsion
233, 252, 395
Rifle recoil
233
Karate blow
235
Nuclear reactors
238
Nuclear collisions 238, 239, 241, 243
Ballistic pendulum, speed measured 240
Distant planet discovery
250, 262
Conveyor belt
253
Car crashworthiness
261
Asteroid danger
262
Force wind exerts
263
Bowling
263
Chapter 10
Acuity of bird’s eye
266
Centrifuge
271
*Biceps, triceps, torque Situps
273, 295, 339 291
Fast mammal
291
Rotating carnival rides 264, 267, 268
Tire iron extension
272
Flywheel, energy
282, 301
Yo-yo
287
Braking forces on a car
288–9
Bicycle odometer
291
Tightrope walking
291
*Total solar eclipse
293
Wrench torque
294
Hammer throw
296
CD rotation frequency
298
Bicycle gears
299
Cue stick, ball roll
300
*Bicycle turn angle
301
Chapter 11
Rotating skaters /divers 302, 304, 330
Neutron star collapse
305, 330
Strange spinning bike wheel 307, 314
xii  APPLICATIONS

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 12
03/07/20 16:04
Automobile wheel balancing Precessing top Gyroscope Hurricanes, cyclones, typhoons *Anticyclonic weather Precession of the equinoxes SUV rollover Baseball bat sweet spot
314–5 317–8
318 321, 394
321 327 328 330
Chapter 12
*Forces in muscles & joints 339, 358, (273)
*What can make an athlete
339
*Forces on the spine and back pain 340
Human balance with loads
342
*Bone fracture
346, 359, 364
Buildings, statics
331–352
Lever, mechanical advantage
334
Balancing a seesaw
335
Cantilever
336
Fracture
345–7
Tragic collapse
346–7, (332)
Trusses and bridges
347–9, 363
Architecture: arches and domes 350–2
*Forces in a dome
352
Chapter 13
Pressure in cells
371
Blood flow
380, 384, 386
Human circulatory system
380
Blood loss to brain, TIA
384
*Air flow in animal burrow
384
Heart disease, artery clogging
386
Walking on water, insect
387
Heart as a pump
388–9
*Blood pressure measurement
389
Blood transfusion
395, 396
Water supply pressure
369
Atmospheric pressure decrease
with elevation
370
Altitude where air pressure is half 370
Finger holds water in straw
371
Hydraulic lift
372
Hydraulic brakes
372
Pressure gauges
372–3
Barometer
373
Suction
374
Hydrometer
377
Continental drift, plate tectonics 378
*Lake level change, rock thrown overboard
378, 390
Helium balloon lift
378
Heating duct
380
Hot-water heating system flow
382
Perfume atomizer
383
Airplane wing lift
383
Sailing upwind
383
Baseball curve
384
Why smoke goes up a chimney
384
Soaps and detergents
387
Pumps
388–9, (374)
Siphon
390
Hydraulic press
393
Rocket thrust
395
Reynolds number
395
Barrel broken by thin liquid column 397
Chapter 14
Spider web oscillations
405
Human leg as pendulum
424
Shock absorbers
399, 415
Unwanted floor vibrations
406
Loudspeaker
406–7
*Pendulum clock Geology
412, 421, 424 412, 415
Measure g with pendulum
412
Earthquake dampers
415
Child on a swing, resonance
417–8
Resonance damage
418
Q-value
419, 425, 896
Bungee jumper
422
*Metronome
424
Natural stride
424
Tall building sway
426
Chapter 15
Echolocation by bats, dolphins,
whales
434
Water waves
428, 435
Sound wave
431, 460 ff
Geology
435, 452, 457
Earthquake waves 435, 437, 450, 453
Square wave
442
*Cell phone signal
451
AM and FM radio wave bending 452
Fish and fisher: internal reflection 456
Seismic reflection: oil prospecting 457
Coffee spill
457
Tsunami
459
Chapter 16
Wide range of human hearing
464
Sensitivity of the ear
467, (466)
Bats use Doppler
479
Doppler blood-flow meter
479, 491
Ultrasound medical imaging
482–3
*Doppler ultrasound imaging
483
Stringed instruments
460, 468–9
Wind instruments
460, 469–72
Piano strings
460, 468, 469
Distance from lightning, seconds 461
Autofocusing camera
462
Loudspeaker output
465
Musical scale
468
Guitar, violin
468, 469, 484, 487
Organ pipes
471–2
Tuning with beats
475–6
Doppler in weather forecasting 480
*Radar speed gun
480
Galaxy redshift
480
Sonic boom; sound barrier
481, 489
Sonar: depth in sea, Earth
“soundings”
481–2, 489
Signal-to-noise ratio
486, 490, 679
Quartz oscillator clock
487
Motion sensor
489
Audio gain
490
Chapter 17
Life under ice
500–1
Molecules in one breath
507, 514
Snorkels are short
515
*Hot air balloon Expansion joints
492, 515 495, 498, 501
Do holes expand?
499
Opening a tight lid
499
Gas tank overflow
500
Highway buckling
501
Closed jars in fires
503
Mass (weight) of air in a room
505
Cold and hot tire pressure
506
Thermostat
509
Pyrex glass
509
*Tape measure inaccuracy
510, 513
Scuba
512, 513, 514, 515
Potato chip bag puff up
513
Chapter 18
KE of molecules in cells
519
Humidity, and comfort
525–6
Chromatography
531
Diffusion in living organisms 531–2, 536
Temperature effect on chemical
reactions
521
Evaporation cools
524, 548
Humidity, weather
526
*Temperature decrease of boiling
water with altitude
526–7
Pressure cooker
535
Chapter 19
Working off Calories
540
Measuring Calorie content 545, 570
Evaporation and body
temperature
548–9, (524)
Body heat: convection by blood 563, 574
Body’s radiative heat loss
564
Room comfort: cool air, warm walls 565
Medical thermography
567
Avoid plants freezing
568
Eating snow makes you colder
571
Heat conduction to skin,
blood capillaries
573
Leaf’s energy absorption
575
Metabolizing fat
575
Cold tile, warm rugs
561
Heat loss through windows
562
Thermal windows (two panes)
562
How clothing insulates
562
R-values of thermal insulation
562
Ocean currents and wind
563
Convective home heating
563
Dark vs. light clothing
564
Radiation from the sun, seasons 566
Astronomy—size of a star
566
Goose down loft
568
Thermos bottle
568
Emergency blanket
568
Air parcels, weather, adiabatic
lapse rate
573
APPLICATIONS  xiii

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 13
03/07/20 16:04
Chapter 20
Biological development, evolution 594
*Trees offsetting CO2 buildup
608
Steam engine
576, 578, 582, 606
Internal combustion engines
578–80, 583–4
Engine efficiency
582–3
Refrigerators, air conditioners 584–6, 603
Heat pump
586–7, 603
*SEER rating Thermal pollution, climate
587 598–600
*Carbon footprint Energy resources
598 599, 605–6
Solar, thermal, wind energy 599, 605
Diesel engine
607, (575)
Stirling cycle
607
Jet engine, Brayton cycle
607
Dehumidifier
608, (537)
Chapter 21
Inside a cell: kinetic theory plus
electrostatic force
631
DNA structure, replication 631–3, 640
Static electricity
609, 610, 635, 640
Photocopiers and printers
619
Electrical shielding, safety
628
Chapter 23
Electrocardiogram
(ECG)
660, 682–3, 779
Dipoles in molecular biology
672
Heart beat, depolarization process 682–3 Common voltages 10-4 V to 108 V 663
Breakdown voltage
666
Lightning rods
666
*Supply voltage, signal voltage
676
*Digital, bits, bytes, binary numbers 676
*Analog-to-digital converter (ADC) 676
*Morse code
676
*Bit-rate,TV transmission 676, 678–9, 682
*Data compression, jpeg
677–8
*Quantization error
677
*Sampling rate, bit depth
677
*Digital-to-analog converter (DAC)
677, 780
*Bandwidth
678
*Noise, bit flips
678–9
*Digital error correction, parity bit 678
*Bit error rate
679
*Signal-to-noise ratio (S/N)
679
*TV and computer monitors
*Digital TV, pixels, subpixels *Flat screens, HD *Addressing pixels *Data stream *Active matrix, TFT, data lines *TV refresh rate
Oscilloscope
679–82 680
680–1 680–1
681 681–2
682 682
*ASCII code
688
Photocell
689
Chapter 24
Capacitor shocks, burns
703
Heart defibrillator
703, 712, 764
Capacitor use as power backup,
surge protector, memory 692, 695
Condenser microphone
695
Computer key
695
Camera flash energy
701
Electrostatic air cleaner
710
Tiny distance measurement
710
Coaxial cable
714, 818, 874, 911
*Dynamic random access memory (DRAM)
716, 857
Chapter 25
Electrical conduction in human
nervous system, neurons
736–8
Action potential
737
Battery construction, terminals 718–9
*Electric cars
720, 744
Battery connections
721, 724
Loudspeaker wire thickness
725
Heating element
726–8
Resistance thermometer
726
Lightning bolt
728, (690, 716)
Household circuits, shorts
729–30
Fuses, circuit breakers
729, 766
Safety—wires getting hot 729, 764–6
Extension cord danger
730
Hair dryer
732
Strain gauge
746
Chapter 26
*Blood sugar phone app
747
Heart pacemaker
764
Electricity dangers to humans 764–6
Ventricular fibrillation
764
Two-speed fan
752–3
Car battery charging
757
*Jump-starting a car, safely
758–9
RC: sawtooth, flashers, wipers 763, 780
Hazards, electric safety
764–6
Proper grounding, plugs
765–6
Leakage current
766
Dangerous downed power line
766
Ammeters, voltmeters,
ohmmeters
767–9
Meter connection, corrections 768–9, 781
*Measurement affects quantity
measured
769
Voltage divider
774
Solar panel
778
Potentiometer and bridge circuits 778–9
Car battery corrosion
780
Digital-to-analog converter (DAC)
780, (677)
Chapter 27
Electromagnetic blood pump
802
Blood flow rate, Hall effect
807
Use of a compass
784
Magnetic declination
784
Maps and true north
784
Aurora borealis Electric motors, DC and AC Loudspeakers and headsets
792 795–6
796
Chapter 28
Coaxial cable
818, 874, 911
Solenoid switches: doorbell, car starter 826
Magnetic circuit breakers
826
Relay (magnetic)
830
Chapter 29
EM blood-flow measurement
845
Induction stove
842
Generators, power plants
846–7
Alternators, in cars
848
Motor overload
849
Eddy-current damping
850, 861
Airport metal detector
850
Transformers, power transmission 851–3
Cell phone charger
852
Car ignition system
852
*Wireless electric power
transmission
854
Inductive charger
854
Magnetic information storage
856
*Semiconductor memory
857–8
*RAM, DRAM
857
*Bit-line & word-line
857
*Writing and reading memory
857
*Volatile and non volatile memory 858
*Flash memory, MOSFET, MRAM 858
Microphone
858
Card reader, magnetic strip
858
Seismograph
859
Ground fault circuit interrupter
(GFCI)
859
Shielded cable
861
Recycling solid waste
861
Chapter 30
*Electric car inductive charging
869
Surge protection
877
Capacitors as filters
884, 896, 897
Loudspeaker cross-over
884
Impedance matching
888
3-phase AC
889
Q-value
896, (419, 425)
Filter circuit
896
Chapter 31
Optical tweezers
916, 923
*TV from the Moon
898, 920, 924
Wireless devices, transmission
898, 917–20
Antennas
911, 919
Phone call time lag
912
*Solar sail
916, 925
Radio and TV
917–9
AM and FM
918
Cell phones, remotes, cable TV,
satellite TV
920
*GPS
924
Solar power use
924
xiv  APPLICATIONS 
GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 14
03/07/20 16:04
Chapter 32
Medical endoscope, bronchoscope,
colonoscope
945
How tall a mirror do you need
930
Seeing yourself in a magnifying
mirror (concave)
936–7
Convex (rearview mirrors)
938
Optical illusions
939, 998
Apparent water depth
939–40, 941
Rainbows
942, 957
Colors underwater
943
Diamonds sparkle
944
Prism binoculars
944
Fiber optic cables
945, 954, 956
*High-frequency trading,
interception
945
Solar cooker
951
Washing machine water level
detector
956
Road reflectors
957
Chapter 33
Human eye
975–8
Fovea, denser in cones
976, 1037
Near- and far-sighted
976–8
Corrective lenses
976–7, 987
Contact lenses
978
Seeing underwater
978
Light microscopes
983–4, 1048
Where your eye can see a
lens image
961
Cameras, film and digital
970–5
*CCD, CMOS sensors, potential well
970–1
*Bayer pixels, Fovean
971
Digital artifacts
971
Camera adjustments, f-stop
971–3
Depth of field
973
*Resolution, compression, JPEG,
raw
973–4
Telephoto, wide angle
975
Optical vs. digital zoom
975
Magnifying glass
979–80
Telescopes
980–2
*Microscopes Lens aberrations
983–4, 1048 984–5
Film projector
989
Pinhole camera
990
Chapter 34
Soap bubbles, oil films,
colors
995, 1004–8
Highway mirages
998
Lens coatings
1008–9
Polarizing sunglasses
1012–13
Liquid crystal displays, TV and
computer screens
1014–5
Sky color
*Lightbulb efficiency, LED Stealth aircraft coating
CD bits, pits & lands
1015–6 1016–7
1022 1024
Chapter 35
Resolution of eye
1035, 1037
Useful magnification
1037
Spectroscopy in biology
1041
X-ray diffraction in biology
1044
Medical imaging: X-rays, CT 1045–7
*Interference microscope *Phase-contrast microscope
Hubble space telescope
1048 1048 1034–5
Telescope and microscope
resolution
1035–7
X-rays
1043–7
Tomography
1045–7
Chapter 36 Space travel Global position system (GPS) Fantasy supertrain Radar speed gun
1067–8 1068–9
1071 1092
Chapter 37
Electron microscope image: blood vessel,
clot, retina, viruses 1095, 1109, (7)
Photosynthesis
1102
Measuring bone density
1103
Electron microscopes (EM),
TEM, SEM
1109, 1151, (1095)
Photocells
1098
Photodiodes, soundtracks
1101
Chapter 38
Scanning tunneling electron
microscope
1151
Atomic force microscope
1151
Chapter 39
Fluorescence analysis
1174–5
Medical uses of lasers, surgery 1178, 1183
Neon lights
1158
Fluorescent lightbulbs
1175
Lasers
1175–9, 1216
Bar code readers
1177
DVD, CD, Blu-ray
1177–8
Holography
1178–9
Chapter 40
Cell energy—ATP
1192
Weak bonds, DNA
1192–4
Protein synthesis
1194–6
*Pulse oximeter
1216
Computer processor chips
1186
Transparent objects
1210
Zener diode voltage regulator
1213–4, 1225
Rectifiers
1214
*Photovoltaic cells *LED displays, bulbs
TV remote
*Solid-state lighting *pn diode laser *OLED, AMOLED displays
Amplifiers
*MOSFET switch *Technology generation
1214–5 1215–6 1215, 1225 1215–6
1216 1216–7
1218 1218–9
1219
Chapter 41
Earliest life
1248
Radiation film badges
1249, 1274
Smoke detector
1237
Radioactive activity and safety 1243–4
Carbon-14 dating
1246–7
Archeological & geological
dating
1246–8
Oldest Earth rocks
1248
Geiger counter
1248
Rubidium-strontium dating
1253
Tritium dating
1254
*Mass excess, mass defect
1254
Chapter 42
Biological radiation damage 1271–6
Radiation dosimetry, RBE
1272–6
Radon exposure
1274, 1276
Natural radioactive background 1274
Radiation exposure, film badge 1274
Radiation sickness
1274
Whole-body dose
1275
Radiation therapy
1276–7
Proton therapy
1277
Radioactive tracers
1277–8
Gamma camera
1278
Medical imaging, PET, SPECT, MRI
1278–82
*Brain PET scan using cell phone 1279
Imaging resolutions compared 1282
Radiation and thyroid
1286
Nuclear reactors, power plants
1256, 1263–5, 1269–71
Breeder reactors
1265
Manhattan project
1266
Nuclear fusion
1266–71
Why stars shine
1267–9
Thermonuclear devices
1269
Fusion energy reactors
1269–71
Chapter 43 Linacs and tumor irradiation
1294
Chapter 44 Stars and galaxies Black holes Big Bang storyline
1323–32 1331, 1337–8
1345–8

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 15
APPLICATIONS  xv
03/07/20 16:04
xvi  PREFACE 
GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 16
Preface
New Stuff!
1. MisConceptual Questions, 10 or 15 at the end of each chapter. The multiplechoice answers include common misconceptions as well as correct responses. Pedagogically, asking students to think, to consider the options, is more effective than just telling them what is valid and what is wrong. (These are in addition to the one at the start of each chapter.
2. Digital is all around us. Yet that word is not always used carefully. In this new edition we have 20 new pages describing the basics from the ground up. Binary numbers, bits and bytes, are introduced in Chapter 23 along with analog-to-digital conversion (ADC), and vice versa, including digital audio and how video screens work. Also information compression, sampling rate, bit depth, pixel addressing, digital transmission and, in later chapters, information storage (RAM, DRAM, flash), digital cameras and their sensors (CCD, CMOS).
3. Gravitational Assist (Slingshot) to accelerate spacecraft (Chapter 8). 4. Magnetic field of a single moving charge, rarely treated (and if it is, maybe
not well), and it shows the need for relativity theory. 5. Seeing yourself in a magnifying mirror (concave), angular magnification and
blurriness with a paradox. Also convex (rearview) mirrors (Chapter 32). 6. Pedagogical clarification on defining potential energy, and energy itself
(Chapter 8), and on hundreds of other topics. 7. The Moon rises an hour later each day (Chapter 6), its phases, periods, and diagram. 8. Efficiency of lightbulbs (Chapter 34). 9. Idealization vs. reality emphasized—such as PV diagrams (Chapter 19) as an
idealized approximation. 10. Many new Problems (' 500) plus new Questions as well as the 500 or so
MisConceptual Questions (point 1 above). 11. Many new worked-out Examples. 12. More math steps included in derivations and Examples. 13. State of a system and state variables clarified (Chapter 17). 14. Contemporary physics: Gravitational waves, LIGO and Virgo, Higgs, WIMPS,
OLEDS and other semiconductor physics, nuclear fusion updates, neutrino-less double beta decay. 15. New SI units (Chapter 1, Chapter 21, Tables). 16. Boiling temperature of water vs. elevation (Chapter 18). 17. Modern physics in earlier classical Chapters (sometimes in Problems): Light-years, observable universe (Chapter 1); optical tweezers (Chapter 4); uranium enrichment (Chapter 5); black holes and curved space, white dwarfs (Chapter 6); crystal structure (Chapter 7); Yukawa potential, Lennard-Jones potential (Chapter 8); neutrons, nuclear reactors, moderator, nuclear collisions, radioactive decay, neutron star collapse (Chapter 9); galaxy redshift (Chapter 16); gas diffusion of uranium (Chapter 18); quarks (Chapter 21); liquid-drop model of nucleus, Geiger counter, Van de Graaff (Chapter 23); transistors (Chapters 23, 29); isotopes, cyclotron (Chapter 27); MOSFET (Chapter 29); semiconductor (camera sensor), photon (Chapter 33); line spectra, X-ray crystallography (Chapter 35). 18. Second law of thermodynamics and heat energy reorganized (Chapter 20). 19. Symmetry emphasized throughout. 20. Uranium enrichment, % needed in reactors, bombs (Chapters 5, 42). 21. Mass excess, mass defect (Chapter 41). 22. The mole, more careful definition (Chapter 17). 23. Liquid-gas ambiguity above critical temperature (Chapter 18). 24. Measurement affects quantity measured, new emphasis.
03/07/20 16:04
25. New Applications: • Ocean Tides (Chapter 6) • Anticyclonic weather (Chapter 11) • Jump starting a car safely (Chapter 26) • Light bulb efficiency (Chapter 34) • Specialty microscopes and contrast (Chapter 35) • Forces on Muscles and Joints (Chapter 12) • Doppler ultrasound imaging (Chapter 16) • Lake level change when rock thrown from boat (Chapter 13) • Skier speed on snow vs. flying through the air (Chapter 5) • Inductive charging (Chapter 29) • Human body internal heat transfer is convection (blood) (Chapter 19) • Blood pressure measurement (Chapter 13) • Sports (lots) • Voltage divider (Chapter 26, Problems) • Flat screen TV (Chapters 23, 34, 40) • Carbon footprint and climate (Chapter 20) • Electrocardiogram (Chapter 23) • Wireless from the Moon unimaginable (Chapter 31) • Why snorkels are short (Chapter 17 Problem) • Electric cars (Chapter 25) • Digital (Chapters 23, 29, 33, 40) includes (in addition to details in point 2 above) quantization error, digital error correction, noise, bit error rate, digital TV data stream, refresh rate, active matrix, thin film transistors, digital memory, bit-line, reading and writing of memory cells (MOSFET), floating gate, volatile and nonvolatile memory, Bayer, JPEG, ASCII code, and more.
Seeing the World through Eyes that Know Physics
I was motivated to write a textbook different from others which typically present physics as a sequence of facts, like a catalog. Instead of beginning formally and dogmatically, I aim to begin each topic with everyday observations and experiences the students can relate to: start with specifics, the real world, and then go to the great generalizations and the more formal aspects of the physics, showing why we believe what we believe. This approach reflects how science is actually practiced.
The aim is to give students a thorough understanding of the basic concepts of physics in all its aspects, from mechanics to modern physics. Also important is to show students how useful physics is in their own everyday lives and in their future professions by means of interesting applications to biology, medicine, engineering, architecture, and more.
Much effort has gone into approaches for the practical techniques of solving problems: worked-out Examples, Problem Solving sections, and Problem Solving Strategies.
Chapter 1 is not a throwaway. It is fundamental to physics to realize that every measurement has an uncertainty, and how significant figures are used. Being able to make rapid estimates is a powerful tool useful for every student, and used throughout the book starting in Chapter 1 (you can estimate the Earth’s radius!).
Mathematics can be an obstacle to students. I have aimed at including all steps in a derivation. Important mathematical tools, such as addition of vectors and vector product, are incorporated in the text where first needed, so they come with a context rather than in a forbidding introductory Chapter. Appendices contain a basic math review, derivatives and integrals, plus some more advanced topics including numerical integration, gravitational field of spherical mass distribution, Maxwell’s equations in differential form, and a Table of selected nuclear isotopes (carefully updated, as are the Periodic Table and the Fundamental Constants found inside the back and front covers).
Some instructors may find this book contains more material than can be covered in their courses. The text offers great flexibility. Sections marked with a star * may be considered optional. These contain slightly more advanced
Versions of this Book
Complete version: 44 Chapters including 9 Chapters of modern physics.
Classic version: 37 Chapters, 35 on classical physics, plus one each on relativity and quantum theory.
3 Volume version: Available separately or packaged together ­
Volume 1: Chapters 1–20 on ­mechanics, including fluids, oscillations, waves, plus heat and thermodynamics.
Volume 2: Chapters 21–35 on electricity and magnetism, plus light and optics.
Volume 3: Chapters 36–44 on modern physics: relativity, quantum theory, atomic physics, condensed matter, nuclear physics, elementary particles, cosmology and astrophysics.
PREFACE  xvii

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 17
03/07/20 16:04
physics material, or material not usually covered in typical courses, or interesting applications; they contain no material needed in later Chapters (except perhaps in later optional Sections). For a brief course, all optional material could be dropped as well as significant parts of Chapters 13, 16, 26, 30, and 35, and selected parts of Chapters 9, 12, 19, 20, 33. Topics not covered in class can be a valuable resource for outside study by students. Indeed, this text can serve as a useful reference for years because of its wide range of coverage.
Thanks
Many physics professors provided input or direct feedback on every aspect of this textbook. They are listed below, and I owe each a debt of gratitude.
Edward Adelson, The Ohio State University Lorraine Allen, United States Coast Guard Academy Zaven Altounian, McGill University Leon Amstutz, Taylor University Kim Arvidsson, Schreiner University Philip S. Baringer, Kansas University Bruce Barnett, Johns Hopkins University Michael Barnett, Lawrence Berkeley Lab Anand Batra, Howard University David Branning, Trinity College Bruce Bunker, University of Notre Dame Wayne Carr, Stevens Institute of Technology Charles Chiu, University of Texas Austin Roger N. Clark, U. S. Geological Survey Russell Clark, University of Pittsburgh Robert Coakley, University of Southern Maine David Curott, University of North Alabama Biman Das, SUNY Potsdam Bob Davis, Taylor University Kaushik De, University of Texas Arlington Michael Dennin, University of California Irvine Kathryn Dimiduk, Cornell University John DiNardo, Drexel University Scott Dudley, United States Air Force Academy John Essick, Reed College Cassandra Fesen, Dartmouth College Leonard Finegold, Drexel University Alex Filippenko, University of California Berkeley Richard Firestone, Lawrence Berkeley Lab Tom Furtak, Colorado School of Mines Gill Gabelmann, Washburn University Gabriel Orebi Gann, University of California Berkeley Edward Gibson, California State University Sacramento John Hamilton, University of Hawai’i – Hilo John Hardy, Texas A&M J. Erik Hendrickson, University of Wisconsin-Eau Claire Charles Hibbard, Lowell High School Dr. Laurent Hodges, Iowa State University David Hogg, New York University Mark Hollabaugh, Normandale Community College Russell Holmes, University of Minnesota Twin Cities William Holzapfel, University of California Berkeley Bob Jacobsen, University of California Berkeley Arthur W. John, Northeastern University David Jones, Florida International University Andrew N. Jordan, University of Rochester Teruki Kamon, Texas A&M Thomas Hemmick, State University of New York Stonybrook Daryao Khatri, University of the District of Columbia Woo-Joong Kim, Seattle University John Kinard, Greenwood High School Jay Kunze, Idaho State University Jim LaBelle, Dartmouth College Andrei Linde, Stanford University M.A.K. Lodhi, Texas Tech Lisa Madewell, University of Wisconsin
Ponn Maheswaranathan, Winthrop University Bruce Mason, University of Oklahoma Mark Mattson, James Madison University Linda McDonald, North Park College Raj Mohanty, Boston University Giuseppe Molesini, Isituto Nazionale di ottica Florence Lisa K. Morris, Washington State University Richard Muller, University of California Berkeley Blaine Norum, University of Virginia Lauren Movatne, Reedley College Alexandria Oakes, Eastern Michigan University Ralph Oberly, Marshall University Michael Ottinger, San Juan College Lyman Page, Princeton Laurence Palmer, University of Maryland Bruce Partridge, Haverford College R. Daryl Pedigo, University of Washington Robert Pelcovitz, Brown University Saul Perlmutter, University of California Berkeley Vahe Peroomian, UCLA Harvey Picker, Trinity College Amy Pope, Clemson University James Rabchuk, Western Illinois University Michele Rallis, Ohio State University Andrew Resnick, Cleveland State University Paul Richards, University of California Berkeley Peter Riley, University of Texas Austin Dennis Rioux, University of Wisconsin Oshkosh John Rollino, Rutgers University Larry Rowan, University of North Carolina Chapel Hill Arthur Schmidt, Northwestern University Cindy Schwarz, Vassar College Peter Sheldon, Randolph-Macon Woman’s College James Siegrist, University of California Berkeley Christopher Sirola, University of Southern Mississippi Earl Skelton, Georgetown University George Smoot, University of California Berkeley Stanley Sobolewski, Indiana University of Pennsylvania Mark Sprague, East Carolina University Michael Strauss, University of Oklahoma Leo Takahashi, Pennsylvania State University Richard Taylor, University of Oregon Oswald Tekyi-Mensah, Alabama State University Ray Turner, Clemson University Som Tyagi, Drexel University David Vakil, El Camino College Robert Webb, Texas A&M Robert Weidman, Michigan Technological University Edward A. Whittaker, Stevens Institute of Technology Lisa M. Will, San Diego City College Suzanne Willis, Northern Illinois University Michael Winokur, University of Wisconsin-Madison Stanley George Wojcicki, Stanford University Mark Worthy, Mississippi State University Edward Wright, UCLA Todd Young, Wayne State College
xviii  PREFACE

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 18
03/07/20 16:04
I owe special thanks to Prof. Bob Davis for much valuable input, and especially for working out all the Problems and producing the Solutions Manual for all Problems, as well as for providing the answers to odd-numbered Problems at the back of the book. Many thanks also to J. Erik Hendrickson who collaborated with Bob Davis on the solutions, and to the team they managed (Michael Ottinger, John Kinard, David Jones, Kristi Hatch, Lisa Will).
I am especially grateful to Profs. Lorraine Allen, Kathryn Dimiduk, Michael Strauss, Cindy Schwarz, Robert Coakley, Robert Pelcovitz, Mark Hollabaugh, Charles Hibbard, and Michael Winokur, who helped root out errors and offered significant improvements and clarifications.
For Chapters 43 and 44 on Particle Physics and Cosmology and Astrophysics, I was fortunate to receive generous input from some of the top experts in the field, to whom I owe a debt of gratitude: Saul Perlmutter, George Smoot, Richard Muller, Alex Filippenko, Paul Richards, Gabriel Orebi Gann, James Siegrist, and William Holzapfel (UC Berkeley), Andreí Linde (Stanford U.), Lyman Page (Princeton), Edward Wright (UCLA), Michael Strauss (University of Oklahoma), and Bob Jacobsen (UC Berkeley).
I also wish to thank many others at the University of California, Berkeley, Physics Department for helpful discussions, and for hospitality. Thanks also to Prof. Tito Arecchi at the Istituto Nazionale di Ottica, Florence, Italy.
Finally, I am grateful to the many people at Pearson Education with whom I worked on this project, especially Jeanne Zalesky and Paul Corey, and the perspicacious editors Margy Kuntz and Andrea Giancoli.
The final responsibility for all errors lies with me. I welcome comments, corrections, and suggestions as soon as possible to benefit students for the next reprint.
email: paper mail:
jeanne.zalesky@pearson.com Jeanne Zalesky Pearson Education 501 Boylston Street Boston, MA 020116
D.G.
About the Author
Doug Giancoli obtained his BA in physics (summa cum laude) from UC Berkeley, his MS in physics at MIT, and his PhD in elementary particle physics back at UC Berkeley. He spent 2 years as a post-doctoral fellow at UC Berkeley’s Virus Lab developing skills in molecular biology and biophysics.
His mentors include Nobel winners Emilio Segrè, Barry Barish, and Donald Glaser.
He has taught a wide range of undergraduate courses, traditional as well as innovative ones, and works to improve his textbooks meticulously, seeking ways to provide a better understanding of physics for students.
Doug loves the outdoors, especially climbing peaks. He says climbing peaks is like learning physics: it takes effort and the rewards are great.

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 19
PREFACE  xix
03/07/20 16:04
xx  PREFACE 
GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 20
Students Advice
HOW TO STUDY 1. Read the Chapter. Learn new vocabulary and notation. Respond to questions and
exercises as they occur. Follow carefully the steps of worked-out Examples and derivations. Avoid time looking at a screen. Paper is better than pixels when it comes to learning and thinking. 2. Attend all class meetings. Listen. Take notes. Ask questions (everyone wants to, but maybe you will have the courage). You will get more out of class if you read the Chapter first. 3. Read the Chapter again, paying attention to details. Follow derivations and worked-out Examples. Absorb their logic. Answer Exercises and as many of the end-of-Chapter Questions as you can, and all MisConceptual Questions. 4. Solve at least 10 to 20 end-of-Chapter Problems, especially those assigned. In doing Problems you may find out what you learned and what you didn’t. Discuss them with other students. Problem solving is one of the great learning tools. Don’ t just look for a formula : it might be the wrong one.
NOTES ON THE FORMAT AND PROBLEM SOLVING 1. Sections marked with a star (*) may be considered optional or advanced. They can
be omitted without interrupting the main flow of topics. No later material depends on them except possibly later starred Sections. They may be fun to read, though. 2. The customary conventions are used: symbols for quantities (such as m for mass) are italicized, whereas units (such as m for meter) are not italicized. Symbols for vectors are shown in boldface with a small arrow above: F5. 3. Few equations are valid in all situations. Where practical, the range of validity of important equations are stated in square brackets next to the equation. The equations that represent the great laws of physics are displayed with a tan background, as are a few other indispensable equations. 4. At the end of each Chapter is a set of Questions you should try to answer. Attempt all the multiple-choice MisConceptual Questions, which are intendend to get common misconceptions “out on the table” by including them as responses (temptations) along with correct answers. Most important are Problems which are ranked as Level I, II, or III, according to estimated difficulty. Level I Problems are easiest, Level II are standard Problems, and Level III are “challenge problems.” These ranked Problems are arranged by Section, but Problems for a given Section may depend on earlier material too. There follows a group of General Problems, not arranged by Section or ranked. Problems that relate to optional Sections are starred (*). Answers to odd-numbered Problems are given at the end of the book. 5. Being able to solve Problems is a crucial part of learning physics, and provides a powerful means for understanding the concepts and principles. This book contains many aids to problem solving: (a) worked-out Examples, including an Approach and a Solution, which should be studied as an integral part of the text; (b) some of the worked-out Examples are Estimation Examples,which show how rough or approximate results can be obtained even if the given data are sparse (see Section 1-6); (c) Problem Solving Strategies placed throughout the text to suggest a step-by-step approach to problem solving for a particular topic : but the basics remain the same; most of these “Strategies” are followed by an Example that is solved by explicitly following the suggested steps; (d) special problem-solving Sections; (e) “Problem Solving” marginal notes which refer to hints within the text for solving Problems; (f) Exercises within the text that you should work out immediately, and then check your response against the answer given at the bottom of the last page of that Chapter; (g) the Problems themselves at the end of each Chapter. 6. Conceptual Examples pose a question which hopefully starts you to think about a response. Give yourself a little time to come up with your own response before reading the Response given. 7. Math review, plus additional topics, are found in Appendices. Useful data, conversion factors, and math formulas are found inside the front and back covers.
03/07/20 16:04
USE OF COLOR
Vectors
A general vector
resultant vector (sum) is slightly thicker
components of any vector are dashed
Displacement
(
S
D,
rS)
Velocity (vS)
Acceleration (aS)
S
Force (F )
Force on second object
or third object in same gure
Momentum (pS or mvS)
S
Angular momentum (L)
Angular velocity (VS )
Torque (TS)
Electric
eld
(
S
E
)
Magnetic eld (BS )
Electricity and magnetism
Electric circuit symbols
Electric eld lines Equipotential lines Magnetic eld lines Electric charge (+) Electric charge (–)
+ or + or
Wire, with switch S
S
Resistor
Capacitor
Inductor
Battery
Ground
Optics
Light rays Object
Real image (dashed)
Virtual image (dashed and paler)
Other
Energy level (atom, etc.) Measurement lines
Path of a moving object
Direction of motion or current
1.0 m

GIAN_PSE5_CH_FM_Classic_i-xxii_c.indd 21
PREFACE  xxi
03/07/20 16:04
This page intentionally left blank
Image of the Earth from out in space. The sky appears black because there are so few molecules to reflect light. (Why the sky appears blue to us on Earth has to do with scattering of light by molecules of the atmosphere, as discussed in Chapter 34.) Note the storm off the coast of Mexico. Important physics is covered in this first Chapter, including measurement uncertainty and how to make an estimate. For example, we can determine the radius of the Earth without going out in space, but just by being near a lake or bay.
C
r
ha p te
1 Introduction,
Measurement, Estimating
Chapter-Opening Questions—Guess now! 1.  How many cm3 are in 1.0 m3?
(a) 10.  (b) 100.  (c) 1000.  (d) 10,000.  (e) 100,000.  (f) 1,000,000.
2. Suppose you wanted to actually measure the radius of the Earth, at least roughly, rather than taking other people’s word for what it is. Which response below ­describes the best approach?
(a) Use an extremely long measuring tape. (b) It is only possible by flying high enough to see the actual curvature of the Earth. (c) Use a standard measuring tape, a stepladder, and a large smooth lake. (d) Use a laser and a mirror on the Moon or on a satellite. (e) Give up; it is impossible using ordinary means.
[We start each Chapter with a Question : sometimes two. Try to answer right away. Don’t worry about getting the right answer now : the idea is to get your preconceived notions out on the table. If they are misconceptions, we expect them to be cleared up as you read the Chapter. You will get another chance at the Question later in the Chapter when the appropriate material has been covered. These Chapter-Opening Questions will also help you see the power and usefulness of physics.]
Contents
1–1 How Science Works 1–2 Models, Theories, and Laws 1–3 Measurement and ­Uncertainty;
Significant Figures 1–4 Units, Standards, and
the SI System 1–5 Converting Units 1–6 Order of Magnitude:
Rapid Estimating *1–7 Dimensions and ­Dimensional
Analysis
1
GIAN_PSE5_CH01_001-019_ca.indd 1
01/07/20 16:17
P hysics is the most basic of the sciences. It deals with the behavior and structure of matter. The field of physics is usually divided into classical physics which includes motion, fluids, heat, sound, light, electricity and magnetism; and
­modern physics which includes the topics of relativity, atomic structure, condensed
matter, nuclear physics, elementary particles, and cosmology and astrophysics. We
will cover all these topics in this book, beginning with motion (or mechanics, as it
is often called) and ending with the most recent results in our study of the cosmos.
An understanding of physics is wonderfully useful for anyone making a
career in science or technology. Engineers, for example, must know how to
calculate the forces within a structure to design it so that it remains standing
(Fig. 1 9 1a). Indeed, in Chapter 12 we will see a worked-out Example of how a
(a)
simple physics c­ alculation : or even intuition based on understanding the physics
of forces : would have saved hundreds of lives (Fig. 1 9 1b). We will see many
examples in this book of how physics is useful in many fields, and in everyday life.
1–1  How Science Works
(b)
Figure 1 – 1  (a) This bridge over the River Tiber in Rome was built 2000 years ago and still stands. (b) The Hartford Civic Center collapsed in 1978, just two years after it was built.
c a u ti o n Science is not static.
It changes and develops
There is a real physical world out there. We could just walk through it, not thinking much about it. Or, we can instead examine it carefully. That is what scientists do. The aim of science is the search for order in our observations of the physical world so as to provide a deeper picture or description of this world around us. Sometimes we just want to understand how things work.
Some people seem to think that science is a mechanical process of collecting facts and devising theories. But it is not so simple. Science is a creative activity, and in many ways resembles other creative activities of the human mind.
One important aspect of science is observation of events (which great writers and artists also do), and includes the design and carrying out of experiments. But observation and experiment require imagination, because scientists can never include everything in a description of what they observe. In other words, scientists must make judgments about what is relevant in their observations and experiments.
Consider, for example, how two great minds, Aristotle (384 9 322 b.c.) and ­Galileo (15649 1 642), interpreted motion along a horizontal surface. Aristotle noted that objects given an initial push along the ground (or on a level tabletop) always slow down and stop. Consequently, Aristotle argued, the natural state of an object is to be at rest. Galileo, in his reexamination of horizontal motion in the 1600s, had the idea that friction is a kind of force like a push or a pull; and he imagined that if friction could be eliminated, an object given an initial push along a horizontal surface would continue to move indefinitely without stopping. He concluded that for an object to be in motion was just as natural as for it to be at rest. By inventing a new approach, Galileo founded our modern view of motion (Chapters 2, 3, and 4), and he did so with a leap of the imagination. Galileo made this leap conceptually, without actually eliminating friction.
Observation, with careful experimentation and measurement, is one side of the scientific process. The other side is the invention or creation of theories to explain and order the observations. Theories are never derived directly from observations. Observations may help inspire a theory, and theories are accepted or rejected based on the results of observation and experiment.
Theories are inspirations that come from the minds of humans. For example, the idea that matter is made up of atoms (the atomic theory) was not arrived at by direct observation of atoms. Rather, the idea sprang from creative minds. The theory of relativity, the electromagnetic theory of light, and Newton’s law of universal gravitation were likewise the result of human imagination.
The great theories of science may be compared, as creative achievements, with great works of art or literature. But how does science differ from these other creative activities? One important difference is that science requires testing of its ideas or theories to see if their predictions are borne out by experiment.
But theories are not “proved” by testing. First of all, no m­ easuring instrument is perfect, so exact confirmation is not possible. Furthermore, it is not possible to test a theory in every single possible circumstance. Hence a t­heory cannot be absolutely verified.
2  CHAPTER 1  Introduction, Measurement, Estimating
GIAN_PSE5_CH01_001-019_ca.indd 2
01/07/20 16:17
Indeed, the history of science tells us that long-held theories can often be replaced by new ones.
1–2  Models, Theories, and Laws
When scientists are trying to understand a particular aspect of the physical world, they often make use of a model. A model, in the scientist’s sense, is a kind of analogy or mental image of the phenomena in terms of something we are familiar with. One example is the wave model of light. We cannot see waves of light as we can water waves. But it is valuable to think of light as made up of waves because experiments indicate that light behaves in many respects as water waves do.
The purpose of a model is to give us an approximate mental or visual p­ icture :  something to hold on to : when we cannot see what actually is happening in the real world. Models often give us a deeper understanding: the analogy to a known system (for instance, water waves in the above example) can suggest new experiments to perform and can provide ideas about what other related phenomena might occur.
You may wonder what the difference is between a theory and a model. Usually a model is relatively simple and provides a structural similarity to the phenomena being studied. A theory is broader, more detailed, and can give ­quantitatively testable predictions, often with great precision.
It is important not to confuse a model or a theory with the real world and the phenomena themselves. Theories are descriptions of the physical world, and they are made up by us. Theories are ­invented : usually by very smart people.
Scientists give the title law to certain concise but general statements about how nature behaves (that energy is conserved, for example). Sometimes the statement takes the form of a relationship or equation between quantities (such as Newton’s second law,  F = ma).
To be called a law, a statement must be found experimentally valid over a wide range of observed phenomena. For less general statements, the term p­ rinciple is often used (such as Archimedes’ principle). We use “theory” to describe a more general picture of a large group of phenomena.
Scientific laws are different from political laws, which are prescriptive: they tell us how we ought to behave. Scientific laws are descriptive: they do not say how nature should behave, but rather are meant to describe how nature does behave. As with theories, laws cannot be tested in the infinite variety of cases possible. So we cannot be sure that any law is absolutely true. We use the term “law” when its validity has been tested over a wide range of situations, and when any limitations and the range of validity are clearly understood.
Scientists normally do their research as if the accepted laws and theories were true. But they are obliged to keep an open mind in case new information should alter the validity of any given law or theory. In other words, laws of physics, or the “laws of nature”, represent our descriptions of reality and are not inalterable facts that last forever. Laws are not lying there in nature, waiting to be discovered. We humans, the brightest humans, invent the laws using observations and intuition as a basis. And we hope our laws provide a good description of nature, and at a minimum give us a reliable approximation of how nature really behaves.
c a u ti o n
Theories and laws are NOT discovered. They are invented
1–3 Measurement and Uncertainty;
Significant Figures
In the quest to understand the world around us, scientists seek to find relationships among physical quantities that can be measured.
Uncertainty
Reliable measurements are an important part of physics. But no measurement is absolutely precise. There is an uncertainty associated with every measurement. Among the most important sources of uncertainty, other than blunders, are the ­limited accuracy of every measuring instrument and the inability to read
SECTION 1–3  Measurement and Uncertainty; Significant Figures  3
GIAN_PSE5_CH01_001-019_ca.indd 3
01/07/20 16:17
Figure 1 – 2  Measuring the width of a board with a centimeter ruler. The uncertainty is about { 1 mm.
P r o b l em S o l v in g Significant figures rule:
Number of significant figures in final result should be same as
the least significant input value
an instrument (such as a ruler) beyond some fraction of the smallest division shown. For example, if you were to use a centimeter ruler to measure the width of a board (Fig. 1 9 2), the result could be claimed to be precise to about 0.1 cm (1 mm), the smallest division on the ruler, although half of this value might be a valid claim as well. The reason is that it is difficult for the observer to estimate (or interpolate) between the smallest divisions. Furthermore, the ruler itself may not have been manufactured to an accuracy very much better than this.
When giving the result of a measurement, it is important to state the ­estimated uncertainty in the measurement. For example, the width of a board might be written as  8.8 { 0.1 cm.  The {0.1 cm (“plus or minus 0.1 cm”) represents the estimated uncertainty in the measurement, so that the actual width most likely lies between 8.7 and 8.9 cm. The percent uncertainty is the ratio of the uncertainty to the measured value, multiplied by 100. For example, if the measurement is 8.8 and the uncertainty about 0.1 cm, the percent uncertainty is
0.1 8.8 * 100% L 1%, where L means “is approximately equal to.” Often the uncertainty in a measured value is not specified explicitly. In such cases, scientists follow a general rule that
uncertainty in a numerical value is assumed to be one or a few units in the last digit specified.
For example, if a length is given as 5.6 cm, the uncertainty is assumed to be about 0.1 cm or 0.2 cm, or possibly 0.3 cm. It is important in this case that you do not write 5.60 cm, for this implies an uncertainty on the order of 0.01 or 0.02 cm; it assumes that the length is probably between about 5.58 cm and 5.62 cm, when actually you believe it is between about 5.4 and 5.8 cm.
Significant Figures
The number of reliably known digits in a number is called the number of ­significant figures. Thus there are four significant figures in the number 23.21 cm and two in the number 0.062 cm (the zeros in the latter are merely place holders that show where the decimal point goes). The number of significant figures may not always be clear. Take, for example, the number 80. Are there one or two significant figures? We need words here: If we say it is roughly 80 km between two cities, there is only one significant figure (the 8) since the zero is merely a place holder. If there is no suggestion that the 80 is a rough approximation, then we can often assume (as we will in this book) that it has two significant figures: so it is 80 km within an accuracy of about 1 or 2 km. If it is precisely 80 km, to within { 0.1 or { 0.2 km, then we need to write 80.0 km (three significant figures).
When specifying numerical results, you should avoid the temptation to keep more digits in the final answer than is justified: see boldface statement above. For example, to calculate the area of a rectangle 11.3 cm by 6.8 cm, the result of multip­ lication would be 76.84 cm2. But this answer can not be accurate to the implied 0.01 cm2 uncertainty. Why? Because (using the outer limits of the assumed uncertainty for each measurement) the result could be between  11.2 cm * 6.7 cm = 75.04 cm2  and  11.4 cm * 6.9 cm = 78.66 cm2.  At best, we can quote the answer as 77 cm2, which implies an uncertainty of about 1 or 2 cm2. The other two digits (in the number 76.84 cm2) must be dropped (rounded off) because they are not significant. As a rough general significant figures rule,
the final result of a multiplication or division should have no more digits than the numerical value with the fewest significant figures.
In our example, 6.8 cm has the least number of significant figures, namely two. Thus the result 76.84 cm2 needs to be rounded off to 77 cm2.
Exercise A  The area of a rectangle 4.5 cm by 3.25 cm is correctly given by (a) 14.625 cm2;  (b) 14.63 cm2;  (c) 14.6 cm2;  (d) 15 cm2.
4  CHAPTER 1  Introduction, Measurement, Estimating
GIAN_PSE5_CH01_001-019_ca.indd 4
01/07/20 16:17
When adding or subtracting numbers, the final result should contain no more decimal places than the number with the fewest decimal places. For example, the result of subtracting 0.57 from 3.6 is 3.0 (not 3.03). Similarly 36 + 8.2 = 44,  not 44.2.
Be careful not to confuse significant figures with the number of decimal places. Significant figures are related to the expected uncertainty in any measured quantity.
P r o b l em S o l v in g
Significant figures when adding and subtracting
Exercise B  For each of the following numbers, state the number of significant figures and the number of decimal places: (a) 1.23; (b) 0.123; (c) 0.0123.
Keep in mind when you use a calculator that all the digits it produces may not be significant. When you divide 2.0 by 3.0, the proper answer is 0.67, and not 0.666666666 as calculators give (Fig. 1 9 3a). Digits should not be quoted in a result unless they are truly significant figures. However, to obtain the most accurate result, you should normally keep one or more extra significant figures throughout a calculation, and round off only in the final result. (With a calculator, you can keep all its digits in intermediate results.) Calculators can also give too few significant figures. For example, when you multiply  2.5 * 3.2,  a calculator may give the answer as simply 8. See Fig. 1 9 3b. But the answer is ­accurate to two significant figures, so the proper answer is 8.0.†
C a u ti o n Calculators err with significant figures
P r o b l em S o l v in g Report only the proper number of significant figures in the final result. But keep extra digits during the calculation
Figure 1 – 3   These two calculators show the wrong number of significant figures. In (a), 2.0 was divided by 3.0. The correct final result should be stated as 0.67. In (b), 2.5 was multiplied by 3.2. The correct result is 8.0.
(a)
(b)
Conceptual  EXAMPLE  1 – 1 Significant figures.  Using a protractor (Fig. 1 9 4), you measure an angle to be 30°. (a) How many significant figures should you quote in this ­measurement? (b) Use a calculator to find the cosine of the angle you measured.
Response (a) If you look at a protractor, you will see that the precision with which you can measure an angle is about one degree (certainly not 0.1°). So you can quote two significant figures, namely 30° (not 30.0°). (b) If you enter cos 30° in your calculator, you will get a n­ umber like 0.866025403. But the angle you entered is known only to two significant figures, so its cosine is correctly given by 0.87; you must round your answer to two significant figures.
Note  Trigonometric functions, like cosine, are reviewed in Appendix A.
Figure 1 – 4   Example 1 9 1. A protractor used to measure an angle.
EXERCISE C  Do 0.00324 and 0.00056 have the same number of significant figures?
Scientific Notation
We commonly write numbers in “powers of ten,” or “scientific” notation : for instance 36,900 as 3.69 * 104,  or 0.0021 as  2.1 * 10-3.  One advantage of scientific notation is that it allows the number of significant figures to be clearly expressed. For example, it is not clear whether 36,900 has three, four, or five significant figures. With powers of ten notation the ambiguity can be avoided: if the number is known to three significant figures, we write 3.69 * 104,  but if it is known to four, we write  3.690 * 104.
†Be careful also about other digital read-outs. If a digital bathroom scale shows 85.6, do not assume the uncertainty is {0.1 or {0.2; the scale was likely manufactured with an accuracy of perhaps only 1% or so: that is, {1 or {2. For digital scientific instruments, also be careful: the instruction manual should state the accuracy.
SECTION 1–3  Measurement and Uncertainty; Significant Figures  5
GIAN_PSE5_CH01_001-019_ca.indd 5
01/07/20 16:17
EXERCISE D  Write each of the following in scientific notation and state the number of ­significant figures for each: (a) 0.0258, (b) 42,300, (c) 344.50.
Percent Uncertainty versus Significant Figures
The significant figures rule is only approximate, and in some cases may underestimate the accuracy (or uncertainty) of the answer. Suppose for example we divide 97 by 92:
97 92 = 1.05 L 1.1. Both 97 and 92 have two significant figures, so the rule says to give the answer as 1.1. Yet the numbers 97 and 92 both imply an uncertainty of {1 if no other uncertainty is stated. Both  92 { 1  and  97 { 1  imply an uncertainty of about 1%  (1>92 L 0.01 = 1%).  But the final result to two significant figures is 1.1, with an implied uncertainty of {0.1, which is an uncertainty of  0.1>1.1 L 0.1 L 10%.  In this case it is better to give the answer as 1.05 (which is three significant figures). Why? Because 1.05 implies an uncertainty of {0.01 which is  0.01>1.05 L 0.01 L 1%,  just like the uncertainty in the original numbers 92 and 97. SUGGESTION: Use the significant figures rule, but consider the % uncertainty too, and add an extra digit if it gives a more realistic estimate of uncertainty.
Approximations
Much of physics involves approximations, often because we do not have the means to solve a problem precisely. For example, we may choose to ignore air resistance or friction in doing a Problem even though they are present in the real world, and then our calculation is only an estimate or approximation. In doing Problems, we should be aware of what approximations we are making, and be aware that the precision of our answer may not be nearly as good as the number of significant figures given in the result.
Accuracy versus Precision
There is a technical difference between “precision” and “accuracy.” Precision in a strict sense refers to the repeatability of the measurement using a given instrument. For example, if you measure the width of a board many times, getting results like 8.81 cm, 8.85 cm, 8.78 cm, 8.82 cm (interpolating between the 0.1 cm marks as best as possible each time), you could say the measurements give a precision a bit better than 0.1 cm. ­Accuracy refers to how close a measurement is to the true value. For example, if the ruler shown in Fig. 1 9 2 was manufactured with a 2% error, the accuracy of its measurement of the board’s width (about 8.8 cm) would be about 2% of 8.8 cm or about {0.2 cm. Estimated uncertainty is meant to take both accuracy and precision into account.
1–4 Units, Standards, and the SI System
The measurement of any quantity is made relative to a particular standard or unit, and this unit must be specified along with the numerical value of the quantity. For example, we can measure length in British units such as inches, feet, or miles, or in the metric system in centimeters, meters, or kilometers. To specify that the length of a particular object is 18.6 is insufficient. The unit must be given, because 18.6 meters is very different from 18.6 inches or 18.6 millimeters.
For any unit we use, such as the meter for distance or the second for time, we need to define a standard which defines exactly how long one meter or one second is. It is important that standards be chosen that are readily reproducible so that a­ nyone needing to make a very accurate measurement can refer to the standard in the laboratory and communicate results with other scientists.
6  CHAPTER 1  Introduction, Measurement, Estimating
GIAN_PSE5_CH01_001-019_ca.indd 6
01/07/20 16:17
Length
The first truly international standard was the meter (abbreviated m) established as the standard of length by the French Academy of Sciences in the 1790s. The standard meter was originally chosen to be one ten-millionth of the distance from the Earth’s equator to either pole,† and a platinum rod to represent this length was made. (One meter is, very roughly, the distance from the tip of your nose to the tip of your finger, with arm and hand stretched out horizontally.) In 1889, the meter was defined more precisely as the distance between two finely engraved marks on a particular bar of platinum9 i ridium alloy. In 1960, to provide greater precision and reproducibility, the meter was redefined as 1,650,763.73 wavelengths of a particular orange light emitted by the gas krypton-86.
In 1983 the meter was again redefined, this time in terms of the speed of light (whose best measured value in terms of the older definition of the meter was 299,792,458 m>s, with an uncertainty of 1 m>s). The new definition reads: “The meter is the length of path traveled by light in vacuum during a time interval of 1>299,792,458 of a second.” The new definition of the meter has the effect of giving the speed of light the exact value of 299,792,458 m>s. [The newer definitions provided greater precision than the 2 marks on the old platinum bar.]
British units of length (inch, foot, mile) are now defined in terms of the meter. The inch (in.) is defined as exactly 2.54 centimeters (cm;  1 cm = 0.01 m). One foot is exactly 12 in., and 1 mile is 5280 ft. Other conversion factors are given in the Table on the inside of the front cover of this book. Table 1 9 1 below presents some typical lengths, from very small to very large, rounded off to the nearest power of 10. (We call this rounded off value the order of magnitude.) See also Fig. 1 9 5. (Note that the abbreviation for inches (in.) is the only one with a period, to distinguish it from the word “in”.) [The nautical mile = 6076 ft = 1852 km is used by ships on the open sea and was originally defined as 1>60 of a degree latitude on Earth’s surface. A speed of 1 knot is 1 nautical mile per hour.]
Time
The standard unit of time is the second (s). For many years, the second was defined as 1>86,400 of a mean solar day (24 h>day * 60 min>h * 60 s>min = 86,400 s>day).  The standard second can be defined more precisely in terms of the frequency of radiation emitted by cesium atoms when they pass between two particular states. ­[Specifically, one second is the time required for 9,192,631,770 periods of this radiation. This number was chosen to keep “one second” the same as in the old definition.] There are, by definition, 60 s in one minute (min) and 60 minutes in one hour (h). Table 1 9 2 presents a range of time intervals, rounded off to the nearest power of 10.
†Modern measurements of the Earth’s circumference reveal that the intended length is off by about one-fiftieth of 1%. Not bad!
New definition of the meter Figure 1 – 5   Some lengths: (a) viruses (about 10-7 m long) attacking a cell; (b) Mt. Everest’s height is on the order of 104 m (8850 m, to be precise).
(a)
(b)
TABLE 1 – 1 Some Typical Lengths or Distances
(order of magnitude)
Length (or Distance)
Meters (approximate)
Neutron or proton (diameter) Atom (diameter) Virus [see Fig. 1 9 5a] Sheet of paper (thickness) Finger width Football field length Height of Mt. Everest [see Fig. 1 9 5b] Earth diameter Earth to Sun Earth to nearest star Earth to nearest galaxy Earth to farthest galaxy visible
10-15 m 10-10 m 10-7 m 10-4 m 10-2 m 102 m 104 m 107 m 1011 m 1016 m 1022 m 1026 m
TABLE 1 – 2 Some Typical Time Intervals
(order of magnitude)
Time Interval
Seconds (approximate)
Lifetime of very unstable subatomic particle
Lifetime of radioactive elements Lifetime of muon Time between human heartbeats One day One year Human life span Length of recorded history Humans on Earth Life on Earth Age of Universe
10-23 s
10-22 s to 1028 s 10-6 s
100 s ( = 1 s) 105 s 3 * 107 s 2 * 109 s 1011 s 1014 s 1017 s
4 * 1017 s
SECTION 1–4  Units, Standards, and the SI System  7
GIAN_PSE5_CH01_001-019_ca.indd 7
01/07/20 16:17
TABLE 1 – 3 Some Masses
Object
Kilograms (approximate)
Electron Proton, neutron DNA molecule Bacterium Mosquito Plum Human Ship Earth Sun Galaxy
10-30 kg 10-27 kg 10-17 kg 10-15 kg 10-5 kg 10-1 kg 102 kg 108 kg 6 * 1024 kg 2 * 1030 kg 1041 kg
P r o b l em S o l v in g Always use a consistent set of units
TABLE 1 – 4  Metric (SI) Prefixes
Prefix Abbreviation Value
yotta
Y
zetta
Z
exa
E
peta
P
tera
T
giga
G
mega
M
kilo
k
hecto
h
deka
da
deci
d
centi
c
milli
m
micro†
m
nano
n
pico
p
femto
f
atto
a
zepto
z
yocto
y
1024 1021 1018 1015 1012 109 106 103 102 101 10-1 10-2 10-3 10-6 10-9 10-12 10-15 10-18 10-21 10-24
†m is the Greek letter “mu.”
8  CHAPTER 1
Mass
The standard unit of mass is the kilogram (kg). The standard mass has been, since 1889, a particular platinum 9 iridium cylinder, kept at the International Bureau of Weights and Measures near Paris, France, whose mass is defined as exactly 1 kg. A range of masses is presented in Table 1 9 3. [For practical purposes, a 1 kg mass weighs about 2.2 pounds on Earth.]
1 metric ton is 1000 kg. In the British system of units, 1 ton is 2000 pounds. When dealing with atoms and molecules, we usually use the unified atomic mass unit (u or amu). In terms of the kilogram,
1 u = 1.6605 * 10-27 kg. (Precise values of this and other numbers are given inside the front cover.) The density of a uniform object is its mass divided by its volume, commonly expressed in kg>m3.
Unit Prefixes
In the metric system, the larger and smaller units are defined in multiples of 10
from the standard unit, and this makes calculation particularly easy. Thus 1 kilo-
me­ter
(km)
is
1000 m,
1 centimeter
is
1 100
m,
1 millimeter
(mm)
is
1 1000
m
or
1 10
cm,
and
so
on.
The
prefixes
“centi-,”
“kilo-,”
and
others
are
listed
in
Table
1 9 4
and can be applied not only to units of length but to units of volume, mass, or
any other unit. For example, a centiliter (cL) is 1010 liter (L), and a kilogram (kg) is 1000 grams (g). An 8.2-megapixel camera has a detector with 8,200,000 pixels
(individual “picture elements”).
In common usage,  1 mm (= 10 - 6 m)  is called 1 micron.
Systems of Units
When dealing with the laws and equations of physics it is very important to use a consistent set of units. Several systems of units have been in use over the years. Today the most important is the Système International (French for International System), which is abbreviated SI. In SI units, the standard of length is the meter, the standard for time is the second, and the standard for mass is the kilogram. This ­system used to be called the MKS (meter-kilogram-second) system.
A second metric system is the cgs system, in which the centimeter, gram, and second are the standard units of length, mass, and time, as abbreviated in the title. The British engineering system (although more used in the U.S. than Britain) has as its standards the foot for length, the pound for force, and the second for time.
We use SI units almost exclusively in this book, although we often define the cgs and British units when a new quantity is introduced. In the SI, there have traditionally been seven base quantities, each defined in terms of a standard; seven is the smallest number of base quantities consistent with a full description of the physical world. See Table 1 9 5. All other quantities† can be defined in terms of seven base quantities; see the Table inside the front cover which lists many quantities and their units in terms of base units.
*A New SI
As always in science, new ideas and approaches can produce better precision and closer correspondence with the real world. Even for units and standards.
International organizations on units have proposed further changes that should make standards more readily available and reproducible. To cite one example, the standard kilogram (see above) has been found to have changed slightly in mass (­contamination is one cause).
The new redefinition of SI standards follows the method already used for the meter as being related to the defined value of the speed of light, as we mentioned on page 7 under “Length”. For example, the charge on the electron, e, instead of being a measured value, becomes defined as a certain value (its current value), and the unit of electric charge (the coulomb) follows from that. All units then become based on
†Some exceptions are for angle (radians : see Chapter 10), solid angle (steradian), and sound level (bel or decibel, Chapter 16).
*Some Sections of this book, such as this subsection, may be considered optional at the discretion of the instructor and they are marked with an asterisk (*). See the Preface for more details.
GIAN_PSE5_CH01_001-019_ca.indd 8
01/07/20 16:17
defined fundamental constants like e and the speed of light. Seven is still the number of basic standards. The new definitions maintain the values of the traditional definitions: the “new” meter is the same length as the “old” meter. The new definitions do not change our understanding of what length, time, or mass means.
For us, using this book, the difference between the new SI and the traditional SI is highly technical and does not affect the physics we study. We include the traditional SI because there is some good physics in explaining it. [The Table of Fundamental Constants inside the front cover would look slightly different using the new SI. The value of the charge e on the electron, for example, is defined, and so would have no uncertainty attached to it; instead, our Table inside the front cover includes the traditional SI measured uncertainty (updated) of { 98 * 10-29 C.]
1–5  Converting Units
Any quantity we measure, such as a length, a speed, or an electric current, consists of a number and a unit. Often we are given a quantity in one set of units, but we want it expressed in another set of units. For example, suppose we measure that a shelf is 21.5 inches wide, and we want to express this in centimeters. We must use a conversion factor, which in this case is, by definition, exactly
1 in. = 2.54 cm
or, written another way,
1 = 2.54 cm>in.
Since multiplying by the number one does not change anything, the width of our
shelf, in cm, is
21.5 inches
=
( 21.5
in. )
*
a 2.54
cm b in.
=
54.6 cm.
Note how the units (inches in this case) cancelled out (thin red lines). A Table
­containing many unit conversions is found inside the front cover of this book.
Let’s consider some Examples.
TABLE 1 – 5  Traditional SI Base Quantities
Quantity Unit
Unit Abbreviation
Length
meter
m
Time
second
s
Mass
kilogram
kg
Electric
current ampere
A
Temperature kelvin
K
Amount of
substance mole
mol
Luminous
intensity candela
cd
EXAMPLE 1 – 2 The 8000-m peaks.  There are only 14 peaks whose summits are over 8000 m above sea level. They are the highest peaks in the world (Fig. 1 9 6 and Table 1 9 6) and are referred to as “eight-thousanders.” What is the elevation, in feet, of an elevation of 8000 m?
Approach  We need to convert meters to feet, and we can start with the conversion factor 1 in. = 2.54 cm,  which is exact. That is, 1 in. = 2.5400 cm  to any number of significant figures, because it is defined to be.
FIGURE 1 – 6   The world’s second highest peak, K2, whose summit is considered the most difficult of the “8000-ers.” Example 1 9 2.
P h y sics App l ie d The world’s tallest peaks
Solution  One foot is defined to be 12 in., so we can write
1 ft
=
(12 in. ) ¢ 2.54 cm ≤ in.
=
30.48 cm
=
0.3048 m,
TABLE 1 – 6 The 8000-m Peaks
which is exact. Note how the units cancel (colored slashes). We can rewrite this Peak
Height (m)
equation to find the number of feet in 1 meter: 1 ft
1 m = 0.3048 = 3.28084 ft.
Mt. Everest K2 Kangchenjunga
8850 8611 8586
(We could carry the result to 6 significant figures because 0.3048 is exact, Lhotse
8516
0.304800 g.) We multiply this equation by 8000.0 (to have five significant figures): Makalu
8462
8000.0 m
=
(8000.0 m ) ¢ 3.28084 ft ≤ m
=
26,247 ft.
Cho Oyu Dhaulagiri Manaslu
8201 8167 8156
An elevation of 8000 m is 26,247 ft above sea level.
Nanga Parbat
8125
Note  We could have done the unit conversions all in one line:
8000.0 m
=
(8000.0 m ) ¢ 100 cm ≤ ¢ 1 in. ≤ ¢ 1 ft ≤ 1 m 2.54 cm 12 in.
=
26,247 ft.
Annapurna Gasherbrum I Broad Peak Gasherbrum II
8091 8068 8047 8035
The key is to multiply conversion factors, each equal to one  ( = 1.0000),  and
Shisha Pangma
8013
to make sure which units cancel.
SECTION 1–5  Converting Units  9
GIAN_PSE5_CH01_001-019_ca.indd 9
01/07/20 16:17
TABLE 1 – 6 The 8000-m Peaks
Peak
Height (m)
Mt. Everest K2 Kangchenjunga Lhotse Makalu Cho Oyu Dhaulagiri Manaslu Nanga Parbat Annapurna Gasherbrum I Broad Peak Gasherbrum II Shisha Pangma
8850 8611 8586 8516 8462 8201 8167 8156 8125 8091 8068 8047 8035 8013
Rule of thumb: Floor area in ft 2 is about 10 * area in m2: 100 m2 L 1000 ft 2
P r o b l em S o l v in g Conversion factors = 1
The first two equations in Example 1 9 2 on the previous page show how to change from feet to meters, or meters to feet. For practical purposes
1 m = 3.28 ft L 3.3 ft
which means that we can change any distance or height in meters to feet by multiplying by 3 and adding 10% (0.1). For example, a 3000-m-high peak in feet is 9000 ft + 900 ft L 10,000 ft.
EXERCISE E The names and elevations of the 14 eight-thousand-meter peaks in the world (see Example 1 9 2) are given in Table 1 9 6, repeated here. They are all in the Himalaya mountain range in India, Pakistan, Tibet, and China. Determine the elevation of the world’s three highest peaks in feet.
EXAMPLE 1 – 3 Apartment area. You have seen a nice apartment whose
floor area is 880 square feet (ft2). What is its area in square meters?
Approach  We use the same conversion factor,  1 in. = 2.54 cm, but this time we have to use it twice.
Solution  Because  1 in. = 2.54 cm = 0.0254 m, then
1 ft2 = (12 in.)2(0.0254 m>in.)2 = 0.0929 m2. So
880 ft2 = (880 ft2) (0.0929 m2>ft2) L 82 m2.
Note  As a rule of thumb, an area given in ft2 is roughly 10 times the number of square meters (more precisely, about 10.8*).
Exercise F  One hectare is defined as 1.000 * 104 m2. There are 640 acres in a square
mile. Both units are used for land area. (a) How many acres are in one hectare? (b) What
would be an easy everyday rule-of-thumb conversion factor?
EXAMPLE 1 – 4 Speeds.  Where the posted speed limit is 55 miles per hour (mi>h or mph), what is this speed (a) in meters per second (m>s) and (b) in kilometers per hour (km>h)?
Approach  We again use the conversion factor  1 in. = 2.54 cm, and we recall that there are 5280 feet in a mile and 12 inches in a foot; also, one hour contains  (60 min>h) * (60 s>min) = 3600 s>h.
Solution  (a) We can write 1 mile as
1 mi
=
(5280 ft ) ¢ 12 in. ≤ ¢ 2.54 cm ≤ ¢ 1 m ≤
ft
in. 100 cm
= 1609 m.
We also know that 1 hour contains 3600 s, so
mi 55
h
=
¢ 55 mi ≤ ¢ 1609 m ≤ ¢ 1 h ≤
h
mi 3600 s
=
25
m s
,
where we rounded off to two significant figures.
(b) Now we use  1 mi = 1609 m = 1.609 km;  then
mi 55
h
=
¢ 55 mi ≤ ¢ 1.609 km ≤
h
mi
=
88 km . h
Note  Each conversion factor is equal to one. You can look up most conversion factors in the Table inside the front cover.
P r o b l em S o l v in g
Unit conversion is wrong if units do not cancel
EXERCISE G  Return to the first Chapter-Opening Question, page 1, and answer it again now. Try to explain why you may have answered differently the first time.
When changing units, you can avoid making an error in the use of conversion
factors by checking that units cancel out properly. For example, in our conversion
of
1 mi
to
1609 m
in
Example
1 9 4(a),
if
we
had
incorrectly
used
the
factor
( ) 100 cm 1 m
instead
of
( ) 1 m 100 cm
,
the
centimeter
units
would
not
have
cancelled
out;
we
would
not
have ended up with meters.
10  CHAPTER 1  Introduction, Measurement, Estimating
GIAN_PSE5_CH01_001-019_ca.indd 10
01/07/20 16:17
1–6  Order of Magnitude: Rapid Estimating
This is an exciting and powerful Section that will be useful throughout this book, and in real life. We will see how to make approximate calculations of quantities you may never have dreamed you could do.
Also, we are sometimes interested only in an approximate value for a quantity, maybe because an accurate calculation would take more time than it is worth or requires data that are not available. In other cases, we may want to make a rough estimate in order to check a calculation made on a calculator, to make sure that no blunders were made when the numbers were entered.
A rough estimate can be made by rounding off all numbers to one significant figure and its power of 10, and after the calculation is made, again keeping only one significant figure. Such an estimate is called an order-of-magnitude estimate and can be accurate within a factor of 10, and often better. In fact, the phrase “order of ­magnitude” is sometimes used to refer simply to the power of 10.
P r o b l em S o l v in g How to make a rough estimate
10 m
r = 500 m
(b)
FIGURE 1 – 7   Example 1 9 5. (a) How much water is in this
lake? (Photo is one of the Rae Lakes in the Sierra Nevada
of California.) (b) Model of the lake as a cylinder. [We could go
one step further and estimate the mass or weight of this lake.
We will see later that water has a density of 1000 kg>m3,
so this lake has a mass of about (103 kg>m3) (107 m3) L 1010 kg,
which is about 10 billion kg or 10 million metric tons.
(A metric ton is 1000 kg, about 2200 lb, slightly larger than a
(a)
British ton, 2000 lb.)]
EXAMPLE 1 – 5 ESTIMATE Volume of a lake.  Estimate how much water there is in a p­ articular lake, Fig. 1 9 7a, which is roughly circular, about 1 km across, and you guess it has an average depth of about 10 m.
Approach  No lake is a perfect circle, nor can lakes be expected to have a ­perfectly flat b­ ottom. We are only estimating here. To estimate the volume, we can use a simple model of the lake as a cylinder: we multiply the average depth of the lake times its roughly circular surface area, as if the lake were a cylinder (Fig. 1 9 7b).
Solution  The volume V of a cylinder is the product of its height h times the area of its base:  V = hpr2,  where r is the radius of the circular base.† The radius r is  12 km = 500 m,  so the volume is approximately
V = hpr2 L (10 m) * (3) * (5 * 102 m)2 L 8 * 106 m3 L 107 m3,
where p was rounded off to 3. So the volume is on the order of 107 m3, ten
million cubic meters. Because of all the estimates that went into this calculation,
the order-of-magnitude estimate (107 m3) is probably better to quote than
the 8 * 106 m3 figure.
Note  To express our result in U.S. gallons, we see in the Table on the
inside
front
cover
that
 1 liter
=
10-3 m3
L
1 4
gallon.  Hence,
the
lake
contains
about (8 * 106 m3) (1 gallon>4 * 10 - 3 m3) L 2 * 109  gallons of water.
P h y sics App l ie d
Estimating the volume (or mass) of a lake; see also Fig. 1–7
†Formulas like this for volume, area, etc., are found inside the back cover of this book.
SECTION 1–6  Order of Magnitude: Rapid Estimating  11
GIAN_PSE5_CH01_001-019_ca.indd 11
01/07/20 16:17
FIGURE 1 – 8   Example 1 9 6. Micrometer used for measuring small thicknesses.
EXAMPLE 1 – 6 ESTIMATE Thickness of a sheet of paper.  Estimate the thickness of a page of this book.
Approach  At first you might think that a special measuring device, a ­micro­meter (Fig. 1 9 8), is needed to measure the thickness of one page since an ordinary ruler can not be read so finely. But we can use a trick or, to put it in physics terms, make use of a symmetry: we can make the reasonable assumption that all the pages of this book are equal in thickness.
Solution  We can use a ruler to measure hundreds of pages at once. If you measure the thickness of the first 500 pages of this book (page 1 to page 500), you might get something like 1.5 cm. Note that 500 numbered pages, counted front and back, is 250 separate pieces of paper. So one sheet must have a t­hickness of about
1.5 cm 250 sheets
L
6 * 10-3 cm
=
6 * 10-2 mm,
or less than a tenth of a millimeter (0.1 mm).
It cannot be emphasized enough how important it is to draw a diagram when solving a physics Problem, as the next Example shows.
FIGURE 1 – 9   Example 1 9 7. Diagrams are really useful!
(a) ?
3 m 1.5 m
2 m
(b) x = ?
1.5 m 2 m
16 m 18 m
1.5 m
EXAMPLE 1 – 7 ESTIMATE Height by triangulation.  Estimate the height of the building shown in Fig. 1 9 9, by “triangulation,” with the help of a bus-stop pole and a friend.
Approach  By standing your friend next to the pole, you estimate the height of the pole to be 3 m. You next step away from the pole until the top of the pole is in line with the top of the building, Fig. 1 9 9a. You are 5 ft 6 in. tall, so your eyes are about 1.5 m above the ground. Your friend is taller, and when she stretches out her arms, one hand touches you and the other touches the pole, so you estimate that distance as 2 m (Fig. 1 9 9a). You then pace off the distance from the pole to the base of the building with big, 1-m-long steps, and you get a total of 16 steps or 16 m.
Solution  Now you draw, to scale, the diagram shown in Fig. 1 9 9b using
these measurements. You can measure, right on the diagram, the last side of
the ­triangle to be about  x L 13 or 14 m.  Alternatively, you can use similar triangles to obtain the height x:
1.5 m 2 m
=
x, 18 m
so
x
L
13
1 2
m.
Finally you add in your eye height of 1.5 m above the ground to get your final result: the ­building is about 15 m tall.
EXAMPLE 1 – 8 ESTIMATE Total number of heartbeats. Estimate the total number of beats a typical human heart makes in a lifetime.
Approach  A typical resting heart rate is 70 beats>min. But during exercise it can be a lot higher. A reasonable average might be 80 beats>min.
Solution  One year, in seconds, is  (24 h>d) (3600 s>h) (365 d) L 3 * 107 s.
If an ­average person lives  70 years = (70 yr) (3 * 107 s>yr) L 2 * 109 s,  then
the total number of heartbeats would be about
or 3 billion.
¢ 80
beats min
≤
¢
1 min 60 s
≤
(2
*
109 s)
L
3 * 109,
12  CHAPTER 1  Introduction, Measurement, Estimating
GIAN_PSE5_CH01_001-019_ca.indd 12
01/07/20 16:17
EXAMPLE 1 – 9 ESTIMATE Estimating the radius of Earth. Believe it
or not, you can estimate the radius of the Earth without having to go into space
(see the photograph on page 1). If you have ever been on the shore of a large lake,
you may have noticed that you cannot see the beaches, piers, or rocks at water
level across the lake on the opposite shore. The lake seems to bulge out between
you and the opposite shore : a good clue that the Earth is round. Suppose you
h
climb a stepladder and discover that when your eyes are 10 ft (3.0 m) above the
water, you can just see the rocks at water level on the opposite shore. From a map,
you estimate the distance to the opposite shore as  d L 6.1 km.  Use Fig. 1 9 10
with  h = 3.0 m to estimate the radius R of the Earth.
Approach  We use simple geometry, including the theorem of Pythagoras,  c2 = a2 + b2,  where c is the length of the hypotenuse of any right triangle, and a and b are the lengths of the other two sides.
d Lake
Earth
R
R
Solution  For the right triangle of Fig. 1 9 10, the two sides are the radius of
the Earth R and the distance  d = 6.1 km = 6100 m. The hypotenuse is approximately the length  R + h, where  h = 3.0 m. By the Pythagorean theorem,
R2 + d2 L (R + h)2 L R2 + 2hR + h2.
We solve algebraically for R, after cancelled R2 on both sides:
R
L
d2 - h2 2h
=
(6100 m)2 - (3.0 m)2 6.0 m
= 6.2 * 106 m
= 6200 km.
Note  Precise measurements give 6380 km. But look at your achievement! With a few simple rough measurements and simple geometry, you made a good
­estimate of the Earth’s radius. You did not need to go out in space, nor did you need a very long measuring tape.†
Exercise h  Return to the second Chapter-Opening Question, page 1, and answer it again now. Try to explain why you may have answered differently the first time.
Another type of estimate, this one made famous by Enrico Fermi (1901 9 1954, Fig. 1 9 11), was to show his students how to estimate the number of piano tuners in a city, such as Chicago or San Francisco. To get a rough order-of-magnitude estimate of the number of piano tuners today in San Francisco, a city of about 800,000 inhabitants, we can proceed by estimating the number of functioning pianos, how often each piano is tuned, and how many pianos each tuner can tune. To estimate the number of pianos in San Francisco, we note that certainly not everyone has a piano. A guess of 1 family in 3 having a piano would correspond to 1 piano per 12 persons, assuming an average family of 4 persons.
As an order of magnitude, let’s say 1 piano per 10 people. This is certainly more reasonable than 1 per 100 people, or 1 per every person, so let’s proceed with the estimate that 1 person in 10 has a piano, or about 80,000 pianos in San Francisco. Now a piano tuner needs an hour or two to tune a piano. So let’s ­estimate that a tuner can tune 4 or 5 pianos a day. A piano ought to be tuned every 6 months or a year : let’s say once each year. A piano tuner tuning 4 pianos a day, 5 days a week, 50 weeks a year can tune about 1000 pianos a year. So San Francisco, with its (very) roughly 80,000 pianos, needs about 80 piano tuners. This is, of course, only a rough estimate.‡ It tells us that there must be many more than 10 piano tuners, and surely not as many as 1000.
Center of Earth FIGURE 1 – 10   Example 1 9 9, but not to scale. You can just barely see rocks at water level on the opposite shore of a lake 6.1 km wide if you stand on a stepladder.
FIGURE 1 – 11   Enrico Fermi. Fermi contributed significantly to both theoretical and experimental physics, a feat almost unique in modern times.
P r o b l em S o l v in g Estimating how many piano tuners there are in a city
†As a teenager I had a summer job washing dishes at a camp located 350 m above famous Lake Tahoe in California. Starting the drive down to Lake Tahoe, the beaches across the lake were visible. But approaching the level of Lake Tahoe, the beaches across the lake were no longer visible! I realized that Lake Tahoe was bulging up in the middle, blocking the view. (“The Earth is round.”)
‡A search on the internet (done after this calculation) reveals over 50 listings. Each of these listings may employ more than one tuner, but on the other hand, each may also do repairs as well as tuning. In any case, our estimate is reasonable.
SECTION 1–6  Order of Magnitude: Rapid Estimating  13
GIAN_PSE5_CH01_001-019_ca.indd 13
01/07/20 16:17
*1–7 Dimensions and
Dimensional Analysis
When we speak of the dimensions of a quantity, we are referring to the type of
base units that make it up. The dimensions of area, for example, are always length
squared, abbreviated [L2] using square brackets; the units can be square meters,
square feet, cm2, and so on. Velocity, on the other hand, can be measured in units
of km>h, m>s, or mi>h, but the dimensions are always a length [L] divided by a
time [T ]: that is, [ L>T ].
The formula for a quantity may be different in different cases, but the dimen-
sions remain the same. For example, the area of a triangle of base b and height h is 
A
=
1 2
bh, 
whereas
the
area
of
a
circle
of
radius
r
is
 A
=
pr 2. The
formulas
are
­different in the two cases, but the dimensions of area are always [L2].
Dimensions can be used as a help in working out relationships, a procedure
referred to as dimensional analysis. One useful technique is the use of dimensions
to check if a relationship is incorrect. Note that we add or subtract quantities only
if they have the same dimensions (we don’t add centimeters and hours); and the
quantities on each side of an equals sign must have the same dimensions. (In numer-
ical calculations, the units must also be the same on both sides of an equation.)
For
example,
suppose
you
derived
the
equation
 v
=
v0
+
1 2
at
2, 
where
v
is
the velocity of an object after a time t, v0 is the object’s initial velocity, and the
object ­undergoes an acceleration a. Let’s do a dimensional check to see if this
equation could be correct or is surely incorrect. Note that numerical factors, like
the
1 2
here,
do
not
affect
dimensional
checks.
We
write
a
dimensional
equation
as
follows, ­remembering that the dimensions of velocity are [L>T ] and (as we shall
see in Chapter 2) the dimensions of acceleration are [L>T 2]:
BLR T
≟
BLR T
+
B
L T2
R
[
T
2
]
≟
BLR T
+
[L].
The dimensions are incorrect: on the right side, we have the sum of quantities
whose dimensions are not the same. Thus we conclude that an error was made
in the derivation of the original equation.
A dimensional check can only tell you when a relationship is wrong. It can
not tell you if it is completely right. For example, a dimensionless numerical factor
(such
as
1 2
or
2p)
could
be
missing.
Dimensional analysis can also be used as a quick check on an equation you
are not sure about. For example,  consider a simple pendulum of length ℓ. Suppose
that you can’t remember whether the equation for the period T (the time to make
one back-and-forth swing) is T = 2p1ℓ>g  or T = 2p1g>ℓ,  where g is the accel-
eration due to gravity and, like all accelerations, has dimensions [L>T 2]. (Do not
worry about these formulas : the correct one will be derived in Chapter 11; what
we are concerned about here is a person’s recalling whether it contains ℓ>g or g>ℓ.)
A dimensional check shows that the former (ℓ>g) is correct:
[T ]
=
[L]
B [L>T 2]
=
2[T 2]
=
[T ],
whereas the latter (g>ℓ) is not:
[L>T 2]
1
1
[T ] ≠ C [L] = B [T 2] = [T ] ⋅
The constant 2p has no dimensions and so can’t be checked using d­ imensions. Further uses of dimensional analysis are found in Appendix D.
*Some Sections of this book, such as this one, may be considered optional at the discretion of the ­instructor, and they are marked with an asterisk (*). See the Preface for more details.
14  CHAPTER 1  Introduction, Measurement, Estimating
GIAN_PSE5_CH01_001-019_ca.indd 14
01/07/20 16:17
EXAMPLE 1 – 10 Planck length.  The smallest meaningful measure of length is
called the “Planck length,” and is defined in terms of three fundamental ­constants
# in nature: the speed of light  c = 3.00 * 108 m>s,  the gravitational constant 
G = 6.67 * 10-11 m3>kg s2,  and Planck’s constant  h = 6.63 * 10-34 kg ⋅ m2>s.  The Planck length lP (l is the Greek letter “lambda”) is given by the following combination of these three constants:
lP
=
Gh . A c3
Show that the dimensions of lP are length [L], and find the order of magnitude of lP .
Approach  We rewrite the above equation in terms of dimensions. The
­dimensions of c are [L>T ], of G are [ L3>MT 2 ], and of h are [ ML2>T ].
Solution  The dimensions of lP are
[ L3>MT 2 ] [ ML2>T ]
C
[ L3>T 3 ]
= 2[L2 ] = [L]
which is a length. Good. The value of the Planck length is
# # lP
=
Gh B c3
=
(6.67 * 10-11 m3>kg s2) (6.63 * 10-34 kg m2>s)
C
(3.00 * 108 m>s)3
L
4 * 10-35 m,
which is on the order of 10-34 or 10-35 m.
Note  Some recent theories (Chapters 43 and 44) suggest that the smallest ­particles (quarks, leptons) have sizes on the order of the Planck length, 10-35 m.
These ­theories also suggest that the “Big Bang,” with which the Universe is believed
to have begun, started from an initial size on the order of the Planck length.
Summary
[The Summary that appears at the end of each Chapter in this book gives a brief overview of the main ideas of the Chapter. The S­ ummary cannot serve to give an understanding of the material, which can be accomplished only by a detailed reading of the ­Chapter.]
Physics, like other sciences, is a creative endeavor. It is not simply a collection of facts. Important theories are created with the idea of explaining observations. To be accepted, theories are tested by comparing their predictions with the results of actual experiments. Note that, in general, a theory cannot be “proved” in an absolute sense.
Scientists often devise models of physical phenomena. A model is a kind of picture or analogy that helps to describe the phenomena in terms of something we already know about. A theory, often developed from a model, is usually deeper and more complex than a simple model.
A scientific law is a concise statement, often expressed in the form of an equation, which quantitatively describes a wide range of phenomena.
Questions
1. What are the merits and drawbacks of using a person’s foot as a standard? Consider both (a) a particular person’s foot, and (b) any person’s foot. Keep in mind that it is advantageous that fundamental standards be accessible (easy to compare to), invariable (do not change), indestructible, and reproducible.
2. What is wrong with this road sign: Memphis 7 mi (11.263 km)?
3. Why is it incorrect to think that the more digits you include in your answer, the more accurate it is?
Measurements play a crucial role in physics, but can never be perfectly precise. It is important to specify the uncertainty of a measurement either by stating it directly using the { notation, and>or by keeping only the correct number of significant figures.
Physical quantities are always specified relative to a particular standard or unit, and the unit used should always be stated. The commonly accepted set of units today is the Système ­International (SI), in which the standard units of length, mass, and time are the meter, kilogram, and second.
When converting units, check all conversion factors for ­correct cancellation of units.
Making rough, order-of-magnitude estimates is a very useful technique in science as well as in everyday life.
[*The dimensions of a quantity refer to the combination of base quantities that comprise it. Velocity, for example, has dimensions of [length>time] or [L>T ]. Working with only the dimensions of the various quantities in a given relationship : this technique is called dimensional analysis : makes it possible to check a relationship for correct form.]
4. For an answer to be complete, units need to be specified. Why? 5. You measure the radius of a wheel to be 4.16 cm. If you
­multiply by 2 to get the diameter, should you write the result as 8 cm or as 8.32 cm? Justify your answer. 6. Express the sine of 30.0° with the correct number of significant figures. 7. List assumptions useful to estimate the number of car mechanics in (a) San Francisco, (b) your hometown, and then make the estimates.
Questions  15
GIAN_PSE5_CH01_001-019_ca.indd 15
01/07/20 16:17
MisConceptual Questions
[List all answers that are valid.]
1. The laws of physics (a) are permanent and unalterable. (b) are part of nature and are waiting to be discovered. (c) can change, but only because of evidence that convinces the community of physicists. (d) apply to physics but not necessarily to chemistry or other fields. (e) were basically complete by 1900, and have undergone only minor revisions since. (f ) are accepted by all major world countries, and cannot be changed without international treaties.
2. How should we write the result of the following calculation,
being careful about significant figures? (3.84 s) (37 m>s) + (5.3 s) (14.1 m>s) =
(a) 200 m. (b) 210 m. (c) 216.81 m.
(d) 217 m. (e) 220 m.
3. Four students use different instruments to measure the
length of the same pen. Which measurement implies the
greatest precision?
(a) 160.0 mm. 
(d) 0.00016 km.
(b) 16.0 cm. 
(e) Need more
(c) 0.160 m. 
information.
4. The number 0.0078 has how many significant figures?
(a) 1.
(c) 3.
(b) 2.
(d) 4.
5. How many significant figures does  1.362 + 25.2 have?
(a) 2.
(c) 4.
(b) 3.
(d) 5.
6. Accuracy represents (a) repeatability of a measurement, using a given i­nstrument. (b) how close a measurement is to the true value. (c) an ideal number of measurements to make. (d) how poorly an instrument is operating.
7. Precision represents
(a) repeatability of a measurement, using a given i­nstrument.
(b) how close a measurement is to the true value.
(c) an ideal number of measurements to make.
(d) how poorly an instrument is operating.
8. To convert from ft2 to yd2, you should
(a) multiply by 3.
(d) multiply by 1>9.
(b) multiply by 1>3.
(e) multiply by 6.
(c) multiply by 9.
(f ) multiply by 1>6.
9. Which is not true about an order-of-magnitude estimation?
(a) It gives you a rough idea of the answer.
(b) It can be done by keeping only one significant figure.
(c) It can be used to check if an exact calculation is reasonable.
(d) It may require making some reasonable assumptions in
order to calculate the answer.
(e) It will always be accurate to at least two significant f­igures.
*10. [ L2 ] represents the dimensions for which of the following?
(a) cm2.
(c) m2.
(b) square feet.
(d) All of the above.
Problems
[The Problems at the end of each Chapter are ranked I, II, or III according to estimated difficulty, with (I) Problems being ­easiest. Level III are meant as challenges for the best students. The Problems are arranged by Section, meaning that the reader should have read up to and including that Section, but not only that Section : Problems often depend on earlier material. Next is a set of “General Problems” not arranged by Section and not ranked.]
1 – 3 Measurement, Uncertainty, Significant Figures
(Note: In Problems, assume a number like 6.4 is accurate to {0.1; and 950 is accurate to 2 significant figures ({10) unless 950 is said to be “precisely” or “very nearly” 950, in which case assume  950 { 1.)
1. (I) How many significant figures do each of the ­following numbers have: (a) 777, (b) 81.60, (c) 7.03, (d) 0.03, (e) 0.0086, (f ) 6465, and (g) 8700?
2. (I) Write the following numbers in powers of 10 notation: (a) 5.859, (b) 21.8, (c) 0.0068, (d) 328.65, (e) 0.219, (f ) 444.
3. (I) Write out the following numbers in full with the correct number of zeros: (a) 8.69 * 105,  (b) 9.1 * 103, (c) 2.5 * 10-1,  (d) 4.76 * 102,  and (e) 3.62 * 10-5.
4. (II) What is the percent uncertainty in the measurement  3.25 { 0.35 m?
5. (II) Time intervals measured with a physical stopwatch typically have an uncertainty of about 0.2 s, due to human reaction time at the start and stop moments. What is the percent uncertainty of a hand-timed measurement of (a) 4.5 s, (b) 45 s, (c) 4.5 min?
6. (II)Add (9.2 * 103 s) + (6.3 * 104 s) + (0.008 * 106 s).
7. (II) Multiply  4.079 * 102 m by  0.057 * 10-1 m, taking into account significant figures.
8. (II) What, approximately, is the percent uncertainty for a measurement given as 1.27 m2?
9. (II) For small angles u, the numerical value of sin u is approximately the same as the numerical value of tan u. Find the largest angle for which sine and tangent agree to within two significant figures.
10. (II) A report stated that “a survey of 215 students found that 37.2% had bought a sugar-rich soft drink the day before.” (a) How many students bought a soft drink? (b) What is wrong with the original statement?
11. (II) A watch manufacturer claims that its watches gain or lose no more than 9 seconds in a year. How accurate are these watches, expressed as a percentage?
12. (III) What is the area, and its approximate uncertainty, of a circle of radius  5.1 * 104 cm?
13. (III) What, roughly, is the percent uncertainty in the volume of a spherical beach ball of radius  r = 0.64 { 0.04 m?
1 – 4 and 1 – 5 Units, Standards, SI, Converting Units
14. (I) Write the following as full (decimal) numbers without prefixes on the units: (a) 286.6 mm, (b) 74 mV, (c) 430 mg, (d) 47.2 ps, (e) 22.5 nm, (f ) 2.50 gigavolts.
16  CHAPTER 1  Introduction, Measurement, Estimating
GIAN_PSE5_CH01_001-019_ca.indd 16
01/07/20 16:17
15. (I) Express the following using the prefixes of Table 1 9 4: (a)  3 * 106 volts, (b)  2 * 10-6 meters, (c)  5 * 103 days, (d) 18 * 102 bucks, and (e) 9 * 10-7 seconds.
16. (I) Determine your own height in meters, and your mass in kg.
17. (II) To the correct number of significant figures, use the information inside the front cover of this book to determine the ratio of (a) the surface area of Earth compared to the surface area of the Moon, (b) the volume of Earth compared to the volume of the Moon.
18. (II) Would a driver traveling at 15 m>s in a 35 mi>h zone be exceeding the speed limit? Why or why not?
19. (II) The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this in powers of 10 in (a) years, (b) seconds.
20. (II) The Sun, on average, is 93 million miles from Earth. How many meters is this? Express (a) using powers of 10, and (b) using a metric prefix (km).
21. (II) Express the following sum with the correct number of significant figures:  1.90 m + 142.5 cm + 6.27 * 105 mm.
22. (II) A typical atom has a diameter of about 1.0 * 10-10 m.  (a) What is this in inches? (b) Approximately how many atoms are along a 1.0-cm line, assuming they just touch?
23. (II) Determine the conversion factor between  (a) km>h and mi>h, (b) m>s and ft>s, and (c) km>h and m>s.
24. (II) What is the conversion factor between (a) ft2 and yd2, (b) m2 and ft2?
25. (II) A light-year is the distance light travels in one year
(at speed  =  2.998 * 108 m>s).  (a) How many meters are
there in 1.00 light-year? (b) An astronomical unit (AU) is the ­average distance from the Sun to Earth,  1.50 * 108 km.  How many AU are there in 1.00 light-year?
26. (II) How much longer (percentage) is a one-mile race than a 1500-m race (“the metric mile”)?
27. (II) How many wavelengths of orange krypton-86 light (Section 1 9 4) would fit into the thickness of one page of this book? See Example 1 9 6.
28. (II) Using the French Academy of Sciences’ original definition of the meter, calculate Earth’s circumference and radius in those meters. Give % error relative to today’s accepted values (inside front cover).
29. (II) A passenger jet uses about 12 liters of fuel per km of flight. What is that value expressed as miles per gallon?
30. (II) American football uses a field that is 100.0 yd long, whereas a soccer field is 100.0 m long. Which field is longer, and by how much (give yards, meters, and percent)?
31. (II) (a) How many seconds are there in 1.00 year? (b) How many nanoseconds are there in 1.00 year? (c) How many years are there in 1.00 second?
32. (II) Use Table 1 9 3 to estimate the total number of protons or neutrons in (a) a bacterium, (b) a DNA molecule, (c) the human body, (d) our Galaxy.
33. (II) The diameter of the planet Mercury is 4879 km. (a) What is the surface area of Mercury? (b) How many times larger is the surface area of the Earth?
34. (III) A standard baseball has a circumference of approximately 23 cm. If a baseball had the same mass per unit ­volume (see Tables in Section 1 9 4) as a neutron or a proton, about what would its mass be?
1 – 6 Order-of-Magnitude Estimating (Note:  Remember that for rough estimates, only round numbers are needed both as input to calculations and as final results.)
35. (I) Estimate the order of magnitude (power of 10) of: (a) 3200, (b) 86.30 * 103,  (c) 0.076, and (d) 15.0 * 108.
36. (II) Estimate how many books can be shelved in a c­ ollege library with 6500 m2 of floor space. Assume 8 shelves high, having books on both sides, with corridors 1.5 m wide. Assume books are about the size of this one, on average.
37. (II) Estimate how many hours it would take to run (at 10 km>h) across the U.S. from New York to California.
38. (II) Estimate the number of liters of water a human drinks in a lifetime.
39. (II) Estimate the number of cells in an adult human body, given that a typical cell has a diameter of about 10 mm, and the human body has a density of about 1000 kg>m3 .
40. (II) Estimate how long it would take one person to mow a football field using an ordinary home lawn mower (Fig. 1 9 12). (State your assumptions, such as the mower moves with a 1@km>h speed, and has a 0.5-m width.)
Figure 1 – 12   Problem 40. 41. (II) Estimate the number of gallons of gasoline consumed by the total of all automobile drivers in the U.S., per year. 42. (II) Estimate the number of dentists (a) in San Francisco and (b) in your town or city. 43. (II) Estimate how many kilograms of laundry soap are used in the U.S. in one year (and therefore pumped out of washing machines with the dirty water). Assume each load of laundry takes 0.1 kg of soap. 44. (II) How big is a ton (1000 kg)? That is, what is the volume of something that weighs a ton? To be specific, estimate the diameter of a 1-ton rock, but first make a wild guess: will it be 1 ft across, 3 ft, or the size of a car? [Hint: Rock has mass per volume about 3 times that of water, which is 1 kg per
liter (103 cm3) or 62 lb per cubic foot.]
45. (II) A hiking trail is 270 km long through varying terrain. A group of hikers cover the first 49 km in two and a half days. Estimate how much time they should allow for the rest of the trip.
46. (II) Estimate how many days it would take to walk around the circumference of the Earth, assuming 12 h walking per day at 4 km>h.
47. (II) Estimate the number of jelly beans in the jar of Fig. 1 9 13.
Figure 1 – 13   Problem 47. Estimate the number of jelly beans in the jar.
Problems  17
GIAN_PSE5_CH01_001-019_ca.indd 17
01/07/20 16:17
48. (II) Estimate the number of bus drivers (a) in Washington,
D.C., and (b) in your town.
49. (III) You are in a hot air balloon, 300 m above the flat
Texas plains. You look out toward the horizon. How far out can you see : that is, how far is your horizon? The Earth’s
radius is about 6400 km.
50. (III) I agree to hire you for 30 days. You can decide between
two methods of payment: either (1) $1000 a day, or (2) one
penny on the first day, two pennies on the second day and
continue to double your daily pay each day up through day
30. Use quick estimation to make your decision, and justify it.
51. (III) The rubber worn from tires mostly enters the atmosphere
as particulate pollution. Estimate how much rubber (in kg) is
put into the air in the United States every year. To get started,
a good estimate for a tire tread’s depth is 1 cm when new, and rubber has a mass of about 1200 kg per m3 of volume.
52. (III) Many sailboats are docked at a marina 4.4 km away on
the opposite side of a lake. You stare at one of the sailboats
because, when you are lying flat at the water’s edge, you
can just see its deck but none of the side of the sailboat.
You then go to that sailboat on the other side of the
lake and ­measure that the deck is 1.5 m above the level of the water. Using Fig. 1 9 14, where   h = 1.5 m, estimate the radius R of the Earth.
d
Lake
h
Earth
Figure 1 – 14   Problem 52. You see a sailboat across a lake (not to scale). R is the radius of the Earth. Because of the curvature of the Earth, the water “bulges out” between you and the boat.
RR Earth center
53. (III) You are lying on a beach, your eyes 20 cm above the sand. Just as the Sun sets, fully disappearing over the ­horizon, you immediately jump up, your eyes now 150 cm above the sand, and you can again just see the top of the Sun. If you count the number of seconds  (= t)  until the Sun fully disappears again, you can estimate the Earth’s radius. But for this Probl­em, use the known radius of the Earth to calculate the time t.
*1 – 7 Dimensions
* 54. (I) What are the dimensions of density, which is mass per ­volume?
*55. (II) The speed v of an object is given by the equation  v = At3 - Bt,  where t refers to time. (a) What are the dimensions of A and B? (b) What are the SI units for the
constants A and B?
* 56. (II) Three students derive the following equations in which
x refers to distance traveled, v the speed, a the acceleration
(m>s2), t the time, and the subscript zero (0) means a quantity
at time  t = 0.  Here are their equations:  (a) x = vt2 + 2at,
(b)
x
=
v0 t
+
1 2
at 2,
 and
(c)
x
=
v0
t
+
2at 2.
 Which of
these
could
possibly be c­ orrect according to a dimensional check, and why?
* 57. (II) (a) Show that the following combination of the three fundamental constants of nature that we used in Example 1 9 10 (that is G, c, and h) forms a quantity with the dimensions of time:
tP
=
Gh . A c5
This quantity, tP , is called the Planck time and is thought to be the earliest time, after the creation of the Universe, at
which the currently known laws of physics can be applied. (b) Estimate the order of magnitude of tP using values given inside the front cover (or Example 1 9 10).
General Problems
58. Global positioning satellites (GPS) can be used to determine your position with great accuracy. If one of the satellites is 20,000 km from you, and you want to know your position to{2 m, what percent uncertainty in the distance is required? How many significant figures are needed in the distance?
59. One mole of atoms consists of  6.02 * 1023 individual atoms. If a mole of atoms were spread uniformly over the Earth’s surface, how many atoms would there be per square meter?
60. Computer chips (Fig. 1 9 15) can be etched on circular silicon wafers of thickness 0.300 mm that are sliced from a solid cylindrical silicon crystal of length 25 cm. If each wafer can hold 750 chips, what is the maximum number of chips that can be produced from one entire cylinder?
Figure 1 – 15   Problem 60. The wafer held by the hand is shown below, enlarged and ­illuminated by colored light. Visible are rows of integrated circuits (chips).
61. If you used only a keyboard to enter data, how many years would it take to fill up a hard drive in a computer
that can store 1.0 terabytes  (1.0 * 1012 bytes)  of data?
Assume 40-hour work weeks, and that you can type 150 characters per minute, and that one byte is one keyboard character.
62. An average family of four uses roughly 1200 L (about
300 gallons) of water per day  (1 L = 1000 cm3).  How much
depth would a lake lose per year if it covered an area of 60 km2 with uniform depth and supplied a local town with a population of 40,000 people? Consider only population uses, and neglect evaporation, rain, creeks and rivers.
63. A certain compact disc (CD) contains 783.216 megabytes of digital information. Each byte consists of exactly 8 bits. When played, a CD player reads the CD’s information at a constant rate of 1.4 megabits per second. How many ­minutes does it take the player to read the entire CD?
64. An angstrom (symbol Å) is a unit of length, defined as 10-10 m, which is on the order of the diameter of an atom. (a) How many nanometers are in 1.0 angstrom? (b) How many femtometers or fermis (the common unit of length in nuclear physics) are in 1.0 angstrom? (c) How many ­angstroms are in 1.0 m? (d) How many angstroms are in 1.0 light-year (see Problem 25)?
18  CHAPTER 1  Introduction, Measurement, Estimating
GIAN_PSE5_CH01_001-019_ca.indd 18
01/07/20 16:17
65. A typical adult human lung contains about 300 million tiny cavities called alveoli. Estimate the average diameter of a single alveolus.
66. Hold a pencil in front of your eye at a position where its blunt end just blocks out the Moon (Fig. 1 9 16). Make appropriate measurements to estimate the diameter of the Moon, given that the Earth 9 Moon distance is  3.8 * 105 km.
72. Determine the percent uncertainty in u, and in sin u, when
(a) u = 15.0° { 0.5°,  (b) u = 75.0° { 0.5°.
73. Jim stands beside a wide river and wonders how wide it
is. He spots a large rock on the bank directly across from
him. He then walks upstream 85 strides and judges that
the angle between him and
the rock, which he can still
see, is now at an angle of
30° downstream (Fig. 1 9 17).
Jim measures his stride to be
about 0.8 m long. Estimate
the width of the river.
30°
Figure 1 – 16   Problem 66. How big is the Moon?
67. A storm dumps 1.0 cm of rain on a city 5 km wide
and 7 km long in a 2-h period. How many metric tons  
(1 metric ton = 103 kg)  of water fell on the city? (1 cm3 of water has a mass of  1 g = 10-3 kg.)  How many gallons
of water was this?
68. Greenland’s ice sheet covers over  1.7 * 106 km2  and is
approximately 2.5 km thick. If it were to melt completely then
by
how
much
would
you
expect
the
ocean
to
rise?
Assume
2 3
of Earth’s surface is ocean. See Tables inside front and back
covers.
69. Noah’s ark was ordered to be 300 cubits long, 50 cubits wide, and 30 cubits high. The cubit was a unit of measure equal to the length of a human forearm, elbow to the tip of the ­longest finger. Express the dimensions of Noah’s ark in
meters, and estimate its volume (m3).
70. One liter (1000 cm3) of oil is spilled onto a smooth lake. If
the oil spreads out uniformly until it makes an oil slick just
one molecule thick, with adjacent molecules just touching,
estimate the diameter of the oil slick. Assume the oil molecules have a diameter of  2 * 10-10 m.
71. If you walked north along one of Earth’s lines of longitude until you had changed latitude by 1 minute of arc (there are 60 minutes per degree), how far would you have walked (in miles)? This distance is a nautical mile (page 7).
Figure 1 – 17  Problem 73.
85 Strides
74. Make a rough estimate of the volume of your body (in m3).
75. Estimate the number of plumbers in San Francisco.
76. Estimate the ratio (order of magnitude) of the mass of a human to the mass of a DNA molecule. [Hint: Check the Tables in this Chapter.]
77. The following formula estimates an average person’s lung
capacity V (in liters, where  1 L = 103 cm3):
V = 4.1H - 0.018A - 2.7,
where H and A are the person’s height (in meters) and age (in years), respectively. In this formula, what are the units of the numbers 4.1, 0.018, and 2.7?
78. The density of an object is defined as its mass divided by its volume. Suppose a rock’s mass and volume are measured to be 6 g and 2.8325 cm3. To the correct number of significant figures, determine the rock’s density (mass>volume).
79. Recent findings in astrophysics suggest that the observable universe can be modeled as a sphere of radius R = 13.7 * 109 light-years = 13.0 * 1025 m with an average total mass density of about 1 * 10-26 kg>m3. Only about 4% of total mass is due to “ordinary” matter (such as protons, neutrons, and electrons). Estimate how much ordinary matter (in kg) there is in the observable universe. (For the light-year, see Problem 25.)
Answers to Exercises
A: (d). B: All three have three significant figures; the number of
decimal places is (a) 2, (b) 3, (c) 4. C: No: they have three and two, respectively. D: (a) 2.58 * 10-2, 3;  (b) 4.23 * 104, 3 (probably);
(c) 3.4450 * 102, 5.
E: Mt. Everest, 29,035 ft; K2, 28,251 ft; Kangchenjunga, 28,169 ft.
F:
(a)
2.47 acres
in
1
hectare;
(b)
2
1 2
or
even
just
2 acres
in
1 hectare.
G: (f ) 1,000,000; that is, one million.
H: (c).
GIAN_PSE5_CH01_001-019_ca.indd 19
General Problems  19
01/07/20 16:17
A space shuttle has released a parachute to reduce its speed quickly. The directions of the shuttle’s velocity and acceleration are shown by the green (v5) and gold (a5) arrows.
Motion is described using the concepts of velocity and acceleration. In the case shown here, the velocity v5 is to the right, in the direction of motion. The acceleration a5 is in the opposite direction from the velocity v5, which means the object is slowing down.
We examine in detail motion with constant acceleration, including the vertical motion of objects falling under gravity.
a
v
C
ha p te
2
CONTENTS
2–1 Reference Frames and Displacement
2–2 Average Velocity 2–3 Instantaneous Velocity 2–4 Acceleration 2–5 Motion at Constant
Acceleration 2–6 Solving Problems 2–7 Freely Falling Objects *2–8 Variable Acceleration;
Integral Calculus
20

GIAN_PSE5_CH02_020-053_ca.indd 20
r
Describing Motion: Kinematics in One Dimension
CHAPTER-OPENING QUESTION—Guess now!
[Don’t worry about getting the right answer now : you will get another chance later in the Chapter. See also page 1 of Chapter 1 for more explanation.]
Two small heavy balls have the same diameter but one weighs twice as much as the other. The balls are dropped from a second-story balcony at the exact same time. The time to reach the ground below will be:
(a) twice as long for the lighter ball as for the heavier one. (b) longer for the lighter ball, but not twice as long. (c) twice as long for the heavier ball as for the lighter one. (d) longer for the heavier ball, but not twice as long. (e) nearly the same for both balls.
T he motion of objects : baseballs, automobiles, joggers, and even the Sun and Moon : is an obvious part of everyday life. It was not until the sixteenth and seventeenth centuries that our modern understanding of motion was established. Many individuals contributed to this understanding, particularly Galileo Galilei (1564 9 1642) and Isaac Newton (1642 9 1727).
The study of the motion of objects, and the related concepts of force and energy, form the field called mechanics. Mechanics is customarily divided into two parts: ­kinematics, which is the description of how objects move, and dynamics, which deals with force and why objects move as they do. This Chapter and the next deal with kinematics.
01/07/20 16:18
For now we only discuss objects that move without rotating (Fig. 2 9 1a). Such motion is called translational motion. In this Chapter we will be concerned with describing an object that moves along a straight-line path, which is one-­ dimensional translational motion. In Chapter 3 we will describe translational motion in two (or three) dimensions along paths that are not straight. (Rotation, shown in Fig. 2 9 1b, is discussed in Chapters 10 and 11.)
We will often use the concept, or model, of an idealized particle which is considered to be a mathematical point with no spatial extent (no size). A point particle can undergo only translational motion. The particle model is useful in many real situations where we are interested only in translational motion and the object’s size is not significant. For example, we might consider a billiard ball, or even a spacecraft traveling toward the Moon, as a particle for many purposes.
2–1  Reference Frames and Displacement
Any measurement of position, distance, or speed must be made with respect to a reference frame, or frame of reference. For example, while you are on a train traveling at 80 km>h, suppose a person walks past you toward the front of the train at a speed of, say, 5 km>h (Fig. 2 9 2). This 5 km>h is the person’s speed with respect to the train as frame of reference. With respect to the ground, that person is moving at a speed of  80 km>h + 5 km>h = 85 km>h.  It is always important to specify the frame of reference when stating a speed. In everyday life, we usually mean “with respect to the Earth” without even thinking about it, but the reference frame must be specified whenever there might be confusion.
(a)
(b)
FIGURE 2 – 1   A falling pinecone undergoes (a) pure translation; (b) it is rotating as well as translating.
FIGURE 2 – 2   A person walks toward the front of a train at 5 km>h. The train is moving at 80 km>h with respect to the ground, so the walking person’s speed, relative to the ground, is 85 km>h.
When specifying the motion of an object, it is important to specify not only the speed but also the direction of motion. Often we can specify a direction by using north, east, south, and west, and by “up” and “down.” In physics, we often draw a set of coordinate axes, as shown in Fig. 2 9 3, to represent a frame of reference. We can always place the origin 0, and the directions of the x and y axes, as we like for convenience. The x and y axes are always perpendicular to each other. The origin is where  x = 0,  y = 0.  Objects positioned to the right of the origin of ­coordinates (0) on the x axis have an x coordinate which we almost always choose to be positive; objects at points to the left of 0 have a negative x ­coordinate. The position along the y axis is usually considered positive when above 0, and negative when below 0, although the reverse convention can be used if convenient. Any point on the xy plane can be specified by giving its x and y coordinates. In three dimensions, a z axis perpendicular to the x and y axes is added.
For one-dimensional motion, we often choose the x axis as the line along which the motion takes place. Then the position of an object at any moment is given by its x coordinate. If the motion is vertical, as for a dropped object, we usually use the y axis.
FIGURE 2 – 3   Standard set of xy coordinate axes, sometimes called “rectangular coordinates.” [Also called Cartesian coordinates, after René Descartes (1596 9 1650), who invented them.]
+ y
- x
0
+ x
- y
SECTION 2–1  Reference Frames and Displacement  21

GIAN_PSE5_CH02_020-053_ca.indd 21
01/07/20 16:18
CAUTION The displacement may not equal the
total distance traveled
y
70 m x
West 0 40 m 30 m East
Displacement
FIGURE 2 – 4   A person walks 70 m east, then 30 m west. The total distance traveled is 100 m (path is shown dashed in black); but the displacement, shown as a solid blue arrow, is 40 m to the east.
FIGURE 2 – 5   The arrow represents the displacement  x2 - x1 .  Distances are in meters.
y
x1
x2
0 10 20 30 40 x
Distance (m)
FIGURE 2 – 6   For the displacement ∆x = x2 - x1 = 10.0 m - 30.0 m,  the displacement vector points left.
y
x2
x1
∆x 0 10 20 30 40 x
Distance (m)
We need to make a distinction between the distance an object has traveled and its displacement, which is defined as the change in position of the object.
That is, displacement is how far the object is from its starting point. To see the distinction between total distance and displacement, imagine a person walking 70 m to the east and then turning around and walking back (west) a distance of 30 m (see Fig. 2 9 4). The total distance traveled is  70 m + 30 m = 100 m,  but the displacement is only 40 m since the person is now only 40 m from the starting point.
Displacement is a quantity that has both magnitude and direction. Such quantities are called vectors, and are represented by arrows in diagrams. For example, in Fig. 2 9 4, the blue arrow represents the displacement whose magnitude is 40 m and whose direction is to the right (east).
We will deal with vectors more fully in Chapter 3. For now, we deal only with motion in one dimension, along a line. In this case, vectors which point in one direction will be positive (usually to the right along the x axis). Vectors that point in the opposite direction will have a negative sign in front of their magnitude.
Consider the motion of an object over a particular time interval. Suppose that at some initial time, call it t1 , the object is on the x axis at the position x1 in the coordinate system shown in Fig. 2 9 5. At some later time, t2 , suppose the object has moved to position x2 . The displacement of our object is  x2 - x1 ,  and is represented by the arrow pointing to the right in Fig. 2 9 5. It is convenient to write
∆x = x2 - x1 ,
where the symbol ∆ (Greek letter delta) means “change in.” Then ∆x means “the change in x,” or “change in position,” which is in fact the displacement. The change in any quantity means the final value of that quantity, minus the initial value. Suppose x1 = 10.0 m  and  x2 = 30.0 m,  as in Fig. 2 9 5. Then
∆x = x2 - x1 = 30.0 m - 10.0 m = 20 .0 m,
so the displacement is 20.0 m in the positive direction, Fig. 2 9 5. Now consider an object moving to the left as shown in Fig. 2 9 6. Here the object,
a person, starts at  x1 = 30.0 m  and walks to the left to the point  x2 = 10.0 m.  In this case her displacement is
∆x = x2 - x1 = 10.0 m - 30.0 m = - 20.0 m,
and the blue arrow representing the vector displacement points to the left. For one-dimensional motion along the x axis, a vector pointing to the right is positive, whereas a vector pointing to the left has a negative sign.
EXERCISE A  An ant starts at  x = 20 cm  on a piece of graph paper and walks along the x axis to  x = - 20 cm.  It then turns around and walks back to  x = - 10 cm.  Determine (a) the ant’s displacement and (b) the total distance traveled.
2–2  Average Velocity
An important aspect of the motion of a moving object is how fast it is moving : its
speed or velocity.
The term “speed” refers to how far an object travels in a given time interval,
regardless of direction. If a car travels 240 kilometers (km) in 3 hours (h), we
say its average speed was 80 km>h. In general, the average speed of an object is
defined as the total distance traveled along its path divided by the time it takes to
travel this distance:
average speed
=
distance traveled . time elapsed
(2 – 1)
The terms “velocity” and “speed” are often used interchangeably in ordinary language. But in physics we make a distinction between the two. Speed is simply a positive number, with units. Velocity, on the other hand, is used to signify both the magnitude (numerical value) of how fast an object is moving and also the direction in which it is moving. Velocity is therefore a vector.
22  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 22
01/07/20 16:18
There is a second difference between speed and velocity: namely, the average
velocity is defined in terms of displacement, rather than total distance traveled:
average velocity
=
displacement time elapsed
=
final position - initial position . time elapsed
Average speed and average velocity have the same magnitude when the
motion is all in one direction. In other cases, they may differ: recall the walk we described earlier, in Fig. 2 9 4, where a person walked 70 m east and then 30 m
west. The total distance traveled was  70 m + 30 m = 100 m,  but the displacement was 40 m. Suppose this walk took 70 s to complete. Then the average speed was:
distance time elapsed
=
100 m 70 s
=
1.4 m>s.
The magnitude of the average velocity, on the other hand, was:
displacement time elapsed
=
40 m 70 s
=
0.57 m>s.
In everyday life, we are usually interested in average speed. If this second equation
on average velocity seems strange, we will see its usefulness in the next Section.
To discuss one-dimensional motion of an object in general, suppose that at
some moment in time, call it t1 , the object is on the x axis at position x1 in a coordinate system, and at some later time, t2 , suppose it is at position x2 . The elapsed time (= change in time) is  ∆t = t2 - t1 .  During this time interval the displacement of our object is  ∆x = x2 - x1 . Then the average velocity, defined as the displacement divided by the elapsed time, can be written
v
=
x2 - x1 t2 - t1
=
∆x , ∆t
[average velocity]  (2 – 2)
where v stands for velocity and the bar ( ) over the v is a standard symbol
meaning “average.”
It is always important to choose (and state) the elapsed time, or time interval, t2 - t1 ,  the time that passes during our chosen period of observation.
EXAMPLE 2 – 1 Runner’s average velocity. The position of a runner is plotted as moving along the x axis of a coordinate system. During a 3.00-s time interval, the runner’s position changes from  x1 = 50.0 m  to  x2 = 30.5 m,  as shown in Fig. 2 9 7.  What is the runner’s average velocity?
APPROACH  We want to find the average velocity, which is the displacement divided by the elapsed time.
SOLUTION  The displacement is
∆x = x2 - x1 = 30.5 m - 50.0 m = -19.5 m.
In this case the displacement is negative.
The elapsed time, or time interval, is given as  ∆t = 3.00 s.  The average velocity (Eq. 2 9 2) is
v
=
∆x ∆t
=
- 19.5 m 3.00 s
=
- 6.50 m>s.
The displacement and average velocity are negative: that is, the runner is moving to the left along the x axis, as indicated by the arrow in Fig. 2 9 7. The runner’s average velocity is 6.50 m>s to the left.
CAUTION
Average speed is not necessarily equal to the magnitude of the average velocity
CAUTION Time interval = elapsed time
FIGURE 2 – 7   Example 2 9 1. A person runs from  x1 = 50.0 m  to  x2 = 30.5 m. The displacement is - 19.5 m.
y
Finish Start (x2) (x1)
∆x
0
x 10 20 30 40 50 60
Distance (m)
For one-dimensional motion in the usual case of the +x axis to the right, if x2 is less than x1 , then the object is moving to the left, and  ∆x = x2 - x1 is less than zero. The sign of the displacement, and thus of the average velocity,
indicates the direction: the average velocity is positive for an object moving to the
right along the x axis and negative when the object moves to the left. The direction
of the average velocity is always the same as the direction of the displacement.
PROBLEM SOLVING
+ or - sign can signify the direction for linear motion
SECTION 2–2  Average Velocity  23

GIAN_PSE5_CH02_020-053_ca.indd 23
01/07/20 16:18
EXAMPLE 2 – 2 Distance a cyclist travels. How far can a cyclist travel in 2.5 h along a straight road if her average velocity is 18 km>h?
APPROACH  We want to find the distance traveled, which in this case equals the displacement ∆x, so we solve Eq. 2 9 2 for ∆x. SOLUTION  In Eq. 2 9 2,  v = ∆x>∆t,  we multiply both sides by ∆t and obtain
∆x = v ∆t = (18 km>h) (2.5 h) = 45 km.
EXAMPLE 2 – 3 Car changes speed.  A car travels at a constant 50 km>h for 100 km. It then speeds up to 100 km>h and is driven another 100 km. What is the car’s average speed for the 200-km trip?
APPROACH  At 50 km>h, the car takes 2.0 h to travel 100 km. At 100 km>h, it takes only 1.0 h to travel 100 km. We use the definition of average velocity, Eq. 2 9 2.
SOLUTION  Average velocity (Eq. 2 9 2) is
v
=
∆x ∆t
=
100 km + 100 km 2.0 h + 1.0 h
=
67 km>h.
NOTE  Averaging the two speeds,  (50 km>h + 100 km>h)>2 = 75 km>h,  gives a wrong answer. Can you see why? You must use the definition of v, Eq. 2 9 2.
2–3  Instantaneous Velocity
FIGURE 2 – 8   Car speedometer showing mi>h in white, and km>h
in orange.
If you drive a car along a straight road for 150 km in 2.0 h, the magnitude of your average velocity is 75 km>h. It is unlikely, though, that you were moving at precisely 75 km>h at every instant. To describe this situation we need the
concept of instantaneous velocity, which is the velocity at any instant of time. (Its magnitude is the number, with units, indicated by a speedometer, Fig. 2 9 8.)
More precisely, the ­instantaneous velocity at any moment is defined as the average velocity over an infinitesimally short time interval. That is, Eq. 2 9 2 is to be evalu­ ated in the limit of ∆t becoming extremely small, approaching zero. We can write the definition of instantaneous velocity, v, for one-dimensional motion as
v
=
lim
∆t S 0
∆x ∆t
.
[instantaneous velocity]  (2 – 3)
The notation lim∆tS0 means the ratio ∆x>∆t is to be evaluated in the limit of ∆t approaching zero. But we do not simply set  ∆t = 0  in this definition, for then ∆x would also be zero, and we would not be able to evaluate it. Rather, we consider the ratio ∆x>∆t, as a whole. As we let ∆t approach zero, ∆x approaches zero as well. But the ratio ∆x>∆t approaches some definite value, which is the
instantaneous velocity at a given instant. In Eq. 2 9 3, the limit as  ∆t S 0  is written in calculus notation as dx>dt and is
called the derivative of x with respect to t:
v
=
∆x
lim
∆t S 0
∆t
=
dx . dt
(2 – 4)
This equation is the definition of instantaneous velocity for one-dimensional motion.
For instantaneous velocity we use the symbol v, whereas for average velocity we use v, with a bar above. In the rest of this book, when we use the term “velocity” it will refer to instantaneous velocity. When we want to speak of the average velocity, we will make this clear by including the word “average.”
Note that the instantaneous speed always equals the magnitude of the instantaneous velocity. Why? Because as the time interval becomes infinitesimally small (∆t S 0), an object has no time to change speed or direction, and so the distance traveled and the magnitude of the ­displacement have to be the same.
24  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 24
01/07/20 16:18
Velocity (km>h)
If an object moves at a uniform (that is, constant) velocity during a partic-
ular time interval, then its instantaneous velocity at any instant is the same as its
average velocity (see Fig. 2 9 9a). But in many situations this is not the case. For example, a car may start from rest, speed up to 50 km>h, remain at that velocity for a time, then slow down to 20 km>h in a traffic jam, and finally stop at its
destination after traveling a total of 15 km in 30 min. This trip is plotted on the
graph of Fig. 2 9 9b. Also shown on the graph is the average velocity (dashed line), which is  v = ∆x>∆t = 15 km>0.50 h = 30 km>h.
To better understand instantaneous velocity, let us consider a graph of the ­position versus time (x vs. t) of a particle moving along the x axis, as shown in
Fig. 2 9 10. (Note that this is different from showing the “path” of a particle moving in two dimensions on an x vs. y plot.) The particle is at position x1 at time t1 , and at position x2 at time t2 .   P1 and P2 represent these two points on the graph. A straight line
drawn from point P1 (x1 , t1) to point P2 (x2 , t2) forms the hypotenuse of a right triangle whose sides are ∆x and ∆t . The ratio ∆x>∆t is the slope of the straight line P1 P2 . But ∆x>∆t is also the average velocity of the particle during the time interval  ∆t = t2 - t1 . Therefore, we conclude that the average velocity of a particle during any time interval  ∆t = t2 - t1  is equal to the slope of the straight line (or chord) connecting the two points (x1 , t1) and (x2 , t2) on an x vs. t graph.
Consider now a time ti, intermediate between t1 and t2 , at which time the particle is at xi (Fig. 2 9 11). The slope of the straight line P1 Pi is less than the slope of P1 P2 in this case. Thus the average velocity during the time interval  ti - t1  is less than during the time interval  t2 - t1 .
Now let us imagine that we take the point Pi in Fig. 2 9 11 to be closer and closer to point P1 . That is, we let the interval  ti - t1 ,  which we now call ∆t , become smaller and smaller. The slope of the line connecting the two points becomes
closer and closer to the slope of a line tangent to the curve at point P1 . The average velocity (equal to the slope of the chord) thus approaches the slope of
the tangent at point P1 . The definition of the instantaneous velocity (Eq. 2 9 3) is the limiting value of the average velocity as ∆t approaches zero. Thus the instantaneous velocity equals the slope of the tangent to the x vs. t curve at that point
(which we can simply call “the slope of the curve” at that point).
Because the velocity at any instant equals the slope of the tangent to the x vs. t graph at that instant, we can obtain the velocity at any instant from such a
graph. For example, in Fig. 2 9 12 (which shows the same curve as in Figs. 2 9 10 and
2 9 11), the slope continually increases as our object moves from x1 to x2 , so the velocity is increasing. For times after t2 , however, the slope begins to decrease and in fact reaches zero (so  v = 0)  where x has its maximum value, at point P3 in Fig. 2 9 12. Beyond this point, the slope is negative, as for point P4 . The velocity is therefore negative, which makes sense since x is now decreasing : the particle
is moving to the left on a standard xy plot, toward decreasing values of x.
If an object moves with constant velocity over a particular time interval, its instantaneous velocity is equal to its average velocity. The graph of x vs. t
in this case will be a straight line whose slope equals the velocity. The curve
of Fig. 2 9 10 has no straight sections, so there are no time intervals when the
velocity is constant.
x
FIGURE 2 – 12   Same x vs. t curve as in Figs. 2 9 10 and 2 9 11, but here showing the slope at four different points: At P3 , the slope is zero, so  v = 0.  At P4 the slope is negative, so  v 6 0.
x2 x1 P1
0 t1
P3
P2
P4
t
t2
t3
60
40
20
0
0 0.1 0.2 0.3 0.4 0.5
(a)
Time (h)
Velocity (km>h)
60
40
Average velocity
20
0
0 0.1 0.2 0.3 0.4 0.5
(b)
Time (h)
FIGURE 2 – 9   Velocity of a car as a function of time: (a) at constant velocity; (b) with velocity varying in time.
FIGURE 2 – 10   Graph of a particle’s position x vs. time t. The slope of the straight line P1 P2 represents the average velocity of the particle during the time interval ∆t = t2 - t1 .
x
x2
P2
∆x = x2 - x1
x1 P1 ∆t = t2 - t1
0 t1
t2
t
FIGURE 2 – 11  Same position vs.
time curve as in Fig. 2–10, but including an intermediate time ti. Note that the average velocity over the time interval  ti - t1  (which is the slope of P1 Pi) is less than the average velocity over the time interval  t2 - t1 .  The slope of the thin line tangent
to the curve at point P1 equals the instantaneous velocity at time t1 .
x
x2
P2
xi
Pi
x1
P1 tangent
at
P1
0 t1
ti t2
t
SECTION 2–3  Instantaneous Velocity  25

GIAN_PSE5_CH02_020-053_ca.indd 25
01/07/20 16:18
EXERCISE B  What is your speed at the instant you turn around to move in the opposite direction? (a) Depends on how quickly you turn around; (b) always zero; (c) always negative; (d) none of the above.
The derivatives of various functions are studied in calculus courses, and you can find a review in this book in Appendix B. The derivatives of polynomial functions (which we use a lot) are:
d dt
( Ct n )
= nCt n - 1
and
dC dt = 0,
where C is any constant.
FIGURE 2 – 13   Example 2 9 4.
(a) Engine traveling on a straight track. (b) Graph of x vs. t: x = At2 + B.
x (m)
0 10 20 30 40 50 60
x1
x2
(a)
x (m)
Tangent at P2 whose slope is v2 = 21.0 m>s
60
50
Slope of straight line
40 v = 16.8 m>s
30
20
P1
P2 ∆x = 33.6 m
10
∆t = 2.00 s
0
t (s)
0 123456
(b)
EXAMPLE 2 – 4 Given x as a function of t.  A jet engine moves along an
experimental track (which we call the x axis) as shown in Fig. 2 9 13a. We will
treat the engine as if it were a particle. Its position as a function of time is given by the equation  x = At2 + B, where  A = 2.10 m>s2  and  B = 2.80 m,  and this equation is plotted in Fig. 2 9 13b. (a) Determine the displacement of the engine during the time interval from  t1 = 3.00 s  to  t2 = 5.00 s. (b) Determine the average velocity during this time interval. (c) Determine the magnitude of the instantaneous velocity at  t = 5.00 s.
APPROACH  (a) We substitute values for t1 and t2 in the given equation for x to obtain x1 and x2 . (b) The average velocity can be found from Eq. 2 9 2. (c) To find the instantaneous velocity, we take the derivative of the given x equation with respect to t using the formulas given above.
SOLUTION  (a) At  t1 = 3.00 s,  the position (point P1 in Fig. 2 9 13b) is
x1
=
At
2 1
+
B
=
(2.10 m>s2) (3.00 s)2 + 2.80 m
=
21.7 m.
At  t2 = 5.00 s,  the position (P2 in Fig. 2 9 13b) is
x2 = (2.10 m>s2) (5.00 s)2 + 2.80 m = 55.3 m.
The displacement is thus
x2 - x1 = 55.3 m - 21.7 m = 33.6 m.
(b) The magnitude of the average velocity can then be calculated as
v
=
∆x ∆t
=
x2 - x1 t2 - t1
=
33.6 m 2.00 s
=
16.8 m>s.
This equals the slope of the straight line joining points P1 and P2 shown in Fig. 2 9 13b.
(c) The instantaneous velocity at  t = t2 = 5.00 s  equals the slope of the tangent to the curve at point P2 shown in Fig. 2 9 13b. We could measure this slope off the graph to obtain v2 . But we can calculate v more precisely for any time t, using the given formula
x = At2 + B,
which is the engine’s position x as a function of time t. We take the derivative of x with respect to time (see formulas at top of this page):
v
=
dx dt
=
d dt
( At 2
+
B)
=
2At.
We are given  A = 2.10 m>s2, so for  t = t2 = 5.00 s, v2 = 2At = 2(2.10 m>s2) (5.00 s) = 21.0 m>s.
26  CHAPTER 2  Describing Motion: Kinematics in One Dimension 
GIAN_PSE5_CH02_020-053_ca.indd 26
01/07/20 16:19
2–4 Acceleration
An object whose velocity is changing is said to be accelerating. For instance, a car whose velocity increases in magnitude from zero to 80 km>h is accelerating.
­Acceleration specifies how rapidly the velocity of an object is changing.
Average Acceleration
Average acceleration is defined as the change in velocity divided by the time taken to make this change:
average acceleration
=
change of velocity . time elapsed
In symbols, the average acceleration over a time interval  ∆t = t2 - t1  during which the velocity changes by  ∆v = v2 - v1  is defined as
a
=
v2 - v1 t2 - t1
=
∆v . ∆t
(2 – 5)
Because velocity is a vector, acceleration is a vector too. But for one-dimensional motion, we need only use a plus or minus sign to indicate acceleration direction relative to a chosen coordinate axis.
EXAMPLE 2 – 5 Average acceleration.  A car accelerates along a straight road from rest to 90 km>h in 5.0 s, Fig. 2 9 14. What is the magnitude of its average
acceleration?
APPROACH  Average acceleration is the change in velocity divided by the elapsed time, 5.0 s. The car starts from rest, so  v1 = 0. The final velocity is v2 = 90 km>h = 90 * 103 m>3600 s = 25 m>s.
SOLUTION  From Eq. 2 9 5, the average acceleration is
a
=
v2 - v1 t2 - t1
=
25 m>s - 0 m>s 5.0 s
=
5.0
m>s s
.
This is read as “five meters per second per second” and means that, on average, the velocity changed by 5.0 m>s during each second. That is, assuming the
acceleration was constant, during the first second the car’s velocity increased from zero to 5.0 m>s. During the next second its velocity increased by another 5.0 m>s, reaching a velocity of 10.0 m>s at  t = 2.0 s, and so on. See Fig. 2 9 14.
t1 = 0 v1 = 0
Acceleration [a = 5.0 m>s2]
at t = 1.0 s v = 5.0 m>s
at t = 2.0 s v = 10.0 m>s
at t = t2 = 5.0 s v = v2 = 25 m>s
FIGURE 2 –  14  Example 2 9 5. The
car is shown at the start with  v1 = 0  at  t1 = 0. The car is shown three more times, at  t = 1.0 s,  t = 2.0 s,  and at the end of our time interval, t2 = 5.0 s. The green arrows represent the velocity vectors, whose
length represents the magnitude of
the velocity at that moment and get
longer with time. The acceleration
vector is the orange arrow, whose
magnitude is constant and is found to equal 5.0 m>s2. Distances are not
to scale.

GIAN_PSE5_CH02_020-053_ca.indd 27
SECTION 2–4  Acceleration  27
01/07/20 16:19
CAUTION Distinguish velocity from acceleration
We almost always write the units for acceleration as m>s2 (meters per second squared) instead of m>s>s. This is possible because:
m>s s
=
m s⋅s
=
m. s2
According to the calculation in Example 2 9 5, the velocity changed on average by 5.0 m>s during each second, for a total change of 25 m>s over the 5.0 s; the average acceleration was 5.0 m>s2.
Note that acceleration tells us how quickly the velocity changes, whereas
velocity tells us how quickly the position changes.
CAUTION If v or a is zero, is the other zero too?
CONCEPTUAL EXAMPLE 2 – 6 Velocity and acceleration. (a) If the velo­ ci­ty of an object is zero, does it mean that the acceleration is zero? (b) If the acceleration is zero, does it mean that the velocity is zero? Think of some examples.
RESPONSE  A zero velocity does not necessarily mean that the acceleration is zero, nor does a zero acceleration mean that the velocity is zero. (a) For example, when you put your foot on the gas pedal of your car which is at rest, the velocity starts from zero but the acceleration is not zero since the velocity of the car changes. (How else could your car start forward if its velocity weren’t changing : that is, accelerating?) (b) As you cruise along a straight highway at a constant velocity of 100 km>h, your acceleration is zero:  a = 0, v ≠ 0.
EXERCISE C  A powerful car is advertised to go from zero to 60 mi>h in 5.4 s. What does this say about the car: (a) it is fast (high speed); or (b) it accelerates well?
at t1 = 0 v1 = 15.0 m>s
Acceleration a = -2.0 m>s2
at t2 = 5.0 s v2 = 5.0 m>s
FIGURE 2 – 15   Example 2 9 7, showing the position of the car at times t1 and t2 , as well as the car’s velocity represented by the green arrows. We calculate that the acceleration vector (orange) points to the left as the car slows down while moving to the right.
FIGURE 2 – 16   The car of
Example 2 9 7, now moving to the left
and decelerating. The acceleration is
a
=
v2 - v1 ∆t
(- 5.0 m>s) - (- 15.0 m>s)
a=
5.0 s
=
- 5.0 m>s + 15.0 m>s 5.0 s
=
+ 2.0 m>s2.
v2 = -5.0 m>s
v1 = -15.0 m>s
a
EXAMPLE 2 – 7 Car slowing down. An automobile is moving to the right
along a straight highway, which we choose to be the positive x axis (Fig. 2 9 15).
Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is  v1 = 15.0 m>s, and it takes 5.0 s to slow down to  v2 = 5.0 m>s,  what was the car’s average acceleration?
APPROACH  We put the given initial and final velocities, and the elapsed time, into Eq. 2 9 5 for a.
SOLUTION  In Eq. 2 9 5, we call the initial time  t1 = 0,  and set  t2 = 5.0 s.  (Note that our choice of  t1 = 0  doesn’t affect the calculation of a because only  ∆t = t2 - t1  appears in Eq. 2 9 5.) Then
a
=
5.0 m>s - 15.0 m>s 5.0 s
=
- 2.0 m>s2.
The negative sign appears because the final velocity is less than the initial velocity. In this case the direction of the acceleration is to the left (in the negative x direction) : 
even though the velocity is always pointing to the right. We say that the acceleration is 2.0 m>s2 to the left, and it is shown in Fig. 2 9 15 as an orange arrow.
“Deceleration”
When an object is slowing down, we sometimes say it is decelerating. In physics, the concept of acceleration is all we need: it can be + or - . But if the word “deceleration” is used, be careful: deceleration does not mean that the acceleration is necessarily negative, as in Example 2 9 7. The velocity of an object moving to the right along the positive x axis is positive; if the object is slowing down (as in Fig. 2 9 15), the acceleration is negative. But the same car moving to the left (decreasing x), and slowing down, has positive acceleration that points to the right, as shown in Fig. 2 9 16. We have a deceleration whenever the magnitude of the velocity is decreasing; thus the velocity and acceleration point in opposite directions when there is deceleration.
28  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 28
01/07/20 16:19
EXERCISE D  A car moves along the x axis. What is the sign of the car’s acceleration if it is moving in the positive x direction with (a) increasing speed or (b) decreasing speed? What is the sign of the acceleration if the car moves in the negative x direction with (c) increasing speed or (d) decreasing speed?
Instantaneous Acceleration
The instantaneous acceleration, a, is defined as the limiting value of the average acceleration as we let ∆t approach zero:
a
=
∆v
lim
∆t S 0
∆t
=
dv . dt
(2 – 6)
This limit, dv>dt, is the derivative of v with respect to t. We will use the term
“acceleration” to refer to the instantaneous value. If we want to discuss the
average ­acceleration, we will always include the word “average.” If we draw a graph of the velocity, v, vs. time, t, as shown in Fig. 2 9 17, then
the average acceleration over a time interval  ∆t = t2 - t1  is represented by the slope of the straight line connecting the two points P1 and P2 in Fig. 2 9 17. [Compare this to the position vs. time graph of Fig. 2 9 10 for which the slope of
the straight line represents the average velocity.] The instantaneous ­acceleration at any time, say t1 , is the slope of the tangent to the v vs. t curve at time t1 , which is also shown in Fig. 2 9 17. In Fig. 2 9 17, as we go from time t1 to time t2 the velocity continually increases, but the acceleration (the rate at which the
velocity changes) is decreasing since the slope of the curve is decreasing.
Example 2 – 8 Acceleration given x (t).  A particle is moving in a straight line so that its position is given by the relation  x = (2.10 m>s2)t2 + (2.80 m),  as in Example 2 9 4. Calculate (a) its average acceleration during the time interval from  t1 = 3.00 s  to  t2 = 5.00 s,  and (b) its instantaneous acceleration as a function of time.
APPROACH  To determine acceleration, we first must find the velocity at t1 and t2 by differentiating x:  v = dx>dt. Then we use Eq. 2 9 5 to find the average acceleration, and Eq. 2 9 6 to find the instantaneous acceleration.
SOLUTION  (a) The velocity at any time t is
v
=
dx dt
=
d dt
[(2.10 m>s2)t2
+
2.80 m ]
=
(4.20 m>s2)t,
as we already saw in Example 2 9 4c. Therefore, at time  t1 = 3.00 s, v1 = (4.20 m>s2) (3.00 s) = 12.6 m>s    and  at  t2 = 5.00 s, v2 = 21.0 m>s.  Therefore,
a
=
∆v ∆t
=
21.0 m>s - 12.6 m>s 5.00 s - 3.00 s
=
4.20 m>s2.
(b) With  v = (4.20 m>s2)t,  the instantaneous acceleration at any time is
a
=
dv dt
=
d dt
[ ( 4.20
m>s2)t ]
=
4.20 m>s2.
The acceleration in this case is constant; it does not depend on time. Figure 2 9 18 shows graphs of (a) x vs. t (the same as Fig. 2 9 13b), (b) v vs. t, which is linearly increasing as calculated above, and (c) a vs. t, which is a horizontal straight line
because  a = constant.
Slope is average acceleration
v
during ∆t = t2 - t1
Slope is
instantaneous
acceleration
at t1 v2
P2
∆v = v2 - v1
v1
P1
∆t = t2 - t1
t
0
t1
t2
FIGURE 2 – 17   A graph of velocity
v vs. time t. The average acceleration
over a time interval  ∆t = t2 - t1  is the slope of the straight line P1 P2 : a = ∆v> ∆t.  The instantaneous acceleration at time t1 is the slope of the v vs. t curve at that instant.
FIGURE 2 – 18   Example 2 9 8. Graphs of (a) x vs. t, (b) v vs. t, and (c) a vs. t for the motion x = At2 + B.  Note that v increases linearly with t and that the
acceleration a is constant. Also, v is the slope of the x vs. t curve, whereas a is the slope of the v vs. t curve.
x (m) 2.10 t 2) m + 2.80 m
60
50
40
30
20 10
x = (
t (s) 0 123456
(a)
v (m>s)
25
20 15 10
v = (4.20 t) m>s
5
t (s) 0 123456
(b)
6 a = 4.20 m>s2
4
a (m>s2)
2
Like velocity, acceleration is a rate. The velocity of an object is the rate at which its displacement changes with time; its acceleration, on the other hand, is the rate at which its velocity changes with time. In a sense, acceleration is a “rate of a rate.”
t (s) 0 123456
(c)
SECTION 2–4  Acceleration  29

GIAN_PSE5_CH02_020-053_ca.indd 29
01/07/20 16:19
This can be expressed in equation form: since  a = dv>dt  and  v = dx>dt,  then
a
=
dv dt
=
d dt
¢
dx dt
≤
=
d2x . dt 2
Here d2x>dt2 is the second derivative of x with respect to time: we first take the derivative of x with respect to time (dx>dt), and then we again take the derivative with respect to time, (d>dt) (dx>dt), to get the acceleration.
EXERCISE E  The position of a particle is given by the following equation:
x = (2.00 m>s3)t3 + (2.50 m>s)t. What is the acceleration of the particle at  t = 2.00 s?  Choose one: (a) 13.0 m>s2;
(b) 22.5 m>s2; (c) 24.0 m>s2; (d) 2.00 m>s2; (e) 21.0 m>s2.
100
Car A
Car B
v (km>h)
0
2
4
6
t (s) 8 10
FIGURE 2 – 19   Example 2 9 9.
CONCEPTUAL EXAMPLE 2 – 9 Analyzing with graphs.  Figure 2 9 19 shows the velocity as a function of time for two cars accelerating from 0 to 100 km>h in a time of 10.0 s. Compare for the two cars: (a) the average acceleration; (b) the instantaneous acceleration; and (c) the total distance traveled.
RESPONSE (a) Average acceleration is ∆v>∆t. Both cars have the same ∆v (100 km>h) and the same ∆t (10.0 s), so the average acceleration is the same for both cars. (b) Instantaneous acceleration is the slope of the tangent to the v vs. t curve. For about the first 4 s, the top curve is steeper than the bottom curve, so car A has a greater instantaneous acceleration during this interval. The bottom curve is steeper during the last 6 s, so car B has the larger acceleration during this period. (c) Except at  t = 0  and  t = 10.0 s,  car A is always going faster than car B. Since it is going faster, it will go farther in the same time. Notice what marvelous information we can get from a graph.
2–5  Motion at Constant Acceleration
We now examine motion in a straight line when the magnitude of the acceleration
is constant. In this case, the instantaneous and average acceleration are equal.
We use the definitions of average velocity and acceleration to derive a set of­
valuable equations that relate x, v, a, and t when a is constant, allowing us
to ­determine any one of these variables if we know the others.
Notation in physics varies from book to book; and different instructors use
different notation. We are now going to change our notation, to simplify it a bit
for our discussion here of motion at constant acceleration. First we choose the
initial time in any discussion to be zero, and we call it t0 . That is,  t1 = t0 = 0. (This is effectively starting a stopwatch at t0 .) We can then let  t2 = t   be the elapsed time. The initial position (x1) and the initial velocity (v1) of an object
will now be represented by x0 and v0 , since they represent x and v at  t = 0. At time t the position and velocity will be called x and v (rather than x2 and v2).
The average velocity during the time interval  t - t0  will be (Eq. 2 9 2)
v
=
∆x ∆t
=
x - x0 t - t0
=
x - x0 t
since we chose  t0 = 0. The acceleration, assumed constant in time, is  a = ∆v>∆t
(Eq. 2 9 5), so
a
=
v - v0 . t
A common problem is to determine the velocity of an object after any elapsed
time t, when we are given the object’s constant acceleration. We can solve such
problems by solving for v in the last equation: first we multiply both sides by t ,
which gives at = v - v0 , and then
v = v0 + at.
[constant acceleration]  (2 – 7)
If an object, such as a motorcycle, starts from rest  (v0 = 0)  and accelerates
30  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 30
01/07/20 16:19
at 4.0 m>s2, then after an elapsed time  t = 6.0 s  its velocity will be v = 0 + at = (4.0 m>s2) (6.0 s) = 24 m>s.
Next, let us see how to calculate the position x of an object after a time t
when it undergoes constant acceleration. The definition of average velocity
(Eq. 2 9 2) is  v = (x - x0)>t,  which we can rewrite by multiplying both sides by t:
x = x0 + vt.
(2 – 8)
Because the velocity increases at a uniform rate, the average velocity, v, will be midway between the initial and final velocities:
v
=
v0 + v . 2
[constant acceleration]  (2 – 9)
(Careful: Equation 2 9 9 is not necessarily valid if the acceleration is not constant.) We combine the last two Equations with Eq. 2 9 7 and find, starting with Eq. 2 9 8,
x = x0 + vt
=
x0
+
¢ v0 + 2
v ≤ t
=
x0
+
¢ v0
+
v0 2
+
at ≤ t
or
x
=
x0
+
v0 t
+
1 2
at
2.
[constant acceleration]  (2 – 10)
Equations 2 9 7, 2 9 9, and 2 9 10 are three of the four most useful equations for motion at constant acceleration. We now derive the fourth equation, which is useful in situations where the time t is not known. We substitute Eq. 2 9 9 into Eq. 2 9 8:
x
=
x0 + vt
=
x0
+
v ¢
+ v0 ≤ t. 2
Next we solve Eq. 2 9 7 for t, obtaining
t
=
v - v0 , a
and substituting this into the previous equation we have
x
=
x0
+
v ¢
+ v0 ≤ ¢ v 2
- v0 ≤ a
=
x0
+
v2 - v02 . 2a
We solve this for v2 and obtain
v2 = v02 + 2a(x - x0),
[constant acceleration]  (2 – 11)
which is the other useful equation we sought. We now have four equations relating position, velocity, acceleration, and time,
when the acceleration a is constant. We collect these kinematic equations for constant acceleration here in one place for further reference (the tan background is used to emphasize their importance):
CAUTION
Average velocity, but only if a = constant
v = v0 + at
x
=
x0
+
v0 t
+
1 2
at
2
v2 = v02 + 2a(x - x0)
v
=
v + v0 . 2
[a = constant] [a = constant] [a = constant]
[a = constant]
(2 – 12a) (2 – 12b) (2 – 12c)
(2 – 12d)
Kinematic equations for constant acceleration (we’ll use them a lot)
These useful equations are not valid unless a is a constant. In many cases we can set  x0 = 0,  and this simplifies the above equations a bit. Note that x represents position (not distance), and that  x - x0  is the displacement, whereas t is the elapsed time.
Equations 2 9 12 are useful also when a is approximately constant, in order to obtain ­reasonable estimates.
SECTION 2–5  Motion at Constant Acceleration  31

GIAN_PSE5_CH02_020-053_ca.indd 31
01/07/20 16:19
PHYSICS APPLIED Airport design
PROBLEM SOLVING Equations 2–12 are valid only when the acceleration is constant, which we
assume in this Example
EXAMPLE 2 – 10 Runway design. You are designing an airport for small planes. One kind of airplane that might use this airfield must reach a speed before takeoff of at least 27.8 m>s (100 km>h), and can accelerate at 2.00 m>s2. (a) If the runway is 150 m long, can this airplane reach the required speed for takeoff? (b) If not, what minimum length must the runway have?
APPROACH  Assuming the plane’s acceleration is constant, we use the kinematic equations for constant acceleration. In (a), we want to find v, and we are given:
Known
Wanted
x0 = 0
v
v0 = 0
x = 150 m
a = 2.00 m>s2
SOLUTION  (a) Of the four kinematic equations on page 31, Eq. 2 9 12c will give us v when we know v0 , a, x, and x0 :
v2 = v02 + 2a(x - x0) = 0 + 2(2.00 m>s2)(150 m) = 600 m2>s2
v = 2600 m2>s2 = 24.5 m>s.
This runway length is not sufficient, because the minimum speed is not reached.
(b) Now we want to find the minimum runway length,  x - x0 ,  for a plane to reach v = 27.8 m>s, given  a = 2.00 m>s2. We again use Eq. 2 9 12c, but rewritten as
(x - x0)
=
v2 - v02 2a
=
(27.8 m>s)2 - 0
2(2.00 m>s2)
=
193 m.
A 200-m runway is more appropriate for this plane.
NOTE  We did this Example as if the plane were a particle, so we round off our answer to 200 m.
PHYSICS APPLIED Car safety : air bags
FIGURE 2 – 20   Example 2 9 11. An air bag deploying on impact.
EXAMPLE 2 – 11 ESTIMATE Air bags. Suppose you want to design an air bag system that can protect the driver at a speed of 100 km>h (60 mph) if the car hits a brick wall. Estimate how fast the air bag must inflate (Fig. 2 9 20) to
effectively protect the driver. How does the use of a seat belt help the driver?
APPROACH  We assume the acceleration is roughly constant, so we can use Eqs. 2 9 12. Both Eqs. 2 9 12a and 2 9 12b contain t, our desired unknown. They both contain a, so we must first find a, which we can do using Eq. 2 9 12c if we know the
distance x over which the car crumples. A rough estimate might be about 1 meter.
We choose the time interval to start at the instant of impact with the car moving at  v0 = 100 km>h, and to end when the car comes to rest  (v = 0) after traveling 1 m.
SOLUTION  We convert the given initial speed to SI units:
100 km>h = 100 * 103 m>3600 s = 28 m>s. We then find the acceleration from
Eq. 2 9 12c:
a
=
-
v02 2x
=
(28 m>s)2 - 2.0 m
=
- 390 m>s2.
This enormous acceleration takes place in a time given by (Eq. 2 9 12a):
t
=
v - v0 a
=
0 - 28 m>s - 390 m>s2
=
0.07 s.
To be effective, the air bag would need to inflate faster than this. What does the air bag do? It spreads the force over a large area of the chest
(to avoid puncture of the chest by the steering wheel). The seat belt keeps the person in a stable position directly in front of the expanding air bag.
32  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 32
01/07/20 16:19
EXERCISE F A car starts from rest and accelerates at a constant 10 m>s2 during a 14-mile (402 m) race. How fast is the car going at the finish line? (a) 8040 m>s; (b) 90 m>s; (c) 81 m>s; (d) 804 m>s.
2–6  Solving Problems
Before doing more worked-out Examples, let us look at how to approach problem solving. First, it is important to note that physics is not a collection of equations to be memorized. Simply searching for an equation that might work can lead you to a wrong result and will not help you understand physics (Fig. 2 9 21). A better approach is to use the following (rough) procedure, which we present as a special “Problem Solving Strategy.” (Other such Problem Solving Strategies will be found throughout the book.)
FIGURE 2 – 21   Read each Chapter of this book, study it by reading it again carefully, and work the Problems using your reasoning abilities.
Problem
Solving
1. Read and reread the whole problem carefully before trying to solve it.
2. Decide what object (or objects) you are going to study, and for what time interval. You can often choose the initial time to be  t = 0.
3. Draw a diagram or picture of the situation, with coordinate axes wherever applicable. [You can place the origin of coordinates and the axes wherever you like to make your calculations easier. You also choose which direction is positive and which is negative. Usually we choose the x axis to the right as positive.]
4. Write down what quantities are “known” or “given,” and then what you want to know (“unknowns”). Consider quantities both at the beginning and at the end of the chosen time interval. You may need to “translate” language into physical terms, such as “starts from rest” means  v0 = 0.
5. Think about which principles of physics apply in this problem. Use common sense and your own experiences. Then plan an approach.
6. Consider which equations (and>or definitions) relate the quantities involved. Before using them, be sure their range of validity includes your ­problem (for example, Eqs. 2 9 12 are valid only when the acceleration is constant). If you find
an applicable ­equation that involves only known quantities and one desired unknown, solve the equation algebraically for the unknown. Sometimes several sequential calculations, or a combination of equations, may be needed. It is often preferable to solve algebraically for the desired unknown before putting in numerical values.
7. Carry out the calculation if it is a numerical problem. Keep one or two extra digits during the calculations, but round off the final answer(s) to the correct number of significant figures (Section 1 9 3).
8. Think carefully about the result you obtain: Is it reasonable? Does it make sense according to your own intuition and experience? A good check is to do a rough estimate using only powers of 10, as discussed in Section 1 9 6. Often it is preferable to do a rough estimate at the start of a numerical problem because it can help you focus your attention on finding a path toward a ­solution.
9. A very important aspect of doing problems is keep­ing track of units. An equals sign implies the units on each side must be the same, just as the numbers must. If the units do not balance, a mistake has been made. This can serve as a check on your solution (but it only tells you if you’re wrong, not if you’re right).  Always use a consistent set of units.
SECTION 2–6  Solving Problems  33 
GIAN_PSE5_CH02_020-053_ca.indd 33
01/07/20 16:19
PROBLEM SOLVING
“Starting from rest” means  v = 0  at  t = 0  [i.e.,  v0 = 0]
a = 2.00 m>s2
a = 2.00 m>s2
x0 = 0 v0 = 0
x = 30.0 m
FIGURE 2 – 22   Example 2 9 12. (Not to scale.)
Known
x0 = 0 x = 30.0 m a = 2.00 m>s2 v0 = 0
Wanted t
PROBLEM SOLVING Check your answer
PROBLEM SOLVING “Unphysical” solutions
EXAMPLE 2 – 12 Acceleration of a car.  How long does it take a 4.0-m-long car
to cross a 26.0-m-wide intersection after the light turns green, if the car accelerates from rest at a constant 2.00 m>s2? The car has to travel  26.0 m + 4.0 m = 30.0 m  to clear the intersection.
APPROACH  We follow the Problem Solving Strategy on the previous page.
SOLUTION
1. Reread the problem. Be sure you understand what it asks for (here, a time interval: “how long does it take”).
2. The object under study is the car. We need to choose the time interval during which we look at the car’s motion: we choose  t = 0,  the initial time, to be
the moment the car starts to accelerate from rest  (v0 = 0);  the time t is the
instant the car has traveled the full 30.0 m of the intersection.
3. Draw a diagram: the situation is shown in Fig. 2 9 22 where the car is shown moving along the positive x axis. We choose  x0 = 0  at the front bumper of the car before it starts to move.
4. The “knowns” and the “wanted” information are shown in the Table in the margin. Note that “starting from rest” means  v = 0  at  t = 0;  that is,  v0 = 0.  The wanted time t is how long it takes the car to travel 30.0 m.
5. The physics: the car, starting from rest (at  t0 = 0),  increases in speed as it
covers more distance. The acceleration is constant, so we can use the kine-
matic equations, Eqs. 2 9 12.
6. Equations: we want to find the time, given the distance and acceleration;
Eq. 2 9 12b  (x
=
x0
+
v0 t
+
1 2
at
2
)
 
is
perfect
since
the
only
unknown
quantity
is t. Setting  v0 = 0  and  x0 = 0  in Eq. 2 9 12b, we have
x
=
1 2
at
2.
We solve for t
t
=
2x . Aa
7. The calculation:
t
=
2x Aa
=
2(30.0 m) D 2.00 m>s2
=
5.48 s.
This is our answer. Note that the units come out correctly.
8. We can check the reasonableness of the answer by doing an alternate calculation: we use our result in step 7 and check if the distance traveled turns out to be 30.0 m. First we find the final velocity (Eq. 2 9 12a),
v = at = (2.00 m>s2) (5.48 s) = 10.96 m>s,
and then find the distance traveled using the definition of average velocity (see Eq. 2 9 8).
x
=
x0 + vt
=
0
+
1 2
( 10.96
m>s
+
0) (5.48 s)
=
30.0 m,
which checks with our given distance. 9. We checked the units in step 7, and they came out correctly (seconds).
NOTE  In steps 6 and 7, when we took the square root, we should have written  t = { 12x>a = {5.48 s.  Mathematically there are two solutions. But the second s­ olution,  t = -5.48 s,  is a time before our chosen time interval and makes no sense physically. We say it is “unphysical” and ignore it.
We explicitly followed the steps of the Problem Solving Strategy in Example 2 9 12. In upcoming Examples, we will use our usual “Approach” and “Solution” to avoid being wordy.
34  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 34
01/07/20 16:19
Travel during reaction time
v = constant = 14 m>s t = 0.50 s a= 0
Travel during braking
v decreases from 14 m>s to zero a = -6.0 m>s2
x FIGURE 2 – 23   Example 2 9 13: stopping distance for a braking car.
EXAMPLE 2 – 13 ESTIMATE Braking distances. Estimate the minimum
stopping distance for a car, which is important for traffic safety and traffic design.
The problem is best dealt with in two parts, two separate time intervals. (1) The first
time interval begins when the driver decides to hit the brakes, and ends when the
foot touches the brake pedal. This is the “reaction time” during which the speed is
constant, so  a = 0. (2) The second time interval is the actual braking period when the vehicle slows down  (a ≠ 0) and comes to a stop.The stopping distance depends on the reaction time of the driver, the initial speed of the car (the final speed is
zero), and the deceleration of the car. For a dry road and good tires, good brakes can decelerate a car at a rate of about 5 m>s2 to 8 m>s2. Calculate the total stopping distance for an initial velocity of 50 km>h  ( = 14 m>s L 31 mi>h) and assume the acceleration of the car is - 6.0 m>s2 (the minus sign appears because the velocity is taken to be in the positive x direction and its magnitude is decreasing). Reaction
time for typical drivers varies from perhaps 0.3 s to about 1.0 s; take it to be 0.50 s.
APPROACH  During the “reaction time,” part (1), the car moves at constant speed of 14 m>s, so  a = 0.  Once the brakes are applied, part (2), the acceleration is  a = -6.0 m>s2  and is constant over this time interval. For both parts a is constant, so we can use Eqs. 2 9 12.
SOLUTION  Part (1). We take  x0 = 0  for the first time interval, when the driver is reacting (0.50 s): the car travels at a constant speed of 14 m>s so  a = 0.  See Fig. 2 9 23 and the Table in the margin. To find x, the position of the car at
t = 0.50 s  (when the brakes are first applied), we cannot use Eq. 2 9 12c because x is multiplied by a, which is zero. But Eq. 2 9 12b works:
x = v0 t + 0 = (14 m>s) (0.50 s) = 7.0 m.
Thus the car travels 7.0 m during the driver’s reaction time, until the instant the
brakes are applied. We will use this result as input to part (2).
Part (2). During the second time interval, the brakes are applied and the car is
brought to rest. The initial position is  x0 = 7.0 m  (result of part (1)), and other variables are shown in the second Table in the margin. Equation 2 9 12a doesn’t
contain x ; Eq. 2 9 12b contains x but also the unknown t. Equation 2 9 12c, v2 - v02 = 2a(x - x0),  is what we want; after setting  x0 = 7.0 m,  we solve for x, the final position of the car (where it stops):
x
=
x0
+
v2 - v02 2a
=
0 - (14 m>s)2
7.0 m + 2( -6.0 m>s2)
=
7.0 m
+
- 196 m2>s2 - 12 m>s2
= 7.0 m + 16 m = 23 m.
The car traveled 7.0 m while the driver was reacting and another 16 m during
the braking period before coming to a stop, for a total distance traveled of 23 m. Figure 2 9 24 shows graphs of (a) v vs. t (we were given that v is constant from  t = 0  until t = 0.50 s,  and after  t = 0.50 s  it decreases linearly to zero), and (b) x vs. t.
NOTE  From the equation above for x, we see that the stopping distance after the
driver hit the brakes  ( = x - x0)  increases with the square of the initial speed,
not just linearly with speed. If you are traveling twice as fast, it takes four times
the distance to stop (when braking at the same rate, a).
PHYSICS APPLIED Car stopping distances
Part 1: Reaction time
Known
Wanted
t = 0.50 s
x
v0 = 14 m>s v = 14 m>s
a=0
x0 = 0
Part 2: Braking
Known
x0 = 7.0 m v0 = 14 m>s v=0 a = - 6.0 m>s2
Wanted x
FIGURE 2 – 24   Example 2 9 13. Graphs of (a) v vs. t and (b) x vs. t.
v (m>s)
14
12
10 8
(a)
6
t = 0.50 s
4
2
t (s)
0 0.5 1.0 1.5 2.0 2.5
x (m)
20
15
(b)
10
5
t = 0.50 s
0
t (s) 0.5 1.0 1.5 2.0 2.5
SECTION 2–6  Solving Problems  35

GIAN_PSE5_CH02_020-053_ca.indd 35
01/07/20 16:19
PROBLEM SOLVING Guess the acceleration
CAUTION Initial assumptions need to be checked out for reasonableness
EXAMPLE 2 – 14 ESTIMATE Two Moving Objects: Police and Speeder.  A car speeding at 150 km>h (over 90 mph) passes a still police car which immediately takes off in hot pursuit. Using simple assumptions, such as that the speeder continues at constant speed, estimate how long it takes the police car to overtake the speeder. Then estimate the police car’s speed at that moment and decide if the assumptions were reasonable.
APPROACH  When the police car takes off, it accelerates, and the simplest
assumption is that its acceleration is constant. This may not be reasonable, but
let’s see what happens. We can estimate the acceleration if we have noticed automobile ads, which claim cars can accelerate from rest to 100 km>h in 5.0 s.
So the average acceleration of the police car could be approximately
aP
=
100 km>h 5.0 s
=
20
km>h s
¢ 1000 m ≤ ¢ 1 h 1 km 3600
≤ s
=
5.6 m>s2.
SOLUTION  We need to set up the kinematic equations to determine the unknown
quantities, and since there are two moving objects, we need two separate sets of
equations. We denote the speeding car’s position by xS and the police car’s position by xP . Because we are interested in solving for the time when the two vehicles arrive at the same position on the road, we use Eq. 2 9 12b for each car (x0 = 0  for both cars):
xS
=
v0S t
+
1 2
aS
t
2
=
(150 km>h)t + 0
=
(42 m>s)t
xP
=
v0P t
+
1 2
aP
t
2
=
0
+
1 2
(
5.6
m
>s2
)
t
2,
where we have set  v0P = 0  and  aS = 0  (speeder assumed to move at constant speed). We want the time when the cars meet, so we set  xS = xP  and solve for t:
(42 m>s)t = (2.8 m>s2)t2.
The solutions are
t=0
and
t=
42 m>s 2.8 m>s2
=
15 s.
The first solution corresponds to the instant the speeder passed the police car. The second solution tells us when the police car catches up to the speeder, 15 s later. This is our answer, but is it reasonable? The police car’s speed at  t = 15 s  is
vP = v0P + aP t = 0 + (5.6 m>s2) (15 s) = 84 m>s
or 300 km>h ( L 190 mi>h). Not reasonable, and highly dangerous.
NOTE  More reasonable is to give up the assumption of constant acceleration. The police car surely cannot maintain constant acceleration at high speed. Also, the speeder, if a reasonable person, would slow down upon hearing the police siren. Figure 2 9 25 shows (a) x vs. t and (b) v vs. t graphs, based on the original assumption of  aP = constant, whereas (c) shows v vs. t for more reasonable assumptions.
x
v
Speeder
FIGURE 2 – 25   Example 2 9 14.
Speeder
Police
t
0
15 s
0
(a)
Police (b)
36  CHAPTER 2  Describing Motion: Kinematics in One Dimension
v Speeder
t 0
Police t
(c)

GIAN_PSE5_CH02_020-053_ca.indd 36
01/07/20 16:19
FIGURE 2 – 26   Painting of Galileo demonstrating to the Grand Duke of Tuscany his argument for the action of gravity being uniform acceleration. He used a wooden inclined plane (center) to slow down the action. A ball rolling down the plane still accelerates. Tiny bells placed at equal distances along the inclined plane would ring at shorter time intervals as the ball “fell,” indicating that the speed was increasing.
2–7  Freely Falling Objects
FIGURE 2 – 27   Multiflash photograph of a falling apple, at equal time intervals. The apple falls farther during each successive interval, which means it is accelerating.
One of the most common examples of uniformly accelerated motion is that of an object allowed to fall freely near the Earth’s surface. That a falling object is accelerating may not be obvious at first. And beware of thinking, as was widely believed before the time of Galileo (Fig. 2 9 26), that heavier objects fall faster than lighter objects and that the speed of fall is proportional to how heavy the object is. Wrong. The speed of a falling object is not proportional to its mass.
Galileo made use of his new technique of imagining what would happen in idealized (simplified) cases. For free fall, he postulated that all objects would fall with the same constant acceleration in the absence of air or other resistance. He showed that this postulate predicts that for an object falling from rest, the distance traveled will be proportional to the square of the time (Fig. 2 9 27); that is, d r t2. We can see this from Eq. 2 9 12b for constant acceleration; but Galileo was the first to deduce this mathematical relation.
To support his claim that falling objects increase in speed as they fall, Galileo made use of a clever argument: a heavy stone dropped from a height of 2 m will drive a stake into the ground much further than if the same stone is dropped from a height of only 0.2 m. Clearly, the stone must be moving faster in the former case.
Galileo claimed that all objects, light or heavy, fall with the same acceleration, at least in the absence of air. If you hold a piece of paper flat and horizontal in one hand, and a heavier object like a baseball in the other, and release them at the same time as in Fig. 2 9 28a, the heavier object will reach the ground first. But if you repeat the experiment, this time crumpling the paper into a small wad, you will find (see Fig. 2 9 28b) that the two objects reach the floor at nearly the same time.
Galileo was sure that air acts as a resistance to very light objects that have a large surface area. But in many ordinary circumstances this air resistance is negligible. In a chamber from which the air has been removed, even light objects like a feather or a horizontally held piece of paper will fall with the same acceleration as any other object (see Fig. 2 9 29). Such a demonstration in vacuum was not possible in Galileo’s time, which makes Galileo’s achievement all the greater. Galileo is often called the “father of modern science,” not only for the content of his science: astronomical discoveries, inertia, free fall; but also for his new methods of doing science: idealization and simplification, mathematization of theory, theories that have testable consequences, experiments to test theoretical predictions.
FIGURE 2 – 28   (a) A ball and a light piece of paper are dropped at the same time. (b) Repeated, with the paper wadded up.
(a)
(b)
FIGURE 2 – 29   A rock and a feather are dropped simultaneously (a) in air, (b) in a vacuum.
Air- lled tube (a)
Evacuated tube (b)
SECTION 2–7  Freely Falling Objects  37

GIAN_PSE5_CH02_020-053_ca.indd 37
01/07/20 16:19
Galileo’s specific contribution to our understanding of the motion of falling objects can be summarized as follows:
at a given location on the Earth and in the absence of air resistance, all objects fall with the same constant acceleration.
PROBLEM SOLVING
You can choose y to be positive either up or down
We call this acceleration the acceleration due to gravity at the surface of the Earth, and we give it the symbol g. Its magnitude is approximately
g = 9.80 m>s2.
[ ] acceleration due to gravity
at surface of Earth
In British units g is about 32 ft>s2. Actually, g varies slightly according to latitude and elevation above sea level on the Earth’s surface, but these variations are so small that we will ignore them for most purposes. (Acceleration of gravity out in space beyond the Earth’s surface is treated in Chapter 6.) Air resistance acts to reduce the speed of a falling object, but this effect is often small, and we will neglect it for the most part. However, air resistance will be noticeable even on a reasonably heavy object if the velocity becomes large.†
Acceleration due to gravity is a vector, as is any acceleration, and its direction is downward toward the center of the Earth.
When dealing with freely falling objects we can make use of Eqs. 2 9 12, where for a we use the value of g given above. Also, since the motion is vertical we will substitute y in place of x, and y0 in place of x0 . We take  y0 = 0  unless otherwise specified. It is arbitrary whether we choose y to be positive in the upward direction or in the downward direction; but we must be consistent about it throughout a problem’s solution.
FIGURE 2 – 30   Example 2 9 15. (a) An object dropped from a tower falls with progressively greater speed and covers greater distance with each successive second. (See also Fig. 2 9 27.) (b) Graph of y vs. t (y is + downward).
Acceleration due to gravity
+y
y = 0 y1 = 4.90 m (After 1.00 s)
y2 = 19.6 m (After 2.00 s)
y3 = 44.1 m (After 3.00 s)
+y
EXERCISE G  Return to the Chapter-Opening Question, page 20, and answer it again now, assuming minimal air resistance. Try to explain why you may have answered differently the first time.
EXAMPLE 2 – 15 Falling from a tower. Suppose that a ball is dropped
(v0 = 0)   from a tower. How far will it have fallen after a time  t1 = 1.00 s,  t2 = 2.00 s,  and  t3 = 3.00 s? Ignore air resistance.
APPROACH  Let us take y as positive downward, so the acceleration is a = g = + 9.80 m>s2. We set  v0 = 0  and  y0 = 0. We want to find the position y of the ball after three different time intervals. Equation 2 9 12b, with x replaced
by y, relates the given quantities (t, a, and v0) to the unknown y.
SOLUTION  We set  t = t1 = 1.00 s  in Eq. 2 9 12b:
y1
=
v0 t1
+
1 2
at
2 1
=
0
+
1 2
at
2 1
=
1 2
(
9.80
m>s2
)
(
1.00
s
) 2
=
4.90 m.
The ball has fallen a distance of 4.90 m during the time interval  t = 0  to t1 = 1.00 s.  Similarly, after 2.00 s  ( = t2),  the ball’s position is
y2
=
1 2
at
2 2
=
1 2
(
9.80
m
>s2
)
(
2.00
s
) 2
=
19.6 m.
Finally, after 3.00 s  ( = t3),  the ball’s position is (see Fig. 2 9 30a)
y3
=
1 2
at
2 3
=
1 2
(
9.80
m
>s2
)
(
3.00
s
) 2
=
44.1 m.
(a) y (m)
40 30 20 10
(b) 0
t (s) 1 23
NOTE  Whenever we say “dropped,” it means  v0 = 0.  Note also the graph of y vs. t (Fig. 2 9 30b): the curve is not straight but bends upward because y is proportional to t2.
NOTE  Because of air resistance, all of the distances in Fig. 2 9 30, y1 , y2 , y3 , would be smaller than shown (and as just calculated); but the difference will be small for a reasonably heavy (but small) object.
†The speed of an object falling in air (or other fluid) does not increase indefinitely. If the object falls far enough, it will reach a maximum velocity called the terminal velocity due to air resistance. (Section 5 9 6 deals directly with air resistance.)
38  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 38
01/07/20 16:19
EXAMPLE 2 – 16 Thrown down from a tower. Suppose the ball in Example 2 9 15 is thrown downward with an initial velocity of 3.00 m>s, instead of being dropped. (a) What then would be its position after 1.00 s and 2.00 s? (b) What would its speed be after 1.00 s and 2.00 s? Compare with the speeds of a dropped ball.
APPROACH  Again we use Eq. 2 9 12b, but now v0 is not zero, it is  v0 = 3.00 m>s. SOLUTION  (a) At  t1 = 1.00 s,  the position of the ball as given by Eq. 2 9 12b is
y
=
v0 t
+
1 2
at
2
=
(3.00 m>s) (1.00 s)
+
1 2
(
9.80
m
>s2
)
(
1.00
s
) 2
=
7.90 m.
At  t2 = 2.00 s  (time interval  t = 0  to  t = 2.00 s),  the position is
y
=
v0 t
+
1 2
at
2
=
(3.00 m>s) (2.00 s)
+
1 2
(
9.80
m
>s2
)
(
2.00
s
) 2
=
25.6 m.
As expected, the ball falls farther each second than when it is dropped with v0 = 0,  Example 2 9 15. (b) The velocity is obtained from Eq. 2 9 12a:
v = v0 + at = 3.00 m>s + (9.80 m>s2) (1.00 s) = 12.8 m>s [at  t1 = 1.00 s] = 3.00 m>s + (9.80 m>s2) (2.00 s) = 22.6 m>s. [at  t2 = 2.00 s]
In Example 2 9 15, when the ball was dropped  (v0 = 0),  the first term (v0) in
these ­equations was zero, so
v = 0 + at = (9.80 m>s2) (1.00 s) = 9.80 m>s      [at  t1 = 1.00 s] = (9.80 m>s2) (2.00 s) = 19.6 m>s.     [at  t2 = 2.00 s]
NOTE  For both Examples 2 9 15 and 2 9 16 the speed increases linearly in time by 9.80 m>s during each second. But the speed of the downwardly thrown ball at any instant is always 3.00 m>s (its initial speed) greater than that of a dropped ball.
FIGURE 2 – 31   An object thrown into the air leaves the thrower’s hand at A, reaches its maximum height at B, and returns to the original position at C. Examples 2 9 17, 2 9 18, 2 9 19, 2 9 20, and 2 9 21.
EXAMPLE 2 – 17 Ball thrown upward. A person throws a ball upward into the air with an initial velocity of 15.0 m>s. Calculate how high it goes. Ignore air
resistance.
APPROACH  We are not concerned here with the throwing action, but only with the motion of the ball after it leaves the thrower’s hand (Fig. 2 9 31) at point A. Let
us choose y to be positive in the upward direction and negative in the downward direction. (This is a different convention from that used in Examples 2 9 15 and 2 9 16, and so illustrates our options.) The acceleration due to gravity is downward and so will have a negative sign,  a = -g = -9.80 m>s2. As the ball rises, its speed decreases until it reaches the highest point (B in Fig. 2 9 31), where its speed is zero
for an instant; then it descends, with increasing speed.
SOLUTION  We consider the time interval from when the ball leaves the throw-
er’s hand until the ball reaches the highest point. To determine the maximum
height, we calculate the position of the ball when its velocity equals zero
(v = 0  at the highest point). At  t = 0  (point A in Fig. 2 9 31) we set  y0 = 0,  v0 = 15.0 m>s,  and  a = - 9.80 m>s2.  At time t (maximum height),  v = 0,  a = -9.80 m>s2,  and we wish to find y. We use Eq. 2 9 12c, replacing x with y:  v2 = v02 + 2ay. We solve this equation for y:
y
=
v2 - v02 2a
=
0 - (15.0 m>s)2
2( -9.80 m>s2)
=
11.5 m.
The ball reaches a height of 11.5 m above the hand.
B(v = 0) +y
g
g
v
v
AC
SECTION 2–7  Freely Falling Objects  39

GIAN_PSE5_CH02_020-053_ca.indd 39
01/07/20 16:19
B(v = 0) +y
g
g
v
v
AC
FIGURE 2 – 31   (Repeated.) An object thrown into the air leaves the thrower’s hand at A, reaches its maximum height at B, and returns to the original position at C. Examples 2 9 17, 2 9 18, 2 9 19, 2 9 20, and 2 9 21.
CAUTION Quadratic equations have two solutions. Sometimes only one
corresponds to reality, sometimes both
CAUTION
(1) Velocity and acceleration are
not always in the same direction;
the acceleration (of gravity)
always points down
(2) a ≠ 0 even at the highest point of

a trajectory
EXAMPLE 2 – 18 Ball thrown upward, II.  In Example 2 9 17, Fig. 2 9 31 (shown here again), how long is the ball in the air before it comes back to the hand?
APPROACH  We need to choose a time interval to calculate how long the ball is in the air before it returns to the hand. We could do this calculation in two parts by first determining the time required for the ball to reach its highest point, and then determining the time it takes to fall back down. However, it is simpler to consider the time interval for the entire motion from A to B to C (Fig. 2 9 31) in one step and use Eq. 2 9 12b. We can do this because y is position (or displacement from the origin), and not the total distance traveled. Thus, at both points A and C we have  y = 0.
SOLUTION  We use Eq. 2 9 12b with  a = -9.80 m>s2  and find
y
=
y0
+
v0 t
+
1 2
at
2
0
=
0
+
(15.0 m>s)t
+
1 2
(
-
9.80
m
>
s2
)
t
2.
This equation can be factored (we factor out one t):
(15.0 m>s - 4.90 m>s2 t) t = 0.
There are two solutions:
t=0
and
t=
15.0 m>s 4.90 m>s2
= 3.06 s.
The first solution  (t = 0)   corresponds to the initial point (A) in Fig. 2 9 31, when the ball was first thrown from  y0 = 0. The second solution,  t = 3.06 s,  corresponds to point C, when the ball has returned to  y = 0. Thus the ball
is in the air 3.06 s.
NOTE  We have ignored air resistance in these last two Examples, which could be significant, so our result is only an approximation to a real, practical situation.
We did not consider the throwing action in these Examples. Why? Because during the throw, the thrower’s hand is touching the ball and accelerating the ball at a rate unknown to us : the acceleration is not g. We consider only the time when the ball is in the air and the acceleration is equal to g.
Every quadratic equation (where the variable is squared) mathematically produces two solutions. In physics, sometimes only one solution corresponds to the real situation, as we saw in Example 2 9 12, in which case we ignore the “unphysical” solution. But here in Example 2 9 18, both solutions to our equation in t2 are physically meaningful:  t = 0  and  t = 3.06 s.
CONCEPTUAL EXAMPLE 2 – 19 Two possible misconceptions. Give examples to show the error in these two common misconceptions: (1) that acceleration and velocity are always in the same direction, and (2) that an object thrown upward has zero acceleration at the highest point (B in Fig. 2 9 31).
RESPONSE  Both are wrong. (1) Velocity and acceleration are not necessarily in the same direction. When the ball in Fig. 2 9 31 is moving upward, its velocity is positive (upward), whereas the acceleration is negative (downward). (2) At the highest point (B in Fig. 2 9 31), the ball has zero velocity for an instant. Is the acceleration also zero at this point? No. The velocity near the top of the arc points upward, then becomes zero for an instant (zero time) at the highest point, and then points downward. Gravity does not stop acting, so   a = -g = -9.80 m>s2  even there. Thinking that  a = 0  at point B would lead to the conclusion that upon reaching point B, the ball would stay there: if the acceleration ( = rate of change of velocity) were zero, the velocity would stay zero at the highest point, and the ball would stay up there without falling. Remember: the acceleration of gravity always points down toward the center of the Earth, even when the object is moving up.
40  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 40
01/07/20 16:19
EXAMPLE 2 – 20 Ball thrown upward, III.  Let us consider again the ball thrown upward of Examples 2 9 17 and 2 9 18, and make more calculations. Calculate (a) how much time it takes for the ball to reach the maximum height (point B in Fig. 2 9 31), and (b) the velocity of the ball when it returns to the thrower’s hand (point C).
APPROACH  Again we assume the acceleration is constant, so we can use Eqs. 2 9 12. We have the maximum height of 11.5 m and initial speed of 15.0 m>s
from Example 2 9 17. Again we take y as positive upward.
SOLUTION  (a) We consider the time interval between the throw  (t0 = 0,  v0 = 15.0 m>s) and the top of the path  (y = +11.5 m, v = 0),  and we want to find t. The acceleration is constant at  a = -g = -9.80 m>s2.  Both Eqs. 2 9 12a and 2 9 12b contain the time t with other quantities known. Let us use Eq. 2 9 12a
with  a = - 9.80 m>s2, v0 = 15.0 m>s,  and  v = 0:
v = v0 + at;
setting  v = 0  gives  0 = v0 + at,  which we rearrange to solve for t:  at = -v0 
or
t
=
-
v0 a
=
15.0 m>s - - 9.80 m>s2 = 1.53 s.
This is just half the time it takes the ball to go up and fall back to its original
position [3.06 s, calculated in Example 2 9 18]. Thus it takes the same time to
reach the maximum height as to fall back to the starting point. We might have
guessed this from the symmetry of the motion. But be careful. When air
­resistance cannot be neglected, the symmetry is no longer perfect.
(b) To find the ball’s velocity when it returns to the hand (point C), we consider
the time interval from the throw  (t0 = 0, v0 = 15.0 m>s)  until the ball’s return to the hand, which occurs at  t = 3.06 s  (as calculated in Example 2 9 18). We use Eq. 2 9 12a again to find v when  t = 3.06 s:
v = v0 + at = 15.0 m>s - (9.80 m>s2)(3.06 s) = -15.0 m>s.
NOTE  The ball has the same speed (magnitude of velocity) when it returns to the starting point as it did initially, but in the opposite direction (this is the meaning of the negative sign). And, as we saw in part (a), the time is the same up as down. Thus the motion is symmetrical about the maximum height.
The acceleration of objects such as rockets and fast airplanes is often given as a multiple of  g = 9.80 m>s2.  For example, an airplane pulling out of a dive (see Fig. 2 9 32) and undergoing 3.00 g’s would have an acceleration
of (3.00)(9.80 m>s2) = 29.4 m>s2 .
PROBLEM SOLVING Acceleration in g’s
FIGURE 2 – 32   Several airplanes, in formation, are just coming out of a downward dive.

GIAN_PSE5_CH02_020-053_ca.indd 41
SECTION 2–7  Freely Falling Objects  41
01/07/20 16:19
PROBLEM SOLVING
Quadratic formula is a very useful tool
B(v = 0) +y
g
g
v
v
EXAMPLE 2 – 21 Ball thrown upward, IV; the quadratic formula.  For the ball in Example 2 9 20, calculate at what time t the ball passes a point 8.00 m above the person’s hand. (See Fig. 2 9 31, repeated below.)
APPROACH  We choose the time interval from the throw  (t0 = 0, v0 = 15.0 m>s)  until the time t (to be determined) when the ball is at position  y = 8.00 m,  using
Eq. 2 9 12b.
SOLUTION  We want to find t, given  y = 8.00 m, y0 = 0, v0 = 15.0 m>s,  and  a = - 9.80 m>s2. We use Eq. 2 9 12b:
y
=
y0
+
v0 t
+
1 2
at
2
8.00 m
=
0
+
(15.0 m>s)t
+
1 2
(
-
9.80
m
>s2
)
t
2.
To solve any quadratic equation of the form  at2 + bt + c = 0,  where a, b, and
c are constants (a is not acceleration here), we use the quadratic formula:
t
=
- b { 2b2 - 4ac . 2a
We rewrite our y equation just above in standard form,  at2 + bt + c = 0:
(4.90 m>s2)t2 - (15.0 m>s)t + (8.00 m) = 0.
So the coefficient a is  4.90 m>s2,  b is  - 15.0 m>s,  and c is 8.00 m. Putting these into the quadratic formula, we obtain
t
=
15.0 m>s { 2(- 15.0 m>s)2 - 4(4.90 m>s2)(8.00 m) , 2(4.90 m>s2)
which gives us  t = 0.69 s  and  t = 2.37 s. Are both solutions valid? Yes, because the ball passes  y = 8.00 m when it goes up  (t = 0.69 s) and again when it comes down  (t = 2.37 s).
NOTE  Figure 2 9 33 shows graphs of (a) y vs. t and (b) v vs. t for the ball thrown upward in Fig. 2 9 31, incorporating the results of Examples 2 9 17, 2 9 18, 2 9 20, and 2 9 21.
AC
FIGURE 2 – 31   (Repeated.) An object thrown into the air leaves the thrower’s hand at A, reaches its maximum height at B, and returns to the original position at C. Examples 2 9 17, 2 9 18, 2 9 19, 2 9 20, and 2 9 21.
FIGURE 2 – 33   Graphs of (a) y vs. t, (b) v vs. t, for a ball thrown upward, Examples 2 9 17, 2 9 18, 2 9 20, and 2 9 21.
12
y = 11.5 m
20
10
15
t = 1.53 s
10
8
t= t=
5
6
0.69 s 2.37 s
0
t = 1.53 s
y (m) v (m>s)
4
2
0
0 0.5 1 1.5 2 2.5 3 3.5
(a)
t (s)
-5
-10
-15
-200 0.5 1 1.5 2 2.5 3 3.5
(b)
t (s)
Example 2 – 22 Ball thrown upward at edge of cliff. Suppose that the person of Examples 2 9 17, 2 9 18, 2 9 20, and 2 9 21 is standing on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below, as shown in Fig. 2 9 34. (a) How long does it take the ball to reach the base of the cliff? (b) What is the total distance traveled by the ball? Ignore air resistance (likely to be significant, so our result is an approximation).
APPROACH  We again use Eq. 2 9 12b, but this time we set  y = -50.0 m,  the
bottom of the cliff, which is 50.0 m below the initial position  (y0 = 0).
42  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 42
01/07/20 16:19
SOLUTION  (a) We use Eq. 2 9 12b with  a = - 9.80 m>s2, v0 = 15.0 m>s, y0 = 0,  and y = -50.0 m:
y
=
y0
+
v0 t
+
1 2
at
2
- 50.0 m
=
0
+
(15.0 m>s)t
-
1 2
(
9.80
m
>
s2
)
t
2.
Rewriting in the standard form we have
(4.90 m>s2)t2 - (15.0 m>s)t - (50.0 m) = 0.
Using the quadratic formula, we find as solutions  t = 5.07 s and  t = -2.01 s.  The first solution,  t = 5.07 s, is the answer we are seeking: the time it takes the ball to rise to its highest point and then fall to the base of the cliff. To rise and fall back to
the top of the cliff took 3.06 s (Example 2 9 18); so it took an additional 2.01 s to fall to
the base. But what is the meaning of the other solution,  t = -2.01 s? This is a time before the throw, when our calculation begins, so it isn’t relevant here. It is outside
our chosen time interval, and so is an unphysical solution just as in Example 2 9 12.
(b) For the total distance traveled, Example 2 9 17 told us that the ball moves
up 11.5 m, falls 11.5 m back down to the top of the cliff; it then falls down
another 50.0 m to the base of the cliff, for a total distance traveled of
2 (11.5 m) + 50.0 m = 73.0 m. Note that the displacement, however, was -50.0 m. Figure 2 9 34b shows the y vs. t graph for this situation.
NOTE  The solution  t = -2.01 s  in part (a) could be meaningful in a different physical situation. Suppose that a person standing on top of a 50.0-m-high cliff sees a rock pass by her at t = 0  moving upward at 15.0 m>s; at what time did the rock leave the base of the cliff, and when did it arrive back at the base of
the cliff? The equations will be precisely the same as for our original Example, and the answers  t = -2.01 s  and  t = 5.07 s  will be the correct answers. Note that we cannot put all the information for a problem into the mathematics, so
we have to use common sense in interpreting results.
EXERCISE H Two balls are thrown from a cliff. One is thrown directly up, the other directly down, each with the same initial speed, and both hit the ground below the cliff. Which ball hits the ground at the greater speed: (a) the ball thrown upward, (b) the ball thrown downward, or (c) both the same? Ignore air resistance.
*2–8 Variable Acceleration;
Integral Calculus
y y = 0
y = -50 m (a)
Hand 10 0
-10
y (m)
-20
-30
-40
t = 5.07 s
-500 1 2 3 4 5 6
Base of cliff
t (s) (b)
FIGURE 2 – 34   Example 2 9 22. (a) A person stands on the edge of a cliff. A ball is thrown upward, and then falls back down past the thrower to the base of the cliff, 50.0 m below. (b) The y vs. t graph.
In this brief optional Section we use integral calculus to derive the kinematic equations for constant acceleration, Eqs. 2 9 12a and b. We also show how calculus can be used when the acceleration is not constant. If you have not yet studied simple integration in your calculus course, you may want to postpone reading this Section until you have. We discuss integration in more detail in Section 7 9 3, where we begin to use it in the physics.
First we derive Eq. 2 9 12a, assuming as we did in Section 2 9 5 that an object has velocity v0 at  t = 0  and a constant acceleration a. We start with the definition of instantaneous acceleration,  a = dv>dt,  which we rewrite as
dv = a dt.
We take the definite integral of both sides of this equation, using the same notation
we did in Section 2 9 5, from  v = v0  at  t0 = 0  to some velocity v at time t:
v
t
3 dv = 3 a dt
v = v0
t0 = 0
which gives, since  a = constant,
v - v0 = at.
This is Eq. 2 9 12a,  v = v0 + at.
*SECTION 2–8  Variable Acceleration; Integral Calculus  43

GIAN_PSE5_CH02_020-053_ca.indd 43
01/07/20 16:19
Next we derive Eq. 2 9 12b starting with the definition of instantaneous velocity, Eq. 2 9 4,  v = dx>dt.  We rewrite this as
dx = v dt or
dx = (v0 + at)dt
where we substituted in Eq. 2 9 12a.
Now we integrate from  x = x0  at  t0 = 0  to an arbitrary position x at time t:
x
t
3 dx = 3 (v0 + at)dt
x = x0
t0 = 0
t
t
x - x0 = 3 v0 dt + 3 at dt
t0 = 0
t0 = 0
x - x0
=
v0 t
+
1 2
at
2
since
v0
and
a
are
constants. This
result
is
just
Eq. 2 9 12b,  x
=
x0
+
v0 t
+
1 2
at
2.
Finally let us use calculus to find velocity and displacement, given an accel-
eration that is not constant but varies in time.
EXAMPLE 2 – 23 Integrating a time-varying acceleration.  An experimental
vehicle starts from rest  (v0 = 0)  at  t = 0  and accelerates for a few seconds at a rate given by  a = (7.00 m>s3)t. What is (a) its velocity and (b) its displacement
2.00 s later?
APPROACH  We cannot use Eqs. 2 9 12 because a is not constant. We integrate the acceleration  a = dv>dt  over time to find v as a function of time; and then integrate  v = dx>dt to get the displacement.
SOLUTION  From the definition of acceleration,  a = dv>dt,  we have
dv = a dt.
We take the integral of both sides from  v = 0  at  t = 0  to velocity v at an arbitrary time t:
v
t
3 dv = 3 a dt
0
0
t
v = 3 (7.00 m>s3)t dt 0
=
(7.00 m>s3) ¢ t2 ≤ 2 t
20
=
(7.00 m>s3) ¢ t2
2
-
0≤
=
(3.50 m>s3)t2.
At  t = 2.00 s, v = (3.50 m>s3)(2.00 s)2 = 14.0 m>s.
(b) To get the displacement, we assume  x0 = 0 and start with  v = dx>dt which we rewrite as  dx = v dt. Then we integrate from  x = 0 at  t = 0 to position x at time t:
x
t
3 dx = 3 v dt
0
0
x
=
2.00 s
3 (3.50 m/s3)t2 dt 0
=
(3.50 m/s3)
t3
2.00 s
2
3 0
=
9.33 m.
In sum, at  t = 2.00 s, v = 14.0 m>s  and  x = 9.33 m.
Some problems in kinematics can be solved using Numerical Integration, which is discussed in Appendix C.
44  CHAPTER 2  Describing Motion: Kinematics in One Dimension 
GIAN_PSE5_CH02_020-053_ca.indd 44
01/07/20 16:19
Summary
[The Summary that appears at the end of each Chapter in this book gives a brief overview of the main ideas of the Chapter. The Summary cannot serve to give an understanding of the material, which can be accomplished only by a detailed reading of the Chapter.]
Kinematics deals with the description of how objects move. The description of the motion of any object must always be given relative to some particular reference frame.
The displacement of an object is the change in position of the object.
Average speed is the distance traveled divided by the elapsed time or time interval, ∆t, the time period over which we choose to make our observations. An object’s average velocity over a particular time interval ∆t is its displacement ∆x during that time interval, divided by ∆t:
v
=
∆x . ∆t
(2 – 2)
The instantaneous velocity, whose magnitude is the same as the instantaneous speed, is defined as the average velocity taken over an infinitesimally short time interval  (∆t S 0):
v
=
∆x
lim
∆t S 0
∆t
=
dx , dt
(2 – 4)
where dx>dt is the derivative of x with respect to t. On a graph of position vs. time, the slope is equal to the
instantaneous velocity.
Acceleration is the change of velocity per unit time. An
object’s average acceleration over a time interval ∆t is
a
=
∆v , ∆t
(2 – 5)
where ∆v is the change of velocity during the time interval ∆t.
Instantaneous acceleration is the average acceleration taken
over an infinitesimally short time interval:
a
=
∆v
lim
∆t S 0
∆t
=
dv . dt
(2 – 6)
On a graph of velocity vs. time, the slope is equal to the
instantaneous acceleration.
If an object has position x0 and velocity v0 at time  t = 0  and moves in a straight line with constant acceleration, the velocity
v and position x at a later time t are related to the acceleration
a, the initial position x0 , and the initial velocity v0 by Eqs. 2 9 12:
v = v0 + at,
x
=
x0
+
v0 t
+
1 2
at
2,
v2 = v02 + 2a(x - x0),
v
=
v + v0 . 2
(2 – 12)
Objects that move vertically near the surface of the Earth,
either falling or having been projected vertically up or down,
move with the constant downward acceleration due to gravity, whose magnitude is  g = 9.80 m>s2 if air resistance can be ignored.
[*Integral calculus can be used to derive the kinematic
Equations 2 9 12 and to solve Problems involving varying accel-
eration. Numerical integration is also a useful tool.]
Questions
1. Does a car speedometer measure speed, velocity, or both? Explain.
2. Can an object have a varying speed if its velocity is constant? Can it have varying velocity if its speed is constant? If yes, give examples in each case.
3. When an object moves with constant velocity, does its average velocity during any time interval differ from its instantaneous velocity at any instant? Explain.
4. If one object has a greater speed than a second object, does the first necessarily have a greater acceleration? Explain, using examples.
5. Compare the acceleration of a motorcycle that accelerates from 80 km>h to 90 km>h with the acceleration of a bicycle that accelerates from rest to 10 km>h in the same time.
6. Can an object have a northward velocity and a southward acceleration? Explain.
7. Can the velocity of an object be negative when its acceleration is positive? What about vice versa? If yes, give examples.
8. Give an example where both the velocity and acceleration are negative.
9. Two cars emerge side by side from a tunnel. Car A is traveling with a speed of 60 km>h and has an acceleration of 40 km>h>min. Car B has a speed of 40 km>h and has an acceleration of 60 km>h>min. Which car is passing the other as they come out of the tunnel? Explain your reasoning.
10. Can an object be increasing in speed as its acceleration decreases? If so, give an example. If not, explain.
11. A baseball player hits a ball straight up into the air. It leaves the bat with a speed of 120 km>h. Ignoring air resistance, how fast would the ball be traveling when the catcher catches it (at the same height it left the bat)? Explain.
12. As a freely falling object speeds up, what is happening to its acceleration : does it increase, decrease, or stay the same? (a) Ignore air resistance. (b) Consider air resistance.
13. You travel from point A to point B in a car moving at a constant speed of 70 km>h. Then you travel the same distance from point B to another point C, moving at a constant speed of 90 km>h. Is your average speed 80 km>h for the entire trip from A to C? Explain why or why not.
14. Can an object have zero velocity and nonzero acceleration at the same time? Give examples.
15. Can an object have zero acceleration and nonzero velocity at the same time? Give examples.
16. Which of these motions is not at constant acceleration: a rock falling from a cliff, an elevator moving from the second floor to the fifth floor making stops along the way, a dish resting on a table? Explain your answers.
17. Discuss two conditions given in Section 2 9 7 for being able to use a constant acceleration of magnitude  g = 9.8 m>s2.  Give an example in which one of these conditions would not be met and would not even be a reasonable approximat­ion of motion. [Hint: Carefully read Section 2 9 7, especially page 38.]
Questions  45

GIAN_PSE5_CH02_020-053_ca.indd 45
01/07/20 16:19
x (m) v (m>s)
18. Describe in words the motion plotted in Fig. 2 9 35 in terms of velocity, acceleration, etc. [Hint: First try to duplicate the motion plotted by walking or moving your hand.]
20
10
00
10
20
30
40
50
t (s)
FIGURE 2 – 35   Question 18.
19. Describe in words the motion of the object graphed in Fig. 2 9 36.
40
30
20
10
0 0 10 20 30 40 50 60 70 80 90 100 110 120 t (s) FIGURE 2 – 36   Question 19.
MisConceptual Questions
[List all answers that are valid, and ignore air resistance.]
1. In which of the following cases does a car have a negative
velocity and a positive acceleration? A car that is traveling
in the
(a) - x direction at a constant 20 m>s.
(b) - x direction increasing in speed.
(c) + x direction increasing in speed.
(d) - x direction decreasing in speed.
(e) + x direction decreasing in speed.
2. At time  t = 0  an object is traveling to the right along the + x axis at a speed of 10.0 m>s with constant acceleration of - 2.0 m>s2. Which statement is true?
(a) The object will slow down, eventually coming to a
­complete stop.
(b) The object cannot have a negative acceleration and be
moving to the right.
(c) The object will continue to move to the right, slowing
down but never coming to a complete stop.
(d) The object will slow down, momentarily stopping, then
pick up speed moving to the left.
3. You drive 4 km at 30 km>h and then another 4 km at 50 km>h.
What is your average speed for the whole 8-km trip?
(a) More than 40 km>h. (c) Less than 40 km>h.
(b) Equal to 40 km>h.
(d) Not enough information.
4. Two cars start from rest and travel a distance d with constant acceleration. The acceleration of car B is four times that of car A. After each has traveled distance d, (a) car B is moving 16 times as fast as car A. (b) car B is moving 8 times as fast as car A. (c) car B is moving 4 times as fast as car A. (d) car B is moving 2 times as fast as car A.
5. A rock is thrown straight up rising to a maximum height before falling back down. (a) The acceleration is constant for the entire trip. (b) The velocity is constant for the entire trip. (c) The magnitudes of both the velocity and acceleration decrease on the way up and increase on the way down. (d) The acceleration is constant for the entire trip except at the top where it is 0.
6. A ball is dropped from the top of a tall building. At the same instant, a second ball is thrown upward from ground level. When the two balls pass one another, one on the way up, the other on the way down, compare the magnit­udes of their acceleration: (a) The acceleration of the dropped ball is greater. (b) The acceleration of the ball thrown upward is greater. (c) The acceleration of both balls is the same. (d) The acceleration changes during the motion, so you ­cannot predict the exact value when the two balls pass each other. (e) The accelerations are in opposite directions.
7. You drop a rock off a bridge. When the rock has fallen 4 m, you drop a second rock. As the two rocks continue to fall, what happens to their velocities? (a) Both increase at the same rate. (b) The velocity of the first rock increases faster than the velocity of the second. (c) The velocity of the second rock increases faster than the velocity of the first. (d) Both velocities stay constant.
8. Two objects are dropped from a bridge, an interval of 1.0 s apart. During the time that both objects continue to fall, their separation (a) decreases at first, but then stays constant. (b) increases at first, but then stays constant. (c) decreases. (d) stays constant. (e) increases.
9. A ball is thrown downward at a speed of 20 m>s. Choosing the + y axis pointing up and neglecting air resistance, which equation(s) would correctly describe the motions? The acceleration due to gravity is  g = 9.8 m>s2  downward. (a) v = (20 m>s) - gt. (b) y = y0 + ( - 20 m>s)t - (1>2)gt2. (c) v2 = (20 m>s)2 - 2g(y - y0). (d) (20 m>s) = (v + v0)>2. (e) All of the above.
46  CHAPTER 2  Describing Motion: Kinematics in One Dimension 
GIAN_PSE5_CH02_020-053_ca.indd 46
01/07/20 16:19
10. A car travels along the x axis with increasing speed. We are not sure if it is moving to the left or to the right. Which of the graphs in Fig. 2 9 37 could possibly represent the motion of the car?
x
x
t (a) x
t (d) x
11. Two objects start at the same place at the same time and move along the same straight line (the x axis). Figure 2 9 38 shows the position x as a function of time t for each object. At point A, what must be true about the motion of the objects? (More than one statement may be correct.) (a) Both have the same instantaneous speed. (b) Both have the same instantaneous velocity. (c) Both are at the same position. (d) Both have traveled the same total distance. (e) With respect to the start, both have the same average velocity.
x
A
t (b) x
t (e)
t (c)
FIGURE 2 – 37   MisConceptual Question 10.
O
t
FIGURE 2 – 38   MisConceptual Question 11.
Problems
[The Problems at the end of each Chapter are ranked I, II, or III according to estimated difficulty, with level I Problems being easiest. Level III are meant as challenges for the best students. The Problems are arranged by Section, meaning that the reader should have read up to and including that Section, but not only that Section : Problems often depend on earlier material. Next is a set of “General Problems” not arranged by Section and not ranked.]
(Note: In Problems, assume a number like 6.4 is accurate to {0.1; and 950 is {10 unless 950 is said to be “precisely” or “very nearly” 950, in which case assume  950 { 1.  See Section 1 9 3.)
2 – 1 to 2 – 3  Speed and Velocity 1. (I) If you are driving 85 km>h along a straight road and you look to the side for 2.0 s, how far do you travel during this inattentive period?
2. (I) What must your car’s average speed be in order to travel 235 km in 2.85 h?
3. (I) A particle at  t1 = - 2.0 s  is at  x1 = 5.2 cm  and at t2 = 3.4 s  is at  x2 = 8.5 cm.  What is its average velocity over this time interval? Can you calculate its average speed from these data? Why or why not?
4. (II) According to a rule-of-thumb, each five seconds between a lightning flash and the following thunder gives the distance to the flash in miles. (a) Assuming that the light from the flash arrives in essentially no time at all, estimate the speed of sound in m>s from this rule. (b) What would be the rule for kilometers?
5. (II) You are driving home from school steadily at 95 km>h for 210 km. It then begins to rain and you slow to 65 km>h. You arrive home after driving 4.5 h. (a) How far is your hometown from school? (b) What was your average speed?
6. (II) A horse trots away from its trainer in a straight line, moving 38 m away in 7.4 s. It then turns abruptly and gallops halfway back in 1.8 s. Calculate (a) its average speed and (b) its average velocity for the entire trip, using “away from the trainer” as the positive direction.
7. (II) A person jogs eight complete laps around a 400-m track in a total time of 14.5 min. Calculate (a) the average speed and (b) the average velocity, in m>s.
8. (II) Every year the Earth travels about 109 km as it orbits the Sun. What is Earth’s average speed in km>h?
9. (II) A car traveling 95 km>h is 310 m behind a truck ­traveling 75 km>h. How long will it take the car to reach the truck?
10. (II) Calculate (a) the average speed and (b) average velocity of a round trip: the outgoing 280 km is covered at 95 km>h, followed by a 1.0-h lunch break, and the return 280 km is covered at 55 km>h.
11. (II) Two locomotives approach each other on parallel tracks. Each has a speed of 155 km>h with respect to the ground. If they are initially 9.5 km apart, how long will it be before they reach each other? (See Fig. 2 9 39.)
v = 155 km>h
9.5 km
v = 155 km>h
FIGURE 2 – 39   Problem 11.
Problems  47 
GIAN_PSE5_CH02_020-053_ca.indd 47
01/07/20 16:19
12. (II) Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spiraling path that starts at radius  R1 = 2.5 cm  and finishes at radius  R2 = 5.8 cm.  The dis­tance between the centers of neighboring spiral-­windings
is 1.6 mm  ( = 1.6 * 10-6 m).  (a) Determine the total length
of the spiraling path. [Hint: Imagine “unwinding” the spiral into a straight path of width 1.6 mm, and note that the ­original spiral and the straight path both occupy the same area.] (b) The CD player adjusts the rotation of the CD so that the player’s readout laser reads along the spiral path at a constant rate of about 1.2 m>s. Estimate the maximum playing time of such a CD.
13. (II) The position of a small object is given by x = 27 + 10t - 2t3,  where t is in seconds and x in meters. (a) Plot x as a function of t from  t = 0  to  t = 3.0 s.  (b) Find the average velocity of the object between 0 and 3.0 s. (c) At what time between 0 and 3.0 s is the ­instantaneous velocity zero?
14. (II) An airplane travels 1900 km at a speed of 720 km>h, and then encounters a tailwind that boosts its speed to 990 km>h for the next 2700 km. What was the total time for the trip? What was the average speed of the plane for this trip? [Hint: Does Eq. 2 9 12d apply?]
15. (II) Two ships need to arrive at a site in the middle of the ocean at the same time. They start out at the same time from positions equally distant from the arrival site. They travel at different velocities but both go in a straight line. The first ship travels at an average velocity of 20 km>h for the first 600 km, 40 km>h for the next 800 km, and 20 km>h for the final 600 km. The second ship can only sail at constant velocity. What is the magnitude of that velocity?
16. (II) The position of an object along a straight tunnel as a function of time is plotted in Fig. 2 9 40. What is its instantaneous velocity (a) at  t = 10.0 s  and (b) at t = 30.0 s?  What is its average velocity (c) between  t = 0  and  t = 5.0 s,  (d) between  t = 25.0 s  and  t = 30.0 s,  and (e) between t = 40.0 s  and  t = 50.0 s?
20
x (m)
10
00
10
20
30
40
50
t (s)
FIGURE 2 – 40   Problems 16, 17, and 18.
17. (II) In Fig. 2 9 40, (a) during what time intervals, if any, is the velocity constant? (b) At what time is the velocity greatest? (c) At what time, if any, is the velocity zero? (d) Does the object move in one direction or in both directions during the time shown?
18. (III) Sketch the v vs. t graph for the object whose displace­ ment as a function of time is given by Fig. 2 9 40.
19. (III) A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound of the ball hitting the pins 2.75 s after the ball is released from his hands. What is the speed of the ball, assuming the speed of sound is 340 m>s?
20. (III) An automobile traveling 95 km>h overtakes a 1.50‑kmlong train traveling in the same direction on a track parallel to the road. If the train’s speed is 75 km>h, how long does it take the car to pass it, and how far will the car have traveled in this time? See Fig. 2 9 41. What are the results if the car and train are traveling in opposite ­directions?
1.50 km
v = 75 km>h
v = 95 km>h FIGURE 2 – 41   Problem 20.
2 – 4 Acceleration 21. (I) A sprinter accelerates from rest to 9.00 m>s in 1.48 s.
What is her acceleration in (a) m>s2; (b) km>h2?
22. (I) A bicyclist in the Tour de France crests a mountain pass as he moves at 15 km>h. At the bottom, 4.0 km farther, his
speed is 65 km>h. Estimate his average acceleration (in m>s2)
while riding down the mountain.
23. (II) A sports car moving at constant velocity travels 120 m in 5.0 s. If it then brakes and comes to a stop in 3.7 s, what is the magnitude of its acceleration (assumed constant) in
m>s2, and in g’s  (g = 9.80 m>s2)?
24. (II) At highway speeds, a particular automobile is capable of an acceleration of about 1.8 m>s2. At this rate, how long does it take to accelerate from 65 km>h to 120 km>h?
25. (II) A car moving in a straight line starts at  x = 0  when  t0 = 0. It passes the point  x = 25.0 m  with a speed of 11.0 m>s at  t = 3.00 s.  It passes the point  x = 385 m  with a speed of 45.0 m>s at  t = 20.0 s.  Find (a) the average velocity, and (b) the average acceleration, between t = 3.00 s  and t = 20.0 s.
26. (II) A particle moves along the x axis. Its position as a function of time is given by  x = 4.8t + 7.3t2,  where t is in seconds and x is in meters. What is the acceleration as a function of time?
27. (II) The position of an object is given by  x = At + Bt2,  where x is in meters and t is in seconds. (a) What are the units of A and B? (b) What is the acceleration as a function of time? (c) What are the velocity and acceleration at t = 6.0 s?  (d) What is the velocity as a function of time if x = At + Bt -3?
28. (II) The position of a race car, which starts from rest at  t = 0  and moves in a straight line, is given as a function of time in the following Table. Estimate (a) its velocity and (b) its acceleration as a function of time. Display each in a Table and on a graph.
t (s) 0 x (m) 0
0.25 0.50 0.75 1.00 1.50 2.00 2.50 0.11 0.46 1.06 1.94 4.62 8.55 13.79
t(s) 3.00 3.50 4.00 4.50 5.00 5.50 6.00 x (m) 20.36 28.31 37.65 48.37 60.30 73.26 87.16
29. (II) A car traveling 25.0 m>s passes a second car which is at rest. When the cars are right next to each other, the first car slows down at a constant rate of 2.0 m>s2 and the second car starts to accelerate at the same constant rate. When will the two cars be next to each other again?
48  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 48
01/07/20 16:19
v (m>s)
30. (II) Figure 2 9 42 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest?
40
30
39. (II) A baseball pitcher throws a baseball with a speed of 43 m>s. Estimate the average acceleration of the ball during
the throwing motion. In throwing the baseball, the pitcher
accelerates it through a displacement of about 3.5 m, from
behind the body to the point where it is
3.5 m
released (Fig. 2 9 44).
20
10
0 0 10 20 30 40 50 60 70 80 90 100 110 120 t (s)
FIGURE 2 – 42   Problem 30.
31. (II) A sports car accelerates approximately as shown in the velocity 9 time graph of Fig. 2 9 43. (The short flat spots in the curve represent manual shifting of the gears.) Estimate the car’s average acceleration in (a) second gear and (b) fourth gear.
50
5th gear
40
4th gear
3rd gear 30
v (m>s)
20 2nd gear
10 1st gear
0
t (s)
0
10 20 30 40
FIGURE 2 – 43   Problem 31. The velocity of a car as a function of time, starting from a dead stop. The flat spots in the curve represent gear shifts.
32. (III)A fugitive tries to hop on a freight train traveling at a constant speed of 5.0 m>s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at  a = 1.4 m>s2  to his maximum speed of 6.0 m>s, which he then maintains.
(a) How long does it take him to catch up to the empty box
car? (b) What is the distance traveled to reach the box car?
FIGURE 2 – 44   Problem 39.
40. (II) A car traveling at 95 km>h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 m. What was the magnitude of the average acceleration of the driver during the collision? Express the answer in terms of “g’s,” where  1.00 g = 9.80 m>s2.
41. (II) A car traveling 85 km>h slows down at a constant 0.50 m>s2 just by “letting up on the gas.” Calculate (a) the distance the car coasts before it stops, (b) the time it takes to stop, and (c) the distance it travels during the first and fifth seconds.
42. (II) Determine the stopping distances for an automobile going a constant initial speed of 95 km>h in the + x direction, and human reaction time of 0.40 s: (a) for an acceleration   a = - 2.5 m>s2;  (b) for a = - 5.5 m>s2.
43. (II) A driver is traveling 18.0 m>s when she sees a red light ahead. Her car is capable of decelerating at a rate of 3.65 m>s2. If it takes her 0.380 s to get the brakes on and she is 24.0 m from the intersection when she sees the light, will she be able to stop in time? How far from the beginning of the intersection will she be, and in which direction?
44. (II) Show that  v = (v + v0) >2  (see Eq. 2 9 12d) is not valid when the acceleration  a = A + Bt,  where A and B are
non-zero constants.
45. (II) An 85-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m>s when it passes a railway worker who is standing 180 m from where the front of the train started. What will be the speed of the last car as it passes the worker? (See Fig. 2 9 45.)
2 – 5 to 2 – 6  Motion at Constant Acceleration 33. (I) A car slows down from 26 m>s to rest in a distance of
88 m. What was its acceleration, assumed constant? 34. (I) A car slows down from 28 m>s to rest in 6.3 s. What was
its (constant) acceleration? 35. (I) A car accelerates from 13 m>s to 22 m>s in 6.5 s. What
was its acceleration? How far did it travel in this time? Assume constant acceleration.
36. (II) A world-class sprinter can reach a top speed (of about 11.5 m>s) in the first 18.0 m of a race. What is the average acceleration of this sprinter and how long does it take her to reach that speed?
37. (II) A car slows down uniformly from a speed of 28.0 m>s to rest in 8.60 s. How far did it travel in that time?
38. (II) In coming to a stop, an old truck leaves skid marks 45 m long on the highway. Assuming a deceleration of 6.00 m>s2, estimate the speed of the truck just before braking.
85 m
v = 18 m>s
FIGURE 2 – 45   Problem 45.
46. (II) A space vehicle accelerates uniformly from 85 m>s at t = 0  to 162 m>s at  t = 10.0 s.  How far did it move between t = 2.0 s  and  t = 7.0 s?
47. (II) A runner hopes to complete the 10,000-m run in less than 30.0 min. After running at constant speed for exactly 27.0 min, there are still 1200 m to go. The runner must then accelerate at 0.20 m>s2 for how many seconds in order to achieve the desired time?
Problems  49

GIAN_PSE5_CH02_020-053_ca.indd 49
01/07/20 16:19
48. (III) Mary and Sally are in a foot race (Fig. 2 9 46). When Mary is 22 m from the finish line, she has a speed of 4.0 m>s and is 5.0 m behind Sally, who has a speed of 5.0 m>s. Sally
thinks she has an easy win and so, during the remaining
portion of the race, slows down at a constant rate of 0.40 m>s2 to the finish line. What constant acceleration does
Mary now need during the remaining portion of the race, if
she wishes to cross the finish line side-by-side with Sally?
Mary
Sally
4.0 m>s 5.0 m>s
Finish
5.0 m
22 m
FIGURE 2 – 46   Problem 48.
49. (III) An unmarked police car traveling a constant 95 km>h is passed by a speeder traveling 135 km>h. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car’s acceleration is 2.60 m>s2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?
50. (III) Assume in Problem 49 that the speeder’s speed is not known. If the police car accelerates uniformly at 2.60 m>s2 and overtakes the speeder after accelerating for 7.00 s, what was the speeder’s speed?
51. (III) A runner completes a 400-meter race in 55.0 s. The 55.0 seconds is made up of a 0.15 s reaction time from the starting sound until the runner starts moving followed by 30.0 m of constant acceleration and then 370 m at constant speed. What are the values of the acceleration and the constant speed?
2 – 7  Freely Falling Objects (neglect air resistance)
52. (I) A stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.25 s. How high is the cliff?
53. (I) Estimate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before “landing.”
54. (I) If a car rolls gently  (v0 = 0) off a vertical cliff, how long
does it take it to reach 55 km>h?
55. (II) A ball player catches a ball 2.6 s after throwing it vertically upward. With what speed did he throw it, and what height did it reach?
56. (II) A baseball is hit almost straight up into the air with a speed of 22 m>s. Estimate (a) how high it goes, and (b) how long it is in the air. (c) What factors make this an estimate?
57. (II) A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
58. (II) The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 120 cm. (a) What is their initial “launch” speed off the ground? (b) How long are they in the air?
59. (II) A stone is thrown vertically upward with a speed of 18.0 m>s. (a) How fast is it moving when it is at a height of 11.0 m? (b) How much time is required to reach this height? (c) Why are there two answers to (b)?
60. (II) For an object falling freely from rest, show that the distance traveled during each successive second increases in the ratio of successive odd integers (1, 3, 5, etc.). (This was first shown by Galileo.) See Figs. 2 9 27 and 2 9 30.
61. (II) If there were no air resistance, how long would it take a free-falling skydiver to fall from a plane at 3800 m to an altitude of 450 m, where she will open her parachute? What would her speed be at 450 m? (In reality, air resistance will restrict her speed to perhaps 150 km>h.)
62. (II) Pelicans tuck their wings and free-fall straight down when diving for fish. Suppose a pelican starts its dive from a height of 16.0 m and cannot change its path once committed. If it takes a fish 0.20 s to perform evasive action, at what minimum height must it spot the pelican to escape? Assume the fish is at the surface of the water.
63. (II) A stone is thrown ver­tically upward with a speed of
15.5 m>s from the edge of a cliff 75.0 m high (Fig. 2 9 47).
(a) How much later
does it reach the bottom
y
of the cliff? (b) What is
its speed just before
hitting? (c) What total distance did it travel?
y = 0
FIGURE 2 – 47   Problem 63.
y = -75 m
64. (II) A rocket rises vertically, from rest, with an acceleration of 3.2 m>s2 until it runs out of fuel at an altitude of 725 m. After this point, its acceleration is that of gravity, downward. (a) What is the velocity of the rocket when it runs out of fuel? (b) How long does it take to reach this point? (c) What maximum altitude does the rocket reach? (d) How much time (total) does it take to reach maximum altitude? (e) With what velocity does it strike the Earth? (f ) How long (total) is it in the air?
65. (II) Suppose you adjust your garden hose nozzle for a fast
stream of water. You point the nozzle vertically upward at
a height of 1.8 m above the
ground (Fig. 2 9 48). When
you quickly turn off the
nozzle, you hear the water
striking the ground next to
you for another 2.5 s. What
is the water speed as it
leaves the nozzle?
1.8 m
FIGURE 2 – 48  Problem 65.
50  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 50
01/07/20 16:19
66. (II) A helicopter is ascending vertically with a constant speed of 6.40 m>s. At a height of 105 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? [Hint: What is v0 for the packa­ ge?]
67. (II) Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 0.83 s after passing his window, 15 m above the sidewalk. (a) How fast are the balloons traveling when they pass Roger’s window? (b) Assuming the balloons are being released from rest, from what height are they being released?
68. (II) A baseball is seen to pass upward by a window with a vertical speed of 13 m>s. If the ball was thrown by a person 18 m below on the street, (a) what was its initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again?
69. (III) A falling stone takes 0.28 s to travel past a window 2.2 m tall (Fig. 2 9 49). From what height above the top of the window did the stone fall?
2.2 m
FIGURE 2 – 49   Problem 69.
To travel this distance took 0.28 s
70. (III) A toy rocket moving vertically upward passes by a 2.0-mhigh window whose base is 8.0 m above the ground. The rocket takes 0.15 s to travel the 2.0 m height of the window. What was the launch speed of the rocket, and how high will it go? Assume the propellant is burned very quickly at blastoff.
71. (III) A ball is dropped from the top of a 55.0-m-high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 24.0 m>s. The stone and ball collide part way up. How far above the base of the cliff does this happen?
*2 – 8  Variable Acceleration; Calculus * 72. (II) Given  v(t) = 25 + 18t,  where v is in m>s and t is in s,
use calculus to determine the total displacement from t1 = 1.3 s  to  t2 = 3.6 s. *73. (III) The acceleration of a particle is given by  a = A1t  where  A = 3.0 m>s5>2 .  At  t = 0, v = 7.5 m>s  and  x = 0.  (a) What is the velocity as a function of time? (b) What is the displacement as a function of time? (c) What are the acceleration, velocity, and displacement at  t = 5.0 s? * 74. (III) Air resistance acting on a falling body can be taken into account by the approximate relation for the ­acceleration:
dv a = dt = g - kv,
where k is a constant. (a) Derive a formula for the velocity of the body as a function of time assuming it starts from rest  (v = 0  at  t = 0).  [Hint: Change variables by setting u = g - kv.] (b) Determine an expression for the terminal velocity, which is the maximum value the velocity reaches.
General Problems
75. The acceleration due to gravity on the Moon is about one-sixth what it is on Earth. If an object is thrown vertically upward on the Moon, how many times higher will it go than it would on Earth, assuming the same initial velocity?
76. A person jumps out a fourth-story window 18.0 m above a firefighter’s safety net. The survivor stretches the net 1.0 m before coming to rest, Fig. 2 9 50. (a) What was the average deceleration experienced by the survivor when she was slowed to rest by the net? (b) What would you do to make it “safer” (that is, to generate a smaller deceleration): would you stiffen or loosen the net? Explain.
FIGURE 2 – 50  Problem 76.
18.0 m 1.0 m
77. A person who is properly restrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the
deceleration does not exceed 30 “g’s”  (1.00 g = 9.80 m>s2).
Assuming uniform deceleration at 30 g’s, calculate the dis­tance over which the front end of the car must be designed to collapse if a crash brings the car to rest from 95 km>h.
78. The position of a ball rolling in a straight line is given by x = 2.0 - 3.6t + 1.7t2,  where x is in meters and t in seconds. (a) What do the numbers 2.0, 3.6, and 1.7 refer to? (b) What are the units of each of these numbers? (c) Determine the position of the ball at  t = 1.0 s,  2.0 s, and 3.0 s. (d) What is the average velocity over the interval  t = 1.0 s  to  t = 3.0 s?
79. In a lecture demonstration, a 3.0-m-long vertical string with ten bolts tied to it at equal intervals is dropped from the ­ceiling of the lecture hall. The string falls on a tin plate, and the class hears the clink of each bolt as it hits the plate. (a) The sounds will not occur at equal time inter­vals. Why? (b) Will the time between clinks increase or decrease as the string falls? (c) How could the bolts be tied so that the clinks occur at equal intervals? (Assume the string is vertical with the bottom bolt touching the tin plate when the string is released.)
80. Two students are asked to find the height of a particular building using a barometer. Instead of using the barometer as an altitude-measuring device, they take it to the roof of the building and drop it off, timing its fall. One student reports a fall time of 2.0 s, and the other, 2.3 s. What % difference does the 0.3 s make for the estimates of the building’s height?
General Problems  51

GIAN_PSE5_CH02_020-053_ca.indd 51
01/07/20 16:19
81. Consider the street pattern shown in Fig. 2 9 51. Each intersection has a traffic signal, and the speed limit is 40 km>h. Suppose you are driving from the west at the speed limit. When you are 10.0 m from the first intersection, all the lights turn green. The lights are green for 13.0 s each. (a) Calculate the time needed to reach the third stoplight. Can you make it through all three lights without stopping? (b) Another car was stopped at the first light when all the lights turned green. It can accelerate at the rate of 2.00 m>s2 to the speed limit. Can the second car make it through all three lights without stopping? By how many seconds would it make it, or not make it?
West
East
88. In putting, the force with which a golfer strikes a ball is planned so that the ball will stop within some small distance of the cup, say 1.0 m long or short, in case the putt is missed. Accomplishing this from an uphill lie (that is, putting the ball downhill, see Fig. 2 9 52) is more difficult than from a downhill lie. To see why, assume that on a particular “green” the ball decelerates constantly at 1.8 m>s2 going downhill, and constantly at 2.6 m>s2 going uphill. Suppose we have an uphill lie 7.0 m from the cup. Calculate the allowable range of initial velocities we may impart to the ball so that it stops in the range 1.0 m short to 1.0 m long of the cup. Do the same for a downhill lie 7.0 m from the cup. What in your results suggests that the downhill putt is more difficult?
Speed limit
Your car
10 m
50 m
40 km>h 70 m
15 m
15 m
15 m
FIGURE 2 – 51   Problem 81.
82. Suppose a car manufacturer tested its cars for front-end collisions by hauling them up on a crane and dropping them from a certain height. (a) Show that the speed just before a car hits the ground, after falling from rest a vertical distance H, is given by 12gH. What height corresponds to a collision at (b) 35 km>h? (c) 95 km>h?
83. A stone is dropped from the roof of a high building. A second stone is dropped 1.50 s later. How far apart are the stones when the second one has reached a speed of 12.0 m>s?
84. A person jumps off a diving board 4.0 m above the water’s surface into a deep pool. The person’s downward motion stops 1.8 m below the surface of the water. Estimate the average deceleration of the person while under the water.
85. A police car at rest is passed by a speeder traveling at a ­constant 140 km>h. The police officer takes off in hot pursuit and catches up to the speeder in 850 m, maintaining a constant acceleration. (a) Qualitatively plot the position vs. time graph for both cars from the police car’s start to the catch-up point. Calculate (b) how long it took the police officer to overtake the speeder, (c) the required police car acceleration, and (d) the speed of the police car at the overtaking point. (e) This last result is unrealistic : so which assumptions do we have to reconsider?
86. Agent Bond is standing on a bridge, 15 m above the road below, and his pursuers are getting too close for comfort. He spots a flatbed truck approaching at 25 m>s, which he measures by knowing that the telephone poles the truck is passing are 25 m apart in this region. The roof of the truck is 3.5 m above the road, and Bond quickly calculates how many poles away the truck should be when he drops down from the bridge onto the truck, making his getaway. How many poles is it?
87. Two children are playing on two trampolines. The first child bounces up one-and-a-half times higher than the second child.The initial speed upwards of the second child is 4.0 m>s. (a) Find the maximum height the second child reaches. (b) What is the initial speed of the first child? (c) How long was the first child in the air?
Uphill lie
Downhill lie
7.0 m
7.0 m Green
FIGURE 2 – 52   Problem 88.
89. A person driving her car at  35 km>h  approaches an intersection just as the traffic light turns yellow. She knows that the yellow light lasts only 2.0 s before turning to red, and she is 28 m away from the near side of the intersection (Fig. 2 9 53). Should she try to stop, or should she speed up to cross the intersection before the light turns red? The intersection is 15 m wide. Her car’s maximum deceleration is - 5.8 m>s2, whereas it can accelerate from 45 km>h to 65 km>h in 6.0 s. Ignore the length of her car and her ­reaction time.
28 m +x
15 m
FIGURE 2 – 53   Problem 89.
90. A car is behind a truck going 18 m>s on the highway. The car’s driver looks for an opportunity to pass, guessing that his car can accelerate at 0.60 m>s2 and that he has to cover the 20-m length of the truck, plus 10 m of extra space at the rear of the truck and 10 m more at the front of it. In the oncoming lane, he sees a car approaching, probably at the speed limit, 25 m>s (55 mph). He estimates that the car is about 500 m away. Should he attempt the pass? Give details.
91. A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 4.1 s later. If the speed of sound is 340 m>s, how high is the cliff?
92. A conveyor belt is used to send burgers through a grilli­ng machine. If the grilling machine is 1.2 m long and the burgers require 2.8 min to cook, how fast must the conveyor belt travel? If the burgers are spaced 25 cm apart, what is the rate of burger production (in burgers>min)?
52  CHAPTER 2  Describing Motion: Kinematics in One Dimension

GIAN_PSE5_CH02_020-053_ca.indd 52
01/07/20 16:19
93. A rock is thrown vertically upward with a speed of 15.0 m>s.
Exactly 1.00 s later, a ball is thrown up vertically along the same path with a speed of 22.0 m>s. (a) At what time will
they strike each other? (b) At what height will the collision occur? (c) Answer (a) and (b) assuming that the order is reversed: the ball is thrown 1.00 s before the rock.
94. Figure 2 9 54 is a position versus time graph for the motion of
an object along the x axis. Consider the time interval from
A to B. (a) Is the object moving in the positive or negative
x direction? (b) Is the object speeding up or slow­ing down?
(c) Is the acceleration of the object positive or negative? Now consider the time interval from D to E. (d) Is the object moving in the positive or negative x direct­ion? (e) Is the object
speed­ing up or slowing down? ( f ) Is the acceleration of the
object positive or negative? (g) Finally, answer these same
30
A 25
B
three questions for
20
the time interval from
E
C to D.
15
x (m)
10
FIGURE 2 – 54   Problem 94.
5
0
C
D t (s)
01 2 34 56
95. In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between station stops. The more stops there are, the slower the train’s average speed. To get an idea of this problem, calculate the time it takes a train to make a 15.0-km trip in two situations: (a) the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends); and (b) the stations are 5.0 km apart (4 stations total). Assume that at each station the train accelerates at a rate of 1.1 m>s2 until it reaches 95 km>h, then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at - 2.0 m>s2. Assume it stops at each intermediate station for 22 s.
96. A race car driver must average 200.0 km>h over the course of a time trial lasting ten laps. If the first nine laps were done at an average speed of 196.0 km>h, what average speed must be maintained for the last lap?
97. A parachutist bails out of an airplane, and freely falls 75 m (ignore air friction). Then the parachute opens, and her acceleration is - 1.5 m>s2 (up). The parachutist reaches the ground with a speed of 1.5 m>s. (a) From how high did she bail out of the plane? (b) How much time did her fall take?
98. You stand at the top of a cliff while your friend stands on a beach below you. You drop a ball from rest and see that she catches it 1.4 s later. Your friend then throws the ball up to you, and it comes to rest just as it reaches your hand. What is the speed with which your friend threw the ball?
99. A robot used in a pharmacy picks up a medicine bottle at  t = 0.  It accelerates at 0.20 m>s2 for 4.5 s , then travels without acceleration for 68 s and finally decelerates at - 0.40 m>s2 for 2.5 s to reach the counter where the pharmacist will take the medicine from the robot. From how far away did the robot fetch the medicine?
100. Bill can throw a ball vertically at a speed 1.5 times faster than Joe can. How many times higher will Bill’s ball go than Joe’s?
101. On an audio compact disc (CD), digital bits of ­information are encoded sequentially along a spiral path. Each bit ­occupies about 0.28 mm. A CD player’s readout laser scans along the spiral’s sequence of bits at a constant rate of about 1.2 m>s as the CD spins. (a) Determine the number N of digital bits that a CD player reads every second. (b) The audio information is sent to each of the two loudspeakers 44,100 times per second. Each of these samplings requires 16 bits, and so you might expect the required bit rate for a CD player to be
N0
=
2 a44,100
samplings b a16 s
bits b sampling
=
1.4 * 106
bits s
,
where the 2 is for the 2 loudspeakers (the 2 stereo­ channels).  Note that N0 is less than the number N of bits actually read per second by a CD player. The excess
number of bits ( = N - N0)  is needed for encoding and
error ­correction. What percentage of the bits on a CD are ­dedicated to encoding and error correction?
102. Figure 2 9 55 shows the position vs. time graph for two bicycles, A and B. (a) Identify any instant at which the two bicycles have the same velocity. (b) Which bicycle has the larger acceleration? (c) At which instant(s) are the bicycles passing each other? Which bicycle is passing the other? (d) Which bicycle has the larger instantaneous velocity? (e) Which bicycle has the larger average velocity?
x
AB
FIGURE 2 – 55 
Problem 102. 0
t
103. You are traveling at a constant speed vM , and there is a car in front of you traveling with a speed vA . You realize that  vM 7 vA ,  so you start slowing down with a constant acceleration a when the distance between you and the
other car is x. What relationship between a and x deter-
mines whether or not you run into the car in front of you?
A nswers to E x ercises
A: (a) displacement  = - 30 cm;  (b) total distance = 50 cm. B: (b). C: (b). D: (a) + ; (b) - ; (c) - ; (d) + .
E: (c). F: (b). G: (e). H: (c).

GIAN_PSE5_CH02_020-053_ca.indd 53
General Problems  53
01/07/20 16:19
This snowboarder flying through
the air is an example of motion
in two dimensions. In the
absence of air resistance, the
path would be a perfect
parabola. The gold arrow
represents the downward
acceleration of gravity, g5.
g5
Galileo analyzed the motion
of objects in two dimensions
under the action of gravity near
the Earth’s surface (now called
“projectile motion”) by
separating its horizontal and
vertical components.
We will discuss how to
manipulate vectors and how to
add them. Besides analyzing
projectile motion, we will also
discuss unit vectors and vector
kinematics, plus see how to work
with relative velocity.
r
ha p te
3 Kinematics in Two or Three Dimensions; Vectors
CONTENTS
3–1 Vectors and Scalars
3–2 Addition of Vectors :  Graphical Methods
3–3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar
3–4 Adding Vectors by Components
3–5 Unit Vectors
3–6 Vector Kinematics
3–7 Projectile Motion
3–8 Solving Problems Involving Projectile Motion
3–9 Relative Velocity
CHAPTER-OPENING QUESTION—Guess now!
[Don’t worry about getting the right answer now : you will get another chance later in the Chapter. See also page 1 of Chapter 1 for more explanation.]
A small heavy box of emergency supplies is dropped from a moving helicopter at point A as it flies at constant speed in a horizontal direction. Which path in the drawing below best describes the path of the box (neglecting wind and air resistance) as seen by a person standing on the ground?
A
B
(a)
(b)
(c)
(d) (e)
C
54  
GIAN_PSE5_CH03_054-084_ca.indd 54
I n Chapter 2 we dealt with motion along a straight line. We now consider the motion of objects that move in paths in two (or three) dimensions. In particular, we discuss an important type of motion known as projectile motion: objects projected outward near the Earth’s surface, such as struck baseballs and golf balls, kicked footballs, and other projectiles. Before beginning our discussion of motion in two dimensions, we will need to discuss vectors, and how to add them.
01/07/20 16:22
3–1  Vectors and Scalars
We mentioned in Chapter 2 that the term velocity refers not only to how fast an object is moving but also to its direction. A quantity such as velocity, which has direction as well as magnitude, is a vector quantity. Other quantities that are also vectors are displacement, force, and momentum. However, many quantities have no direction associated with them, such as mass, time, and temperature. They are specified completely by a number and units. Such quantities are called scalar quantities.
Drawing a diagram of a particular physical situation is always helpful in physics, and this is especially true when dealing with vectors. On a diagram, each vector is represented by an arrow. The arrow is always drawn so that it points in the direction of the vector quantity it represents. The length of the arrow is drawn proportional to the magnitude of the vector quantity. For example, in Fig. 3 9 1, green arrows have been drawn representing the velocity of a car at various places as it rounds a curve. The magnitude of the velocity at each point can be read off Fig. 3 9 1 by measuring the length of the corresponding arrow and using the scale shown (1 cm = 90 km>h).
When we write the symbol for a vector, we will always use boldface type, with a tiny arrow over the symbol. Thus for velocity we write v5. If we are concerned only with the magnitude of the vector, we will write simply v, in italics, as we do for other symbols.
3–2 Addition of Vectors—Graphical
Methods
Because vectors are quantities that have direction as well as magnitude, they must be added in a special way. In this Chapter, we will deal mainly with displacement vectors, for which we now use the symbol D5 , and velocity vectors, v5. But the results apply for acceleration and other vectors we will encounter later.
We use simple arithmetic for adding scalars. Simple arithmetic can also be used for adding vectors if they are in the same direction. For example, if a person walks 8 km east one day, and 6 km east the next day, the person will be 8 km + 6 km = 14 km east of the point of origin. We say that the net or resultant displacement is 14 km to the east (Fig. 3 9 2a). If, on the other hand, the person walks 8 km east on the first day, and 6 km west (in the reverse direction) on the second day, then the person will end up 2 km from the origin (Fig. 3 9 2b), so the resultant displacement is 2 km to the east. In this case, the resultant displacement is obtained by subtraction:  8 km - 6 km = 2 km.
But simple arithmetic cannot be used if the two vectors are not along the same line. For example, suppose a person walks 10.0 km east and then walks 5.0 km north. These displacements can be represented on a graph in which the positive y axis points north and the positive x axis points east, Fig. 3 9 3. On this graph, we draw an arrow, labeled D5 1 , to represent the 10.0-km displacement to the east. Then we draw a second arrow, D5 2 , to represent the 5.0-km displacement to the north. Both vectors are drawn to scale, as shown in Fig. 3 9 3.
Scale for velocity: 1 cm = 90 km>h
FIGURE 3 – 1   Car traveling on a road, slowing down to round the curve. The green arrows represent the velocity vector at each position.
Figure 3 – 2   Combining vectors in one dimension.
Resultant = 14 km (east)
0 8 km
6 km
x (km) East
(a)
Resultant = 2 km (east) 6 km
0 8 km
(b)
x (km) East
West
y (km)
6 4 2
North Result5DanRt
=di5Dsp1la+c5Dem2ent
D5 2
0
u
D5 1
x (km)
2 4 6 8 10 East
South
Figure 3 – 3   A person walks 10.0 km east and then 5.0 km
north. These two displacements are represented by the vectors D5 1 and D5 2, which are shown as arrows. Also shown is the resultant displacement vector, D5 R , which is the vector sum of D5 1 and D5 2. Measurement on the graph with ruler and protractor shows that D5 R has a magnitude of 11.2 km and points at an angle  u = 27° north of east.
SECTION 3–2  Addition of Vectors—Graphical Methods  55

GIAN_PSE5_CH03_054-084_ca.indd 55
01/07/20 16:22
After taking this walk, the person is now 10.0 km east and 5.0 km north of the point of origin. The resultant displacement is represented by the arrow labeled D5 R in Fig. 3 9 3. (The subscript R stands for resultant.) Using a ruler and a protractor, you can measure on this diagram (Fig. 3 9 3 on previous page) that the person is 11.2 km from the origin at an angle  u = 27° north of east. In other words, the resultant displacement vector has a magnitude of 11.2 km and makes an angle  u = 27°  with the positive x axis. The magnitude (length) of D5 R can also be obtained using the theorem of P­ ythagoras in this case, because D1 , D2 , and DR form a right triangle with DR as the h­ ypotenuse. Thus
DR = 2D12 + D22 = 2(10.0 km)2 + (5.0 km)2
= 2125 km2 = 11.2 km.
You can use the Pythagorean theorem only when the vectors are perpendicular
to each other. The resultant displacement vector, D5 R , is the sum of the vectors D5 1 and D5 2 .
That is,
D5 R = D5 1 + D5 2 .
This is a vector equation. An important feature of adding two vectors that are not along the same line is that the magnitude of the resultant vector is not equal to the sum of the magnitudes of the two separate vectors, but is smaller than their sum. That is,
DR … (D1 + D2),
where the equals sign applies only if the two vectors point in the same ­direction. In our example (Fig. 3 9 3),  DR = 11.2 km,  whereas  D1 + D2  equals   15 km, which is the total distance traveled.  Note also that we cannot set D5 R equal to 11.2 km, because we have a vector equation and 11.2 km is only a part of the resultant vector, its magnitude. We could write something like this, though:  D5 R = D5 1 + D5 2 = (11.2 km, 27° N of E).
Figure 3 9 3 illustrates the general rules for graphically adding two vectors together, no matter what angles they make, to get their sum. The rules are as follows:
1. On a diagram, draw one of the vectors : call it D5 1 : to scale. 2. Next draw the second vector, D5 2 , to scale, placing its tail at the tip of the
first vector and being sure its direction is correct.
3. The arrow drawn from the tail of the first vector to the tip of the second vector represents the sum, or resultant, of the two vectors.
The length of the resultant vector represents its magnitude. Note that vectors can
be moved parallel to themselves on paper (maintaining the same length and
angle) to accomplish these manipulations. The length of the resultant can be
FIGURE 3 – 4   If the vectors are added in reverse order, the resultant is the same. (Compare to Fig. 3 9 3.)
measured with a ruler and compared to the scale. Angles can be measured with a protractor. This method is known as the tail-to-tip method of adding vectors.
The resultant is not affected by the order in which the vectors are added.
y (km)
North 6
D5 1
For example, a displacement of 5.0 km north, to which is added a displacement
of 10.0 km east, yields a resultant of 11.2 km and angle  u = 27°  (see Fig. 3 9 4), the same as when they were added in reverse order (Fig. 3 9 3). Thus we can write,
4 2 D5 2
u 5D R= 5D 1+ D5 2
using V5 to represent any type of vector, V51 + V52 = V52 + V51 ,
[commutative property]  (3 – 1a)
West 0
2
4
6
8
x (km) 10 East
which is known as the commutative property of vector addition.
The tail-to-tip method of adding vectors can be extended to three or more
South
vectors. The resultant is drawn from the tail of the first vector to the tip of the last
56  CHAPTER 3  Kinematics in Two or Three Dimensions; Vectors

GIAN_PSE5_CH03_054-084_ca.indd 56
01/07/20 16:22
V51
+
V52
+
V51
=
V53
V5R
V52
FIGURE 3 – 5   The resultant of three vectors: V5R = V51 + V52 + V53 .
V53 one added. An example is shown in Fig. 3 9 5; the three vectors could represent displacements (northeast, south, west) or perhaps three forces. Check for yourself that you get the same resultant no matter in which order you add the three vectors; that is,
(V51 + V52) + V53 = V51 + (V52 + V53), [associative property]  (3 – 1b)
which is known as the associative property of vector addition. A second way to add two vectors is the parallelogram method. It is fully
equivalent to the tail-to-tip method. In this method, the two vectors are drawn starting from a common origin, and a parallelogram is constructed using these two vectors as adjacent sides as shown in Fig. 3 9 6b. The resultant is the diagonal drawn from the common origin. In Fig. 3 9 6a, the tail-to-tip method is shown, and we can see that both methods yield the same result.
V5 1
+
V52 =
V5R
V52 (a) Tail-to-tip
V51
V5R
= V52
V51
(b) Parallelogram
FIGURE 3 – 6   Vector addition by two different methods, (a) and (b). Part (c) is incorrect.
= V52
INCORRECT
(c) Wrong
V51
It is a common error to draw the sum vector as the diagonal running between
the tips of the two vectors, as in Fig. 3 9 6c. This is incorrect: it does not represent the sum of the two vectors. (In fact, it represents their difference,  V52 - V51 ,  as we will see in the next Section.)
CAUTION
Be sure to use the correct diagonal on the parallelogram to get the resultant
CONCEPTUAL EXAMPLE 3 – 1 Range of vector lengths. Suppose two vectors each have length 3.0 units. What is the range of possible lengths for the vector representing the sum of the two?
Response  The sum can take on any value from 6.0  ( = 3.0 + 3.0)  where the vectors point in the same direction, to 0  ( = 3.0 - 3.0)  when the vectors are antiparallel. Magnitudes between 0 and 6.0 occur when the two vectors are at an angle other than 0° and 180°.
EXERCISE A  If the two vectors of Example 3 9 1 are perpendicular to each other, what is the resultant vector length?
3–3  Subtraction of Vectors, and
Multiplication of a Vector by a Scalar
Given a vector V5 , we define the negative of this vector ( -V5 ) to be a vector with
the same magnitude as V5 but opposite in direction, Fig. 3 9 7. Note, however, that no vector is ever negative in the sense of its magnitude: the magnitude of every vector is positive. Rather, a minus sign tells us about its direction.
FIGURE 3 – 7   The negative of a vector is a vector having the same length but opposite direction.
V5
– V5
SECTION 3–3  Subtraction of Vectors, and Multiplication of a Vector by a Scalar  57

GIAN_PSE5_CH03_054-084_ca.indd 57
01/07/20 16:22
Figure 3 – 8   Subtracting two vectors:  V52 - V51 .
V52 –
V51
= V52
+
–V51
–V51
= V52 – V51
V52
FIGURE 3 – 9   Multiplying a vector V5 by a scalar c gives a vector whose magnitude is c times greater and in the same direction as V5 (or opposite direction if c is negative).
V52 = 1.5 V5 V5
V53 = -2.0 V5
We can now define the subtraction of one vector from another: the difference between two vectors  V52 - V51 is defined as
V52 - V51 = V52 + ( - V51).
That is, the difference between two vectors is equal to the sum of the first plus the negative of the second. Thus our rules for addition of vectors can be applied as shown in Fig. 3 9 8 using the tail-to-tip method.
A vector V5 can be multiplied by a scalar c. We define their product so that cV5 has the same direction as V5 and has magnitude cV. That is, multiplication of a vector by a positive scalar c changes the magnitude of the vector by a factor c but doesn’t alter the direction. If c is a negative scalar, the magnitude of the product cV5 is still
0 c 0 V (where 0 c 0 means the magnitude of c), but the direction is precisely opposite
to that of V5 . See Fig. 3 9 9.
EXERCISE B  What does the “incorrect” vector in Fig. 3 9 6c represent? (a) V52 - V51, (b) V51 - V52, (c) something else (specify).
3–4  Adding Vectors by Components
Adding vectors graphically using a ruler and protractor is often not sufficiently
accurate and is not useful for vectors in three dimensions. We discuss now a more
powerful and precise method for adding vectors. But do not forget graphical
methods : they are useful for visualizing, for checking your math, and thus for
getting the correct result. Consider first a vector V5 that lies in a particular plane, and we have chosen an x
and a y axis on this plane. Then V5 can be expressed as the sum of two other vectors,
called the components of the original vector, which are usually chosen to be along
the x and y axes. The process of finding the components is known as resolving the vector into its components. An example is shown in Fig. 3 9 10; the vector V5 could
be a displacement vector that points at an angle  u = 30° north of east, where we have chosen the ­positive x axis to be to the east and the positive y axis north. This vector V5 is resolved into its x and y components by drawing dashed lines
out from the tip (A) of the vector (lines AB and AC), making them perpendicular to
the x and y axes. Then the lines 0B and 0C represent the x and y components of V5 , respectively, as shown in Fig. 3 9 10b. These vector components are written V5x and V5y . In this book we usually show vector components as arrows, like vectors, but dashed. The scalar components, Vx and Vy , are the magnitudes of the vector components, with units, accompanied by a p­ ositive or negative sign depending on
whether they point along the positive or negative x or y axis. As can be seen in Fig. 3 9 10,  V5x + V5y = V5   by the parallelogram method of adding vectors.
Space is made up of three dimensions, and sometimes it is necessary to
resolve a vector into components along three mutually perpendicular directions. In rectangular coordinates the components are V5x , V5y , and V5z .
y
North
Figure 3 – 10   Resolving a vector V5 into its
C
A
components along a chosen set of x and y axes.
V5
The components, once found, themselves
y North
V5y V5
represent the vector. That is, the components contain as much information as the vector itself.
u (= 30°)
x
0
B East
(a)
u (= 30°)
x
0
V5x East
(b)
58  CHAPTER 3  Kinematics in Two or Three Dimensions; Vectors

GIAN_PSE5_CH03_054-084_ca.indd 58
01/07/20 16:22
The use of trigonometric functions for finding the components of a vector is illustrated in Fig. 3 9 11, where a vector and its two components are thought of as making up a right triangle. (See also Appendix A for other details on trigonometric functions and identities.) We then see that the sine, cosine, and tangent are as given in Fig. 3 9 11, where u is the angle V5 makes with the + x axis, measured positive counterclockwise. If we multiply the definition of  sin u = Vy>V by V on both sides, we get
Vy = V sin u. Similarly, from the definition of cos u, we obtain
(3 – 2a)
Vx = V cos u.
(3 – 2b)
Keep in mind that u is chosen (by convention) to be the angle that the vector
makes with the positive x axis, measured positive counterclockwise.
The components of a given vector will be different for different choices of
coordinate axes. It is therefore crucial to specify the choice of coordinate system
when giving the components.
There are two ways to specify a vector in a given coordinate system:
1. We can give its components, Vx and Vy . 2. We can give its magnitude V and the angle u it makes with the positive x axis.
We can shift from one description to the other using Eqs. 3 9 2, and, for the reverse, by using the theorem of Pythagoras† and the definition of tangent:
y
V5 V5y
u
0
V5x
sin
u
=
Vy V
cos
u
=
Vx V
tan
u
=
Vy Vx
V 2 = Vx2 + Vy2
90° x
Figure 3 – 11   Finding the components of a vector using trigonometric functions. The equations are valid only if u is the angle V5 makes with the positive x axis.
V = 2Vx2 + Vy2
(3 – 3a)
Vy
tan u
=
(3 – 3b) Vx
as can be seen in Fig. 3 9 11.
We can now discuss how to add vectors using components. The first step is to resolve each vector into its components, Eqs. 3 9 2. Next we can see, using Fig. 3 9 12, that the addition of any two vectors V51 and V52 to give a resultant  V5R = V51 + V52 ,  implies that
VRx = V1x + V2x VRy = V1y + V2y .
(3 – 4)
That is, the sum of the x components equals the x component of the resultant
vector, and the sum of the y components equals the y component of the resultant
vector, as can be verified by a careful examination of Fig. 3 9 12. Note that we
do not add x components to y components.
If the magnitude and direction of the resultant vector are desired, they can
be obtained using Eqs. 3 9 3.
†In three dimensions, the theorem of Pythagoras becomes  V = 3Vx2 + Vy2 + Vz2 , where Vz is the ­component along the third, or z, axis.
y VRx
VRy 0
5V
R
=
5V
1
+
5V 2 V51
V5 2
V2y
V2x
V1y
Figure 3 – 12   The components of V5R = V51 + V52  are
VRx = V1x + V2x VRy = V1y + V2y .
V1x x
SECTION 3–4  Adding Vectors by Components  59

GIAN_PSE5_CH03_054-084_ca.indd 59
01/07/20 16:22
y North
D5 1 60°
Post 0
office
D5 2
(a) D5 1 y
D2x 0 60°
x East
x
D2y
D5 2
(b)
D5 1 y
0 u D5 2
x
D5 R
(c)
Figure 3 – 13   Example 3 9 2.
(a) The two displacement vectors, D5 1 and D5 2 . (b) D5 2 is resolved into its components. (c) D5 1 and D5 2 are added graphically to obtain the resultant D5 R. The component method of adding the vectors is
explained in the Example.
The components of a given vector depend on the choice of coordinate axes. You can often reduce the work involved in adding vectors by a good choice of axes : for example, by choosing one of the axes to be in the same direction as one of the vectors. Then that vector will have only one nonzero component.
EXAMPLE 3 – 2 Mail carrier’s displacement.  A rural mail carrier leaves the post office and drives 22.0 km in a northerly direction. She then drives in a direction 60.0° south of east for 47.0 km (Fig. 3 9 13a). What is her displacement from the post office?
APPROACH  We choose the positive x axis to be east and the positive y axis to be north, since those are the compass directions used on most maps. The origin of the xy coordinate system is at the post office. We resolve each vector into its x and y components. We add the x components together, and then the y components together, giving us the x and y components of the resultant.
SOLUTION  Resolve each displacement vector into its components, as shown in Fig. 3 9 13b. Since D5 1 has magnitude 22.0 km and points north, it has only a y component:
D1x = 0, D1y = 22.0 km. D5 2 has both x and y components:
D2x = + (47.0 km) (cos 60°) = + (47.0 km) (0.500) = + 23.5 km D2y = - (47.0 km) (sin 60°) = - (47.0 km) (0.866) = - 40.7 km. Notice that D2y is negative because this vector component points along the negative y axis. The resultant vector, D5 R , has components:
DRx = D1x + D2x = 0 km + 23.5 km = + 23.5 km DRy = D1y + D2y = 22.0 km + ( - 40.7 km) = - 18.7 km. This specifies the resultant vector completely:
DRx = 23.5 km, DRy = - 18.7 km.
We can also specify the resultant vector by giving its magnitude and angle using Eqs. 3 9 3:
DR = 2DR2 x + DR2 y = 2(23.5 km)2 + ( - 18.7 km)2 = 30.0 km
tan u
=
DR y DR x
=
- 18.7 km 23.5 km
=
- 0.796.
A calculator with a key labeled inv tan, or arc tan, or tan-1 gives u = tan-1 ( - 0.796) = - 38.5°.  The negative sign means  u = 38.5°  below the x axis, Fig. 3 9 13c. So, the resultant displacement is 30.0 km directed at 38.5° in
a southeasterly ­direction.
Note  Always be attentive about the quadrant in which the resultant vector lies. An ­electronic calculator does not fully give this information, but a good diagram does.
P ro b l e m S o l v ing
Identify the correct quadrant by drawing a careful diagram
As we saw in Example 3 9 2, any component that points along the negative x or y axis gets a minus sign. The signs of trigonometric functions depend on which “quadrant” the angle falls in: for example, the tangent is positive in the first and third quadrants (from 0° to 90°, and 180° to 270°), but negative in the second and fourth quadrants; see Appendix A 9 9, Fig. A 9 6. The best way to keep track of angles, and to check any vector result, is always to draw a vector diagram, like Fig. 3 9 13. A vector diagram gives you something tangible to look at when analyzing a problem, and provides a check on the results.
The following Problem Solving Strategy should not be considered a prescription. Rather it is a summary of things to do to get you thinking and involved in the problem at hand.
60  CHAPTER 3  Kinematics in Two or Three Dimensions; Vectors

GIAN_PSE5_CH03_054-084_ca.indd 60
01/07/20 16:22
Problem
S o l v i n g Adding Vectors
Here is a brief summary of how to add two or more vectors using components:
1. Draw a diagram, adding the vectors graphically by either the parallelogram or tail-to-tip method. 2. Choose x and y axes. Choose them in a way, if possible, that will make your work easier. (For example, choose one axis along the direction of one of the vectors, which then will have only one component.) 3. Resolve each vector into its x and y ­components, showing each component along its appropriate (x or y) axis as a (dashed) arrow. 4. Calculate each component (when not given) using sines and cosines. If u1 is the angle that vector V51 makes with the positive x axis, then:
V1x = V1 cos u1 , V1y = V1 sin u1 .
Pay careful attention to signs: any component that points along the negative x or y axis gets a minus sign.
5. Add the x components together to get the x component of the resultant. Similarly for y:
VRx = V1x + V2x + any others VRy = V1y + V2y + any others.
This is the answer: the components of the resultant vector. Check signs to see if they fit the quadrant shown in your diagram (point 1 above).
6. If you want to know the magnitude and
­direction of the resultant vector, use Eqs. 3 9 3:
VR
=
2V
2 R
x
+
V
2
R y
,
tan u
=
VRy . VR x
The vector diagram you already drew helps to obtain
the correct position (quadrant) of the angle u.
EXAMPLE 3 – 3 Three short trips.  An airplane trip involves three legs, with two stopovers, as shown in Fig. 3 9 14a. The first leg is due east for 620 km; the second leg is southeast (45°) for 440 km; and the third leg is at 53° south of west, for 550 km, as shown. What is the plane’s total displacement?
Approach  We follow the steps in the Problem Solving Strategy above.
Solution 1. Draw a diagram such as Fig. 3 9 14a, where D5 1 , D5 2 , and D5 3 represent the three legs of the trip, and D5 R is the plane’s total displacement. 2. Choose axes: Axes are also shown in Fig. 3 9 14a: x is east, y north.
3. Resolve components: It is imperative to draw a good diagram. The components are drawn in Fig. 3 9 14b. Instead of drawing all the vectors starting from a common origin, as we did in Fig. 3 9 13b, here we draw them “tail-to-tip” style, which is just as valid and may make it easier to see.
4. Calculate the components: D5 1 : D1x = + D1 cos 0° = D1 = 620 km D1y = + D1 sin 0° = 0 km D5 2 : D2x = + D2 cos 45° = + (440 km) (0.707) = + 311 km D2y = - D2 sin 45° = - (440 km) (0.707) = - 311 km D5 3 : D3x = - D3 cos 53° = - (550 km) (0.602) = - 331 km D3y = - D3 sin 53° = - (550 km) (0.799) = - 439 km.
We have given a minus sign to each component that in Fig. 3 9 14b points in the -x or -y direction. The components are shown in the Table in the margin. 5. Add the components: We add the x components together, and we add the y components together to obtain the x and y components of the resultant:
+y North
-x
D5 1
0
u=?
45° D5 2
+x East
D5 R
53°
D5 3
-y
(a) +y
North
-x
D5 1
D2x
0
D5 2 45° D2y
D3x
D3y
53° D5 3
-y
(b)
FIGURE 3 – 14   Example 3 9 3.
+x East
DRx = D1x + D2x + D3x = 620 km + 311 km - 331 km = 600 km DRy = D1y + D2y + D3y = 0 km - 311 km - 439 km = - 750 km.
The x and y components of the resultant are 600 km and - 750 km, and point ­respectively to the east and south. This is one way to give the answer.
6. Magnitude and direction: We can also give the answer as
DR = 2DR2 x + DR2 y = 2(600)2 + ( - 750)2 km = 960 km
tan u
=
DR y DR x
=
- 750 km 600 km
=
- 1.25,
so u = -51°.
Thus, the total displacement has magnitude 960 km and points 51° below
the x axis (south of east), as was shown in our original sketch, Fig. 3 9 14a.
Vector
D5 1 D5 2 D5 3 D5 R
Components
x (km)
y (km)
620 311 - 331 600
0 - 311 - 439 - 750
SECTION 3–4  Adding Vectors by Components  61

GIAN_PSE5_CH03_054-084_ca.indd 61
01/07/20 16:22
y
3–5  Unit Vectors
jˆ
x
kˆ
ˆi
z Figure 3 – 15   Unit vectors Ni, jN, and kN along the x, y, and z axes.
Vectors can be conveniently written in terms of unit vectors. A unit vector is defined
to have a magnitude exactly equal to one (1). (The word “unit” comes from the Latin,
unus, meaning “one”.) It is useful to define unit vectors that point along coordinate
axes, and in an x, y, z rectangular coordinate system these unit vectors are called Ni, jN, and kN . They point, respectively, along the positive x, y, and z axes as shown in Fig. 3 9 15. Like other vectors, Ni, jN, and kN do not have to be placed at the origin, but
can be placed elsewhere as long as the direction and unit length remain unchanged. It is common to write unit vectors with a “hat”: Ni, jN, kN (and we will do so in this
book) as a reminder that each has magnitude of exactly one unit.
Because of the definition of multiplication of a vector by a scalar (Section 39 3 ), the components of a vector V5 can be written  V5x = Vx Ni,  V5y = Vy jN,  and  V5z = Vz kN .  Hence any vector V5 can be written in terms of its components as
V5 = Vx Ni + Vy jN + Vz kN .
(3 – 5)
Unit vectors are helpful when adding vectors analytically by components. For
example, Eq. 3 9 4 can be seen to be true by using unit vector notation for each
vector (which we write for the two-dimensional case, with the extension to three
dimensions being straightforward):
V5R = (VRx) Ni + (VRy) jN = V51 + V52 = (V1x Ni + V1y jN) + (V2x Ni + V2y jN)
= (V1x + V2x) Ni + (V1y + V2y) jN.
Comparing the first line to the third line, we get Eqs. 3 9 4.
Figure 3 – 16   Path of a particle in
the xy plane. At time t1 the particle is at point P1 given by the position vector 5r1 ; at t2 the particle is at point P2 given by the position vector 5r2 . The displacement vector for the time interval  t2 - t1  is  ∆5r = 5r2 - 5r1. The actual distance traveled along the path
between P1 and P2 is D l.
y
P1
∆l
∆5r
P2
5r1 5r2
x 0
EXAMPLE 3 – 4 Using unit vectors. Write the vectors of Example 3 9 2 in unit vector notation, and perform the addition.
Approach  We use the components we found in Example 3 9 2,
D1x = 0, D1y = 22.0 km, and D2x = 23.5 km, D2y = - 40.7 km, and we now write them in the form of Eq. 3 9 5.
Solution  We have D5 1 = 0Ni + 22.0 km jN D5 2 = 23.5 km Ni - 40.7 km jN.
Then the resultant displacement is D5 R = D5 1 + D5 2 = (0 + 23.5) km Ni + (22.0 - 40.7) km jN = 23.5 km Ni - 18.7 km jN.
The components of the resultant displacement, D5 R , are  Dx = 23.5 km and  Dy = - 18.7 km.  The magnitude of  D5 R is  DR = 1(23.5 km)2 + (18.7 km)2 = 30.0 km,  just as in Example 3 9 2.
3–6  Vector Kinematics
We can now extend our definitions of velocity and acceleration in a formal way to two- and three-dimensional motion. Suppose a particle follows a path in the xy plane as shown in Fig. 3 9 16. At time t1 , the particle is at point P1 , and at time t2 , it is at point P2 . The vector 5r1 is the position vector of the particle at time t1 (it represents the displacement of the particle from the origin of the coordinate system). And 5r2 is the position vector at time t2 .
In one dimension, we defined displacement as the change in position of the particle. In the more general case of two or three dimensions, the displacement vector is defined as the vector representing change in position. We call it ∆5r,† where
∆5r = 5r2 - 5r1 . This represents the displacement during the time interval  ∆t = t2 - t1 .
†We used D5 for the displacement vector earlier in the Chapter for illustrating vector addition. The new notation here, ∆5r, emphasizes that it is the difference between two position vectors.
62  CHAPTER 3  Kinematics in Two or Three Dimensions; Vectors

GIAN_PSE5_CH03_054-084_ca.indd 62
01/07/20 16:23
In unit vector notation, we can write 5r1 = x1 Ni + y1 jN + z1 kN ,
(3 – 6a)
where x1 , y1 , and z1 are the coordinates of point P1 . Similarly, 5r2 = x2 Ni + y2 jN + z2 kN .
Hence
∆5r = (x2 - x1) Ni + (y2 - y1) jN + (z2 - z1) kN .
(3 – 6b)
If the motion is along the x axis only, then  y2 - y1 = 0,  z2 - z1 = 0,  and the magnitude of the displacement is  ∆r = x2 - x1 ,  which is consistent with our earlier one-dimensional equation (Section 2 9 1). Even in one dimension, displace-
ment is a vector, as are velocity and acceleration.
The average velocity vector over the time interval  ∆t = t2 - t1 is defined as
average velocity
=
∆5r . ∆t
(3 – 7)
Now let us consider shorter and shorter time intervals : that is, we let ∆t approach zero so that the distance between points P2 and P1 also approaches zero, Fig. 3 9 17a. We define the instantaneous velocity vector as the limit of the average velocity as ∆t approaches zero:
v5
=
∆5r
lim
∆t S 0
∆t
=
d5r . dt
(3 – 8)
The direction of v5 at any moment is along the line tangent to the path at that moment (Fig. 3 9 17b).
Note that the magnitude of the average velocity in Fig. 3 9 16 is not equal to the average speed, which is the actual distance traveled along the path,  D l, divided by ∆t. In some special cases, the average speed and average velocity are equal in magnitude (such as motion along a straight line in one direction), but in general they are not. However, in the limit  ∆t S 0, ∆r always approaches  D l, so the instantaneous speed always equals the magnitude of the instantaneous velocity at any time.
The instantaneous velocity (Eq. 3 9 8) is equal to the derivative of the position vector with respect to time. Equation 3 9 8 can be written in terms of components starting with Eq. 3 9 6a as:
v5
=
d 5r dt
=
dx dt
Ni
+
dy dt
jN
+
dz dt
kN
=
vx Ni + vy jN + vz kN ,
(3 – 9)
where  vx = dx>dt,  vy = dy>dt,  vz = dz>dt are the x, y, and z components of the velocity. Note that  dNi>dt = djN>dt = dkN >dt = 0  since these unit vectors are
constant in both magnitude and direction.
Acceleration in two or three dimensions is treated in a similar way. The
average acceleration vector, over a time interval  ∆t = t2 - t1  is defined as
average acceleration
=
∆v5 ∆t
=
v52 - v51 , t2 - t1
(3 – 10)
where ∆v5 is the change in the instantaneous velocity vector during that time interval:  ∆v5 = v52 - v51 .  Note that v52 in many cases, such as in Fig. 3 9 18a, may not be in the same direction as v51 . Hence the average acceleration vector may be in a different direction from either v51 or v52 (Fig. 3 9 18b). Furthermore, v52 and v51 may have the same magnitude but different directions, and the difference of two
such vectors will not be zero. Hence acceleration can result from either a change
in the magnitude of the velocity, or from a change in direction of the velocity, or
from a change in both.
The instantaneous acceleration vector is defined as the limit of the average acceleration vector as the time interval ∆t is allowed to approach zero:
a5
=
∆v5
lim
∆t S 0
∆t
=
dv5 , dt
(3 – 11)
and is thus the derivative of v5 with respect to t.
y ¢5r
P1 P2
5r1
5r2
x 0
(a)
y
P1
v51
5r1
x
0
(b)
Figure 3 – 17   (a) As we take ∆t and ∆5r smaller and smaller [compare to Fig. 3 9 16] we see that the direction of ∆5r and of the instantaneous velocity (∆5r>∆t, where ∆t S 0) is (b) tangent to the curve at P1.
Figure 3 – 18   (a) Velocity vectors v51 and v52 at instants t1 and t2 for a particle at points P1 and P2, as in Fig. 3 9 16. (b) The direction of the average acceleration is in the direction of  ∆v5 = v52 - v51.
y
P1 5v1
P2
5v2 5r1
5r2
x
0
(a)
5v1
5v2
∆5v
(b)
SECTION 3–6  Vector Kinematics  63

GIAN_PSE5_CH03_054-084_ca.indd 63
01/07/20 16:23
We can write a5 using components:
a5
=
dv5 dt
=
dvx dt
Ni
+
dvy dt
jN
+
dvz dt
kN
= ax Ni + ay jN + az kN ,
(3 – 12a)
where  ax = dvx>dt, etc. Because  vx = dx>dt, then  ax = dvx>dt = d2x>dt2, as we saw in Section 2 9 4. Thus we can also write the acceleration as
a5
=
d2x dt 2
Ni
+
d2y dt 2
jN
+
d2z dt 2
kN .
(3 – 12b)
The instantaneous acceleration will be nonzero not only when the magnitude of the velocity changes, but also if its direction changes. For example, a person riding in a car traveling at constant speed around a curve, or a child riding on a merry-goround, will both experience an acceleration because of a change in the direction of the velocity, even though the speed may be constant. (More on this in Chapter 5.)
In general, we will use the terms “velocity” and “acceleration” to mean the instantaneous values. If we want to discuss average values, we will use the word “average.”
EXAMPLE 3 – 5 Position given as a function of time. The position of a particle as a function of time is given by
5r = [(5.0 m>s)t + (6.0 m>s2)t2 ] Ni + [(7.0 m) - (3.0 m>s3)t3 ] jN,
where r is in meters and t is in seconds. (a) What is the particle’s displacement
between  t1 = 2.0 s and  t2 = 3.0 s? (b) Determine the particle’s instantaneous velocity and acceleration as a function of time. (c) Evaluate v5 and a5 at  t = 3.0 s.
Approach  For (a), we find  ∆5r = 5r2 - 5r1 , inserting  t1 = 2.0 s for finding 5r1 , and t2 = 3.0 s for 5r2 . For (b), we take derivatives (Eqs. 3 9 9 and 3 9 12), and for (c) we substitute  t = 3.0 s into our results in (b).
Solution  (a) We insert  t1 = 2.0 s  into the given equation for 5r: 5r1 = [(5.0 m>s)(2.0 s) + (6.0 m>s2)(2.0 s)2 ] Ni + [(7.0 m) - (3.0 m>s3)(2.0 s)3 ] jN
= (34 m) Ni - (17 m) jN.
Similarly, at  t2 = 3.0 s, 5r2 = (15 m + 54 m) Ni + (7.0 m - 81 m) jN = (69 m) Ni - (74 m) jN.
Thus ∆5r = 5r2 - 5r1 = (69 m - 34 m) Ni + ( - 74 m + 17 m) jN = (35 m) Ni - (57 m) jN.
That is,  ∆x = 35 m, and  ∆y = -57 m. (b) To find velocity, we take the derivative of the given 5r with respect to time,
noting (­Appendix B 9 2) that  d(t2)>dt = 2t, and  d(t3)>dt = 3t2:
v5
=
d 5r dt
=
[5.0 m>s + (12 m>s2)t ] Ni + [0 - (9.0 m>s3)t2 ] jN.
The acceleration is (keeping only two significant figures):
a5
=
dv5 dt
=
(12 m>s2) Ni - (18 m>s3)t jN.
Thus  ax = 12 m>s2 is constant; but  ay = - (18 m>s3)t depends linearly on time,
increasing in magnitude with time in the negative y direction.
(c) We substitute  t = 3.0 s into the equations we just derived for v5 and a5:
v5 = (5.0 m>s + 36 m>s) Ni - (81 m>s) jN = (41 m>s) Ni - (81 m>s) jN
a5 = (12 m>s2) Ni - (54 m>s2) jN.
Their magnitudes at  t = 3.0 s are  v = 2(41 m>s)2 + (81 m>s)2 = 91 m>s, and a = 2(12 m>s2)2 + (54 m>s2)2 = 55 m>s2.
64  CHAPTER 3  Kinematics in Two or Three Dimensions; Vectors

GIAN_PSE5_CH03_054-084_ca.indd 64
01/07/20 16:23
Constant Acceleration
In Chapter 2 we studied the important case of one-dimensional motion for
which the acceleration is constant. In two or three dimensions, if the acceleration vector, a5, is constant in magnitude and direction, then  ax = constant,  ay = constant,  az = constant.  The average acceleration in this case is equal to the instantaneous acceleration at any moment. The equations we derived in Chapter 2 for one dimension, Eqs. 2 9 12a, b, and c, apply separately to
each perpendicular component of two- or three-dimensional motion. In two dimensions we let  v50 = vx0 Ni + vy0 jN  be the initial velocity, and we
apply Eqs. 3 9 6a, 3 9 9, and 3 9 12b for the position vector 5r, velocity v5, and ­acceleration a5. We can then write Eqs. 2 9 12a, b, and c for two dimensions as shown in Table 3 9 1.
TABLE 3 – 1  Kinematic Equations for Constant Acceleration in 2 Dimensions
x component (horizontal)
y component (vertical)
vx = vx0 + ax t
x
=
x0
+
vx 0 t
+
1 2
ax
t
2
vx2 = vx20 + 2ax (x - x0)
(Eq. 2 9 12a) (Eq. 2 9 12b) (Eq. 2 9 12c)
vy = vy0 + ay t
y
=
y0
+
vy 0 t
+
1 2
ay
t
2
vy2 = vy20 + 2ay (y - y0)
The first two of the equations in Table 3 9 1 can be written more formally in vector notation.
v5 = v50 + a5 t
5r
=
5r0
+
v50 t
+
1 2
a5
t
2.
[a5 = constant]  (3 – 13a) [a5 = constant]  (3 – 13b)
Here, 5r is the position vector at any time, and 5r0 is the position vector at  t = 0.  These equations are the vector equivalent of Eqs. 2 9 12a and b. In practical situ-
ations, we usually use the component form given in Table 3 9 1.
3–7 Projectile Motion
In Chapter 2, we studied one-dimensional motion of an object in terms of displacement, velocity, and acceleration, including purely vertical motion of a falling object undergoing acceleration due to gravity. Now we examine the more general translational motion of objects moving through the air in two dimensions near the Earth’s surface, such as a golf ball, a thrown or batted baseball, kicked footballs, and speeding bullets. These are all examples of projectile motion (see Fig. 3 9 19), which we can describe as taking place in two dimensions if there is no wind.
Although air resistance is often important, in many cases its effect can be ignored, and we will ignore it in the following analysis. We will not be concerned now with the process by which the object is thrown or projected. We consider only its motion after it has been projected, and before it lands or is caught : that is, we analyze our projected object only when it is moving freely through the air under the action of gravity alone. Then the acceleration of the object is that due to gravity, which acts downward with magnitude  g = 9.80 m>s2,  and we assume it is constant.†
Galileo was the first to describe projectile motion accurately. He showed that it could be understood by analyzing the horizontal and vertical components of the motion separately. For convenience, we assume that the motion begins at time  t = 0  at the origin of an xy coordinate system (so  x0 = y0 = 0).
†This restricts us to objects whose distance traveled and maximum height above the Earth are small ­compared to the Earth’s radius (6400 km).
(a)
(b) Figure 3 – 19   Photographs of (a) a bouncing ball and (b) a thrown basketball, each showing the characteristic “parabolic” path of projectile motion.
SECTION 3–7  Projectile Motion  65

GIAN_PSE5_CH03_054-084_ca.indd 65
01/07/20 16:23
y 5vx0 x
5a = 5g
Figure 3 – 20   Projectile motion of a small ball projected horizontally with initial velocity v5 = v5x0. The dashed black line represents the path of the object. The velocity vector v5 is in the direction of motion at each point, and thus is tangent to the path. The velocity vectors are green arrows, and velocity components are dashed. (A vertically falling object starting from rest at the same place and time is shown at the left for comparison; vy is the same at each instant for the falling object and the projectile.)
5vx
5vy
Projectile 5v motion
Vertical fall
5vx
5vy
5v
Let us look at a (tiny) ball rolling off the end of a horizontal table with an initial
velocity in the horizontal (x) direction, v5x0 . See Fig. 3 9 20, where an object falling vertically is also shown for comparison. The velocity vector v5 at each instant points
in the direction of the ball’s motion at that instant and is thus always tangent to the
path. Like Galileo, we treat the horizontal and vertical components of the velocity
and acceleration separately, and we apply the kinematic equations (Eqs. 2 9 12) to
the x and y components of the motion.
First we examine the vertical ( y ) component of the motion. At the instant the
ball leaves the table’s top  (t = 0), it has only an x component of velocity. Once the
ball leaves the table (at  t = 0), it experiences a vertically downward acceleration g, the acceleration due to gravity. Thus vy is initially zero  (vy0 = 0)  but increases ­continually in the downward direction (until the ball hits the ground). Let us take y
to be positive upward. Then the acceleration due to gravity is in the -y direction, so 
ay = - g. From Eq. 2 9 12a (using y in place of x) we can write  vy = vy0 + ay t = - gt 
since we set  vy0 = 0. The vertical displacement is given by Eq. 2 9 12b written in terms
of y:   y
=
y0
+
vy 0
+
1 2
ay
t
2 . Given
 y0
=
0,  vy0
=
0, and  ay
=
-g, then  y
=
-
1 2
gt
2.
In the horizontal direction, the acceleration is zero (ignoring air resistance).
With  ax = 0, the horizontal component of velocity, vx , remains constant, equal to
its initial value, vx0 , and thus has the same magnitude at each point on the path. The
horizontal displacement (with  ax
=
0) is given by  x
=
vx0 t
+
1 2
ax
t
2
=
vx0 t. 
The two vector components, v5x and v5y , can be added vectorially at any instant
to obtain the velocity v5 at that time (each point on the path), as shown in Fig. 3 9 20.
One result of this analysis, which Galileo himself predicted, is that an object
projected horizontally will reach the ground in the same time as an object dropped
Figure 3 – 21   Multiple-exposure
vertically. The vertical motions are the same in both cases, as shown in Fig. 3 9 20.
photograph showing positions of two Figure 3 9 21 is a multiple-exposure photograph of an experiment that confirms this.
balls at equal time intervals. One
If an object is projected at an upward angle, as in Fig. 3 9 22, the analysis is
ball was dropped from rest at the same time the other was projected horizontally outward. The vertical position of each ball is seen to be the same at each instant.
similar, but now there is an initial vertical component of velocity, vy0 . Because of the ­downward acceleration of gravity, the upward component of velocity vy gradually decreases with time until the object reaches the highest point on its path, at which
point  vy = 0. The object then moves downward (Fig. 3 9 22) and vy increases in the
downward direction (becoming more negative). As before, vx remains constant.
y Figure 3 – 22   Path of a projectile fired
5vy = 0 at this point
with initial velocity v50 at angle u0 to the horizontal. Path is shown dashed in black,
5v 5vy
5v 5vx
the velocity vectors are green arrows, and velocity components are dashed. The
5vy0 5v0
5vx
acceleration  a5 = dv5>dt is downward. That
is,  a5 = g5 = -g jN where jN is the unit vector
u0
5vy
5v
in the positive y direction. Not shown is
where the projectile hits the ground (at that point projectile motion ceases).
0 5vx0
5a = g5 = -gjˆ
5vx
x
66  CHAPTER 3  Kinematics in Two or Three Dimensions; Vectors
5vy
5v

GIAN_PSE5_CH03_054-084_ca.indd 66
01/07/20 16:23
Conceptual EXAMPLE 3 – 6 Where does the apple land?  A child sits upright in a wagon that moves to the right at constant speed, Fig. 3 9 23. The child extends her hand and throws an apple straight up (from her own point of view, Fig. 3 9 23a), as the wagon moves forward at constant speed. Neglecting air resistance, will the apple land (a) behind the wagon, (b) in the wagon, or (c) in front of the wagon?
Response  The child throws the apple straight up in her own reference frame with initial velocity v5y 0 (Fig. 3 9 23a). But when viewed by someone on the ground, the apple also has an initial horizontal component of velocity equal to the velocity of the wagon, v5x0 . Thus, to a person on the ground, the apple follows the path of a projectile as shown in Fig. 3 9 23b. The apple experiences no horizontal accelera­ tion, so v5x0 stays constant, equal to the speed of the wagon. As the apple follows its arc, the wagon will be directly under the apple at all times because they have the same horizontal velocity. When the apple comes down, it will drop right into the outstretched hand of the child. The answer is (b).
Exercise C  Return to the Chapter-Opening Question, page 54, and answer it again now. Try to explain why you may have answered differently the first time. Describe the role of the helicopter in this example of projectile motion.
5vy0
y
x
(a) Wagon reference frame
5vy0 5v0
5vx0 5vx0
(b) Ground reference frame Figure 3 – 23   Example 3 9 6.
3–8  Solving Problems Involving Projectile Motion
We can simplify Eqs. 2 9 12 (Table 3 9 1) for the case of projectile motion because we can set  ax = 0.  See Table 3 9 2, which assumes y is positive upward, so ay = -g = -9.80 m>s2. If u is chosen relative to the +x axis, as in Fig. 3 9 22, then
vx0 = v0 cos u0 ,  and  vy0 = v0 sin u0 .
In doing Problems involving projectile motion, we must consider a time interval for which our chosen object is in the air, influenced only by gravity. We do not consider the throwing (or projecting) process, nor the time after the object lands or is caught, because then other influences act on the object, and we can no longer set  a5 = g5.
P ro b l e m S o l v ing Choice of time interval
TABLE 3 – 2 Kinematic Equations for Projectile Motion
( y positive upward;  ax = 0, ay = −g = −9.80 m, s2)
Horizontal Motion
Vertical Motion†
(ax = 0, vx = constant)
(ay = −g = constant)
vx = vx0 x = x0 + vx0 t
(Eq. 2 9 12a) (Eq. 2 9 12b) (Eq. 2 9 12c)
vy = vy0 - gt
y
=
y0
+
vy 0 t
-
1 2
gt
2
vy2 = vy20 - 2g (y - y0)
†If y is taken positive downward, the minus ( - ) signs in front of g become plus ( + ) signs.
S o l v i n g Projectile Motion
Our approach to solving Problems in Section 2 9 6 also applies here. Solving projectile motion Problems can require creativity, and cannot be done just by follow­ ing rules. You must avoid just plugging numbers into equations that seem to “work.”
1. As always, read carefully; choose the object (or objects) you are going to analyze.
2. Draw a careful diagram showing what is happening.
3. Choose an origin and an xy coordinate system.
4. Decide on the time interval, which for projectile motion can only include motion under the effect of gravity alone, not throwing or landing. The time interval must be the same for the x and y analyses. The x and y motions are connected by the common time, t.
5. Examine the horizontal (x) and vertical (y) motions separately. If you are given the initial velocity, you may want to resolve it into its x and y components.
6. List the known and unknown quantities, choosing  ax = 0 and  ay = - g or  + g, where  g = 9.80 m>s2,  depending on choice of y positive up or down. Recall: vx never changes throughout the trajectory, and  vy = 0 at the highest point of any trajectory that returns downward. The velocity just before landing is generally not zero.
7. Think for a minute before jumping into the equations. Apply the relevant equations (Table 3 9 2), combining them if necessary.You may need to combine components of a vector to get magnitude and direction (Eqs. 3 9 3).
SECTION 3–8  Solving Problems Involving Projectile Motion  67

Problem
GIAN_PSE5_CH03_054-084_ca.indd 67
01/07/20 16:23
+ y + x
5a = 5g
50.0 m
y = -50.0 m 90.0 m Figure 3 – 24   Example 3 9 7.
Known
Unknown
x0 = y0 = 0
vx 0
x = 90.0 m
t
y = -50.0 m
ax = 0
ay = - g = - 9.80 m>s2
vy 0 = 0
EXAMPLE 3 – 7 Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the clifftop to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.
Approach  We explicitly follow the steps of the Problem Solving Strategy on the previous page.
Solution
1. and 2. Read, choose the object, and draw a diagram. Our object is the motorcycle and driver, taken as a single unit. The diagram is shown in Fig. 3 9 24.
3. Choose a coordinate system. We choose the y direction to be positive
upward, with the top of the cliff as  y0 = 0. The x direction is horizontal with  x0 = 0  at the point where the motorcycle leaves the cliff. 4. Choose a time interval. We choose our time interval to begin  (t = 0) just as the motorcycle leaves the clifftop at position  x0 = 0,  y0 = 0. Our time interval ends just before the motorcycle touches the ground below.
5. Examine x and y motions. In the horizontal (x) direction, the acceleration 
ax = 0, so the velocity is constant. The value of x when the motorcycle reaches the ground is  x = +90.0 m. In the vertical direction, the acceleration is the acceleration due to gravity,  ay = - g = - 9.80 m>s2. The value of y when the motorcycle reaches the ground is  y = -50.0 m. The initial velocity is horizontal and is our unknown, vx0 ; the initial vertical velocity is zero,  vy0 = 0. 6. List knowns and unknowns. See the Table in the margin. Note that in
addition to not knowing the initial horizontal velocity vx0 (which stays constant until landing), we also do not know the time t when the motor-
cycle reaches the ground.
7. Apply relevant equations. The motorcycle maintains constant vx as long as it is in the air. The time it stays in the air is determined by the y motion : when it reaches the ground. So we first find the time using the y motion, and then use this time value in the x equations. To find out how long it takes the motorcycle to reach the ground below, we use Eq. 2 9 12b (Table 3 9 2) for the vertical (y) direction with  y0 = 0  and  vy0 = 0:
y = y0
+
vy0 t
+
1 2
ay
t
2
= 0
+
0
+
1 2
(
-
g
)
t
2
or
y
=
-
1 2
gt
2.
We solve for t and set  y = -50.0 m:
t
=
2y A -g
=
2( -50.0 m) B - 9.80 m>s2
=
3.19 s.
To calculate the needed initial velocity, vx0 , we again use Eq. 2 9 12b, but this time for the ­horizontal (x) direction, with  ax = 0  and  x0 = 0:
x
=
x0
+
vx0 t
+
1 2
ax
t
2
= 0 + vx0 t + 0
or
x = vx0 t.
So the motorcycle needs to leave the clifftop with a speed
vx 0
=
x t
=
90.0 m 3.19 s
=
28.2 m>s,
which is about 100 km>h (roughly 60 mi>h).
Note  In the time interval of the projectile motion, the only acceleration is g in the negative y direction. The acceleration in the x direction is zero.
68  CHAPTER 3  Kinematics in Two or Three Dimensions; Vectors

GIAN_PSE5_CH03_054-084_ca.indd 68
01/07/20 16:23
y
5v0
5vy 0 37.0°
0
5vx 0
5vy = 0 at this point
5v
5v
5a = 5g
5v
x Figure 3 – 25   Example 3 9 8.
EXAMPLE 3 – 8 A kicked football.  A football is kicked at an angle  u0 = 37.0°  with a velocity of 20.0 m>s, as shown in Fig. 3 9 25. Calculate (a) the maximum height, (b) the time of travel before the football hits the ground, and (c) how far away it hits the ground. Assume the ball leaves the foot at ground level, and ignore air resistance, wind, and rotation of the ball.
Approach  This may seem difficult at first because there are so many ­questions. But we can deal with them one at a time. We take the y direction as positive upward, and treat the x and y motions separately. The total time in the air is again determined by the y motion. The x motion occurs at constant velocity. The y component of velocity varies, being positive (upward) initially, decreasing to zero at the highest point, and then becoming negative as the football falls.
Solution  We resolve the initial velocity into its components (Fig. 3 9 25):
vx0 = v0 cos 37.0° = (20.0 m>s) (0.799) = 16.0 m>s vy0 = v0 sin 37.0° = (20.0 m>s) (0.602) = 12.0 m>s.
(a) To find the maximum height, we consider a time interval that begins just after
the football loses contact with the foot until the ball reaches its maximum height.
During this time interval, the acceleration is g downward. At the maximum
height, the velocity is horizontal (Fig. 3 9 25), so  vy = 0;  and this occurs at a
time given by  vy = vy0 - gt  with  vy = 0  (see Eq. 2 9 12a in Table 3 9 2). Thus
t
=
vy 0 g
=
(12.0 m>s)
(9.80 m>s2)
=
1.224 s
L
1.22 s.
From Eq. 2 9 12b, with  y0 = 0,  we can solve for y at this time  t = vy0>g :
y
=
vy0 t
-
1 2
gt
2
=
vy2 0 1 vy2 0 vy2 0 g - 2 g = 2g
=
(12.0 m>s)2
2 (9.80 m>s2)
=
7.35 m.
The maximum height is 7.35 m. [Solving Eq. 2 9 12c for y gives the same result.]
(b) To find the time it takes for the ball to return to the ground, we consider
a different time interval, starting at the moment the ball leaves the foot
(t = 0,  y0 = 0)  and ending just before the ball touches the ground  (y = 0 
again). We can use Eq. 2 9 12b with  y0 = 0 and also set  y = 0 (ground level):
y
=
y0
+
vy0 t
-
1 2
gt
2
0
=
0
+
vy0 t
-
1 2
gt
2.
This equation can be easily factored:
t
(
1 2
gt
-
vy 0 )
=
0.
There are two solutions,  t = 0 (which corresponds to the initial point, y0), and
t
=
2vy 0 g
=
2(12.0 m>s)
(9.80 m>s2)
=
2.45 s,
which is the total travel time of the football. (c) The total distance traveled in the x direction is found by applying Eq. 2 9 12b with x0 = 0, ax = 0, vx0 = 16.0 m>s,  and  t = 2.45 s:
x = vx0 t = (16.0 m>s) (2.45 s) = 39.2 m. Note  In (b) the time needed for the whole trip,  t = 2vy0 >g = 2.45 s, is double the time to reach the highest point, calculated in (a). That is, the time to go up
equals the time to come back down to the same level (ignoring air resistance).
P h y sics A pp l ie d Sports
SECTION 3–8  Solving Problems Involving Projectile Motion  69

GIAN_PSE5_CH03_054-084_ca.indd 69
01/07/20 16:23
Exercise D  In Example 3 9 8, what is (a) the velocity vector at the maximum height, and (b) the acceleration vector at maximum height?
In Example 3 9 8 we treated the football as if it were a particle, ignoring its rotation. We also ignored air resistance. Because air resistance is significant on a football, our results are only estimates (mainly overestimates). We also ignored any wind. If there happened to be a cross-wind, we would need a third axis for such 3-dimensional motion.
EXERCISE E  Two balls are thrown in the air at different angles, but each reaches the same height. Which ball remains in the air longer: the one thrown at the steeper angle or the one thrown at a shallower angle?
d v0 Figure 3 – 26   Example 3 9 9.
Conceptual EXAMPLE 3 – 9 The wrong strategy.  A boy on a small hill aims his water-balloon slingshot horizontally, straight at a second boy hanging from a tree branch a distance d away, Fig. 3 9 26. At the instant the water balloon is released, the second boy lets go and falls from the tree, hoping to avoid being hit. Show that 0 he made the wrong move. (He hadn’t studied physics yet.) Ignore air resistance.
y Response  Both the water balloon and the boy in the tree start falling at the
same
instant, and
in
a
time
t
they
each
fall
the
same
vertical
distance
 y
=
1 2
gt
2, 
much like Fig. 3 9 21. In the time it takes the water balloon to travel the hori-
zontal distance d, the balloon will have the same y position as the falling boy.
Splat. If the boy had stayed in the tree, he would have avoided the h­ umiliation.
Figure 3 – 27   (a) The range R of
a projectile. (b) There are generally
two angles u0 that will give the same range. If one angle is u01, the other is  u0 2 = 90° - u0 1 . Also see Example 3 9 10.
y x0 = 0 y0 = 0
y = 0 again here (where x = R)
u0
x
R
(a)
y 60°
30° (b)
45° x
Level Horizontal Range
The total distance the football traveled in Example 3 9 8 is called the horizontal
range R. We now derive a formula for the range, which applies to a projectile that
lands at the same level it started  (= y0) :  that is,  y(final) = y0  (see Fig. 3 9 27a). Looking back at Example 3 9 8 part (c), we see that  x = R = vx0 t  where (from part b) t = 2vy0>g. Thus
R
=
vx0 t
=
vx
0
¢
2vy g
0
≤
=
2vx0 vy0 g
=
2v02 sin u0 cos u0 , g
[y = y0]
where  vx0 = v0 cos u0 and  vy0 = v0 sin u0 . This can be rewritten, using the trigo­ nometric identity  2 sin u cos u = sin 2u (Appendix A or inside the rear cover):
R
=
v02
sin g
2u0
.
[only if  y(final) = y0]
Note that the maximum range, for a given initial velocity v0 , is obtained when sin 2u takes on its maximum value of 1.0, which occurs for  2u0 = 90°;  so
u0 = 45° for maximum range, and Rmax = v02>g.
The maximum range increases by the square of v0 , so doubling the initial velocity of a projectile increases its maximum range by a factor of 4.
When air resistance is important, the range is less for a given v0 , and the maximum range is obtained at an angle smaller than 45°.
EXAMPLE 3 – 10 Range of a cannon ball. Suppose one of Napoleon’s cannons had a muzzle speed, v0 , of 60.0 m>s. At what angle should it have been aimed (ignore air resistance) to strike a target 320 m away?
Approach  We use the equation just derived for the range,  R = v02 sin 2u0>g,  with R = 320 m.
Solution  We solve for sin 2u0 in the range formula:
Rg (320 m) (9.80 m>s2)
sin 2u0 = v02 =
(60.0 m>s)2
= 0.871.
We want to solve for an angle u0 that is between 0° and 90°, which means 2u0 in this equation can be as large as 180°. Then  sin-1(0.871) = 2u0 = 60.6° is a solution,
70  CHAPTER 3  Kinematics in Two or Three Dimensions; Vectors

GIAN_PSE5_CH03_054-084_ca.indd 70
01/07/20 16:23
so the cannon should be aimed at  u0 = 30.3°. But 2u0 = 180° - 60.6° = 119.4°  is also a solution (see Appendix A 9 9), so u0 can also be  u0 = 59.7°. In general we have two solutions (see Fig. 3 9 27b), which in the present case are given by
u0 = 30.3° or 59.7°.
Either angle gives the same range. Only when  sin 2u0 = 1 (so  u0 = 45°) is there a single solution (that is, both solutions are the same).
The level range formula applies only if takeoff and landing are at the same
height  (y = y0). Example 3 9 11 below considers a case where they are not equal heights  (y ≠ y0).
EXAMPLE 3 – 11 A punt.  Suppose the football in Example 3 9 8 was punted and left the punter’s foot at a height of 1.00 m above the ground. How far did the football travel before hitting the ground? Set  x0 = 0,  y0 = 0.
Approach  The x and y motions are again treated separately. But we cannot use the range formula as in Example 3 9 10 because the range formula is valid only if  y(final) = y0 , which is not the case here. Now we have  y0 = 0, and the football hits the ground where y = -1.00 m (see Fig. 3 9 28). We choose our time interval
to start when the ball leaves his foot  (t = 0,  y0 = 0,  x0 = 0) and end just before the ball hits the ground  (y = -1.00 m). We can get x from Eq. 2 9 12b,  x = vx0 t,  and we saw that  vx0 = 16.0 m>s  in Example 3 9 8. But first we must find t,
the time at which the ball hits the ground, which we obtain from the y motion.
P h y sics A pp l ie d Sports
P ro b l e m S o l v ing Do not use any formula unless you are sure its range of validity fits the problem; the range formula does not apply here because  y ≠ y0
y
y0 = 0 y = -1.00 m
x
Figure 3 – 28   Example 3 9 11: the football leaves the punter’s foot
at  y = 0, and reaches the ground
where  y = - 1.00 m. Ground
Solution  To find t  with  y = -1.00 m and  vy0 = 12.0 m>s (see Example 3 9 8), we use the ­equation
y
=
y0
+
vy0 t
-
1 2
gt
2,
and obtain
-1.00 m = 0 + (12.0 m>s)t - (4.90 m>s2)t2.
We rearrange this equation into standard form  (ax2 + bx + c = 0)  so we
can use the ­quadratic formula:
(4.90 m>s2)t2 - (12.0 m>s)t - (1.00 m) = 0.
The quadratic formula (Appendix A 9 1) gives
t
=
12.0 m>s { 2( - 12.0 m>s)2 - 4(4.90 m>s2) ( - 1.00 m) 2(4.90 m>s2)
= 2.53 s or -0.081 s.
The second solution would correspond to a time prior to our chosen time
interval that begins at the kick  (t = 0),  so it doesn’t apply. With  t = 2.53 s  for the time at which the ball touches the ground, the horizontal distance the ball traveled is (using  vx0 = 16.0 m>s  from E­ xample 3 9 8):
x = vx0 t = (16.0 m>s) (2.53 s) = 40.5 m.
Our assumption in Example 3 9 8 that the ball leaves the foot at ground level gives a result (39.2 m) that is an underestimate of about 1.3 m in the distance our punt traveled. (But Example 3 9 8 would apply for a kickoff or field goal in American football.)
SECTION 3–8  Solving Problems Involving Projectile Motion  71

GIAN_PSE5_CH03_054-084_ca.indd 71
01/07/20 16:23
y vx 0
200 m
“Dropped” (vy0 = 0)
x
Thrown upward? (vy0 7 0)
200 m
Thrown downward? (vy0 6 0)
400 m
(a)
(b)
Figure 3 – 29   Example 3 9 12.
P h y sics A pp l ie d
Reaching a target from a moving helicopter
EXAMPLE 3 – 12 Rescue helicopter drops supplies. A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m>s (250 km>h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped (Fig. 3 9 29a)? (b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position (Fig. 3 9 29b)? (c) With what speed does the package land in the latter case?
Approach  We choose the origin of our xy coordinate system at the initial position of the helicopter, taking +y upward, and use the kinematic equations (Table 3 9 2).
Solution  (a) We can find the time to reach the climbers using the vertical
distance of 200 m. The package is “dropped” so initially it has the velocity of
the helicopter, vx0
=
70 m>s,
vy 0
=
0. Then, since  y
=
-
1 2
gt 2,  we
have
- 2y
-2( -200 m)
t = B g = B 9.80 m>s2 = 6.39 s.
The horizontal motion of the falling package is at constant speed of 70 m>s. So
x = vx0 t = (70 m>s) (6.39 s) = 447 m L 450 m,
assuming the given numbers were good to two significant figures.
(b) We are given  x = 400 m, vx0 = 70 m>s, y = - 200 m, and we want to find vy0 (see Fig. 3 9 29b). Like most problems, this one can be approached in various
ways. Instead of searching for a formula or two, let’s try to reason it out in a simple way, based on what we did in part (a). If we know t, perhaps we can get
vy0 . Since the horizontal motion of the package is at constant speed (once it is released we don’t care what the helicopter does), we have  x = vx0 t,  so
t
=
x vx 0
=
400 m 70 m>s
=
5.71 s.
Now let’s try to use the vertical motion to get  vy0 :
y
=
y0
+
vy0 t
-
1 2
gt
2. Since 
y0 = 0  and  y = - 200 m,  we can solve for vy0 :
vy 0
=
y
+
1 2
gt
2
t
=
- 200 m
+
1 2
(
9.80
m>s2
)
(
5.71
s
) 2
5.71 s
=
- 7.0 m>s.
Thus, in order to arrive at precisely the mountain climbers’ position, the package must be thrown downward from the helicopter with a speed of 7.0 m>s.
(c) We want to know v of the package at  t = 5.71 s. The components are:
vx = vx0 = 70 m>s
vy = vy0 - gt = - 7.0 m>s - (9.80 m>s2) (5.71 s) = - 63 m>s.
So  v = 2(70 m>s)2 + ( - 63 m>s)2 = 94 m>s. (Better not to release the package from such an altitude, or use a parachute.)
72  CHAPTER 3  Kinematics in Two or Three Dimensions; Vectors

GIAN_PSE5_CH03_054-084_ca.indd 72
01/07/20 16:23
Projectile Motion Is Parabolic
We now show that the path followed by any projectile is a parabola, if we can ignore air resistance and can assume that only gravity is acting with g5 constant. To do so, we need to find y as a function of x by eliminating t between the two equations for horizontal and vertical motion (Eq. 2 9 12b in Table 3 9 2), and for simplicity we set  x0 = y0 = 0 :
x = vx0 t
y
=
vy0 t
-
1 2
gt
2.
(a)
From the first equation, we have  t = x>vx0 , and we substitute this into the second
equation to obtain
y
=
¢ vy0 ≤ x vx 0
-
¢
g 2vx2
0
≤
x2.
We see that y as a function of x has the form
(3 – 14)
y = Ax - Bx2,
where A and B are constants for any specific projectile motion. This is the wellknown equation for a parabola. See Figs. 3 9 19 and 3 9 30.
The idea that projectile motion is parabolic was at the forefront of physics research in Galileo’s day. Today we discuss it in Chapter 3 of introductory physics!
3–9 Relative Velocity
We now consider how observations made in different frames of reference are
related to each other. For example, consider two trains approaching one another, each with a speed of 80 km>h with respect to the Earth. Observers on the Earth beside the train tracks will measure 80 km>h for the speed of each of the trains.
Observers on either one of the trains (different frames of reference) will measure a speed of 160 km>h for the other train approaching them.
Similarly, when one car traveling 90 km>h passes a second car traveling in the same direction at 75 km>h, the first car has a speed relative to the second car of 90 km>h - 75 km>h = 15 km>h.
When the velocities are along the same line, simple addition or subtraction
is sufficient to obtain the relative velocity. But if they are not along the same
line, we must make use of vector addition. We emphasize, as mentioned in
Section 2 9 1, that when specifying a velocity, it is important to specify what the
reference frame is.
When determining relative velocity, it is easy to make a mistake by adding
or subtracting the wrong velocities. It is important, therefore, to draw a diagram
and use a careful labeling process. Each velocity is labeled by two subscripts: the first refers to the object, the second to the reference frame in which it has this velocity. For example, suppose a boat heads directly across a river, as shown in Fig. 3 9 31. We let v5BW be the velocity of the Boat with respect to the Water. (This is also what the boat’s velocity would be relative to the shore if the water were still.) Similarly, v5BS is the velocity of the Boat with respect to the Shore, and v5WS is the velocity of the Water with respect to the Shore (this is the river ­current). Note that v5BW is what the boat’s motor produces (against the water), whereas v5BS is equal to v5BW plus the effect of the current, v5WS. Therefore, the v­ elocity of the boat relative to the shore is (see vector diagram, Fig. 3 9 31)
v5BS = v5BW + v5WS .
(3 – 15)
(b)
Figure 3 – 30   Examples of projectile motion: (a) a boy leaping, (b) glowing lava from the volcano Stromboli.
Figure 3 – 31   A boat heads north directly across a river which flows west. Velocity vectors are shown as green arrows:    v5BS = velocity of Boat with
respect to the Shore,
  v5BW = velocity of Boat with respect to the Water,
   v5WS = velocity of Water with respect to the Shore (river current).
As it crosses the river, the boat is dragged downstream by the
current (v5WS).
River current
5vWS
N
WE
S
5vBS
5vBW u
By writing the subscripts using this convention, we see that the inner subscripts (the two W’s) on the right-hand side of Eq. 3 9 15 are the same; also, the outer subscripts on the right of Eq. 3 9 15 (the B and the S) are the same as the two s­ubscripts for the sum vector on the left, v5BS.
SECTION 3–9  Relative Velocity  73

GIAN_PSE5_CH03_054-084_ca.indd 73
01/07/20 16:23
Figure 3 – 32   Derivation of relative velocity equation (Eq. 3 9 15), in this case for a person walking along the corridor in a train. We are looking down on the train and two reference frames are shown: xy on the Earth and x′y′ fixed on the train. We have:
5rPT = position vector of person (P) relative to train (T),
5rPE = position vector of person (P) relative to Earth (E),
5rTE = position vector of train’s coordinate system (T) relative to Earth (E).
y
From the diagram we see that
5rPE = 5rPT + 5rTE .
We take the derivative with respect to time to obtain
d dt
( 5r PE )
=
d dt
( 5r PT )
+
d dt
( 5r TE ) ,
0
or, since  d5r>dt = v5,
v5PE = v5PT + v5TE . This is the equivalent of Eq. 3 9 15 for the present situation (check the subscripts!).
5vTE
5rPE x 5rTE
y¿ 5rPT x¿
0¿
River current
N
W
E
5vWS
S
5vBS u 5vBW
Figure 3 – 33   Example 3 9 13. A boat, in order to go directly across a moving river current, must head upstream.
By following this convention (first subscript for the object, second for the reference
frame), you can write down the correct equation relating velocities in different reference frames.†
Figure 3 9 32 gives a derivation of Eq. 3 9 15,  v5BS = v5BW + v5WS , but using different subscripts.
Equation 3 9 15 is valid in general and can be extended to three or more velocities. For example, if a fisherman on a boat walks with a velocity v5FB rela­tive to the boat, his velocity relative to the shore is  v5FS = v5FB + v5BW + v5WS . The equations involving relative velocity will be correct when adjacent inner subscripts are identical and when
the outermost ones correspond exactly to the two on the velocity on the left of the
equation. But this works only with plus signs (on the right), not minus signs.
It is often useful to remember that for any two objects or reference frames,
A and B, the velocity of A relative to B has the same magnitude, but opposite
direction, as the velocity of B relative to A:
v5BA = - v5AB .
(3 – 16)
For example, if a train is traveling 100 km>h relative to the Earth in a certain
direction, objects on the Earth (such as trees) appear to an observer on the train to be traveling 100 km>h in the opposite direction.
EXAMPLE 3 – 13 Heading upstream. A boat’s speed in still water is vBW = 1.85 m>s. If the boat is to travel north directly across a river whose westward current has speed vWS = 1.20 m>s,  at what upstream angle must the boat head? See Fig. 3 9 33.
Approach  If the boat heads straight across the river, the current will drag the boat downstream (westward). To overcome the river’s current, the boat must have an upstream (eastward) component of velocity as well as a cross-stream (northward) component. Figure 3 9 33 has been drawn with v5BS, the velocity of the Boat relative to the Shore, pointing directly across the river because this is where the boat is supposed to go. Note that  v5BS = v5BW + v5WS .
Solution  Vector v5BW points upstream at angle u as shown. From the diagram,
sin u
=
vWS vBW
=
1.20 m>s 1.85 m>s
=
0.6486.
Thus  u = 40.4°, so the boat must head upstream at a 40.4° angle.
†We thus can see, for example, that the equation  v5BW = v5BS + v5WS  is wrong: the inner subscripts are not the same, and the outer ones on the right do not correspond to the subscripts on the left.
74  CHAPTER 3  Kinematics in Two or Three Dimensions; Vectors 
GIAN_PSE5_CH03_054-084_ca.indd 74
01/07/20 16:23
EXAMPLE 3 – 14 Heading across the river.  The same boat  (vBW = 1.85 m>s) 
now heads directly across the river whose current is still 1.20 m>s. (a) What is the velocity (magnitude and direction) of the boat relative to the shore? (b) If the river is 110 m wide, how long will it take to cross and how far downstream will the boat be then?
Approach  The boat now heads directly across the river and is pulled downstream by the current, as shown in Fig. 3 9 34. The boat’s velocity with respect to the shore, v5BS, is the sum of the boat’s velocity with respect to the water, v5BW , plus the velocity of the water with respect to the shore, v5WS (= the river’s current). Just as before,
v5BS = v5BW + v5WS .
River current 5vWS
5vBS
5vBW u
Solution  (a) Since v5BW is perpendicular to v5WS , we can get vBS using the theorem of ­Pythagoras:
vBS
=
2v
2 BW
+
v
2 WS
=
2(1.85 m>s)2 + (1.20 m>s)2
=
2.21 m>s.
We can obtain the angle (note how u is defined in Fig. 3 9 34) from:
tan u = vWS>vBW = (1.20 m>s)>(1.85 m>s) = 0.6486.
A calculator with a key such as inv tan  or  arc tan  or  tan−1 gives u = tan-1(0.6486) = 33.0°. Note that this angle is not equal to the angle calculated in Example 3 9 13.
(b) To find the travel time for the boat to cross the river, recall the river’s width  D = 110 m, and use the velocity component that points directly across the river,  vBW = D>t.  Solving for t, we get  t = 110 m>(1.85 m>s) = 59.5 s.  Also, the boat will have been ­carried downstream, in this time, a distance
d = vWS t = (1.20 m>s) (59.5 s) = 71.4 m L 71 m.
Figure 3 – 34   Example 3 9 14. A boat heading directly across a river whose current moves at 1.20 m>s.
Figure 3 – 35   Example 3 9 15. 2 v52 E
Note  There is no acceleration in this Example, so the motion involves only
constant velocities (of the boat or of the river).
5v1 E
EXAMPLE 3 – 15 Car velocities at 90°. Two cars approach a street corner at right angles to each other with the same speed of  40.0 km>h ( = 11.11 m>s)  relative to the ground, as shown in Fig. 3 9 35a. What is the relative velocity of car 1 as seen by car 2?
Approach  Figure 3 9 35a shows the situation in a reference frame fixed to the Earth. But we want to view the situation from a reference frame in which car 2 is at rest, and this is shown in Fig. 3 9 35b. In this reference frame (the world as seen by the driver of car 2), the Earth moves toward car 2 with velocity v5E2 (speed of 40.0 km>h), which is of course equal and opposite to v52E, the velocity of car 2 with respect to the Earth (Eq. 3 9 16):
v52E = - v5E2 .
Then the velocity of car 1 as seen by car 2 is (see Eq. 3 9 15)
v512 = v51E + v5E2 .
Solution  Because  v5E2 = - v52E ,  then
v512 = v51E - v52E .
That is, the velocity of car 1 as seen by car 2 is the difference of their velocities,  v51E - v52E, both measured relative to the Earth; see Fig. 3 9 35c. Since the magnitudes of v51E , v52E , and v5E2 are equal  (40.0 km>h = 11.11 m>s),  we see (Fig. 3 9 35b) that v512 points at a 45° angle toward car 2; the speed is
v12 = 2(11.11 m>s)2 + (11.11 m>s)2 = 15.7 m>s ( = 56.6 km>h).
1
(a)
2 At rest
5vE 2
v5E 2
v51 E
v51 2
1
(b) -v52 E
5v1 2 =
v51 E
5v1 E - v52 E
(c)
SECTION 3–9  Relative Velocity  75

GIAN_PSE5_CH03_054-084_ca.indd 75
01/07/20 16:23
Summary
A quantity such as velocity, that has both a magnitude and a direction, is called a vector. A quantity such as mass, that has only a magnitude, is called a scalar. On diagrams, vectors are represented by arrows.
Addition of vectors can be done graphically by placing the tail of each successive arrow at the tip of the previous one. The sum, or resultant vector, is the arrow drawn from the tail of the first vector to the tip of the last vector. Two vectors can also be added using the parallelogram method.
Vectors can be added more accurately by adding their components along chosen axes with the aid of trigonometric functions. A vector of magnitude V making an angle u with the + x axis has components
Vx = V cos u, Vy = V sin u.
(3 – 2)
Given the components, we can find a vector’s magnitude and direction from
V = 2Vx2 + Vy2 ,
tan u
=
Vy . Vx
(3 – 3)
It is often helpful to express a vector in terms of its components along chosen axes using unit vectors, which are vectors of unit length along the chosen coordinate axes; for Cartesian coordinates the unit vectors along the x, y, and z axes are called Ni, jN, and kN .
The general definitions for the instantaneous velocity, v5, and
acceleration, a5, of a particle (in one, two, or three dimensions) are
v5
=
d 5r dt
(3 – 8)
a5
=
dv5 , dt
(3 – 11)
where 5r is the position vector of the particle. The kinematic equa-
tions for motion with constant acceleration can be written for
each of the x, y, and z components of the motion and have the
same form as for one-dimensional motion (Eqs. 2 9 12). Or they
can be written in the more general vector form:
v5 = v50 + a5 t
5r
=
5r0
+
v50 t
+
1 2
a5
t
2.
(3 – 13)
Projectile motion is the motion of an object in the air near
the Earth’s surface under the effect of gravity alone. It can be
analyzed as two separate motions if air resistance can be ignored.
The horizontal component of motion is at constant velocity,
whereas the vertical component is at constant acceleration, g5,
just as for an object falling vertically under the action of gravity.
The velocity of an object relative to one frame of refer-
ence can be found by vector addition if its velocity relative to a
second frame of reference, and the relative velocity of the two
reference frames, are known.
Questions
1. One car travels due east at 40 km>h, and a second car travels north at 40 km>h. Are their velocities equal? Explain.
2. Can you conclude that a car is not accelerating if its speed­ ometer indicates a steady 60 km>h? Explain.
3. Give several examples of an object’s motion in which a great distance is traveled but the displacement is zero.
4. Can the displacement vector for a particle moving in two dimensions be longer than the length of path traveled by the particle over the same time interval? Can it be less? Discuss.
5. During baseball practice, a player hits a very high fly ball and then runs in a straight line and catches it. Which had the greater displacement, the player or the ball? Explain.
6. If  V5 = V51 + V52,  is V necessarily greater than V1 and>or V2 ? Discuss.
7. Two vectors have lengths  V1 = 4.5 km  and  V2 = 5.0 km.  What are the maximum and minimum magnitudes of their vector sum?
8. Can two vectors, of unequal magnitude, add up to give the zero vector? Can three unequal vectors? Under what ­conditions?
9. Can the magnitude of a vector ever (a) equal, or (b) be less than, one of its components?
10. Does the odometer of a car measure a scalar or a vector quantity? What about the speedometer?
11. How could you determine the speed a slingshot imparts to a rock, using only a meter stick, a rock, and the slingshot?
12. In archery, should the arrow be aimed directly at the target? How should your angle of aim depend on the distance to the target?
13. Where in Fig. 3 9 22 is (a) v5 = 0, (b) vy = 0, (c) vx = 0? 14. It was reported in World War I that a pilot flying at an alti-
tude of 2 km caught in his bare hands a bullet fired at the plane! Using the fact that a bullet slows down considerably due to air resistance, explain how this incident occurred.
15. You are on the street trying to hit a friend in his dorm window with a water balloon. He has a similar idea and is aiming at you with his water balloon. You aim straight at each other and throw at the same instant. Do the water balloons hit each other? Explain why or why not.
16. A projectile is launched at an upward angle of 30° to the horizontal with a speed of 30 m>s. How does the horizon­tal component of its velocity 1.0 s after launch compare with its horizontal component of velocity 2.0 s after launch, ignoring air resistance? Explain.
17. A projectile has the least speed at what point in its path?
18. Two cannonballs, A and B, are fired from the ground with identical initial speeds, but with uA larger than uB. (a) Which cannonball reaches a higher elevation? (b) Which stays longer in the air? (c) Which travels farther? Explain.
19. A person sitting in an enclosed train car, moving at constant velocity, throws a ball straight up into the air in her reference frame. (a) Where does the ball land? What is your answer if the car (b) accelerates, (c) decelerates, (d) rounds a curve, (e) moves with constant velocity but is open to the air?
20. If you are riding on a train that speeds past another train moving in the same direction on an adjacent track, it appears that the other train is moving backward. Why?
21. Two rowers, who can row at the same speed in still water, set off across a river at the same time. One heads straight across and is pulled downstream somewhat by the cur­rent. The other one heads upstream at an angle so as to arrive at a point opposite the starting point. Which rower reaches the opposite side first? Explain.
22. If you stand motionless under an umbrella in a rainstorm where the drops fall vertically, you remain relatively dry. However, if you start running, the rain begins to hit your legs even if they remain under the umbrella. Why?
76  CHAPTER 3  Kinematics in Two or Three Dimensions; Vectors

GIAN_PSE5_CH03_054-084_ca.indd 76
01/07/20 16:23
MisConceptual Questions
1. You are adding vectors of length 20 and 40 units. Which of the following choices is a possible resultant magnitude? (a) 0.  (b) 18.  (c) 37.  (d) 64.  (e) 100.
2. The magnitude of a component of a vector must be (a) less than or equal to the magnitude of the vector. (b) equal to the magnitude of the vector. (c) greater than or equal to the magnitude of the vector. (d) less than, equal to, or greater than the magnitude of the vector.
3. You are in the middle of a large field. You walk in a straight line for 100 m, then turn left and walk 100 m more in a straight line before stopping. When you stop, you are 100 m from your starting point. By how many degrees did you turn?
(a) 90°.  (b) 120°.  (c) 30°.  (d) 180°.
(e) This is impossible. You cannot walk 200 m and be only 100 m away from where you started.
4. Which of the following equations correctly expresses the
relation between vectors A5 , B5, and C5, shown in Fig. 3 9 36?
(a) A5 = B5 + C5.
(b) B5 = A5 + C5. (c) C5 = A5 + B5.
A5
(d) A5 + B5 + C5 = 0.
C5
Figure 3 – 36  
MisConceptual Question 4.
B5
5. A car is driven at a constant speed of 10.0 m>s around a
circle
of
radius
20.0
m.
As
it
goes
1 4
of
the
way
around,
(a) the magnitude of the average velocity is 0.
(b) the magnitude of the average velocity is 10 m>s.
(c) the magnitude of the average velocity is between 0 and
10 m>s.
(d) the magnitude of the average velocity is greater than
10 m>s.
6. A bullet fired horizontally from a rifle begins to fall (a) as soon as it leaves the barrel. (b) after air friction reducPes its speed. (c) not at all if air resistance is ignored.
7. A baseball player hits a ball that soars high into the air.
After the ball has left the bat, and while it is traveling
upward (at point P in Fig. 3 9 37), what is the direction of accelera-
P
tion? Ignore air resistance.
(a)
(b)
(c)
Figure 3 – 37  
MisConceptual
(a) Ques(tbio)n 7.
(c)
8. One ball is dropped vertically from a window. At the same
instant, a second ball is thrown horizontally from the same
window. Which ball has the greater speed at ground level? (a) The dropped ball. (b) The thrown ball. (c) Neither : they both have the same speed on impact. (d) It depends on how hard the ball was thrown.
9. Two balls having different speeds roll off the edge of a hori-
zontal table at the same time. Which hits the floor sooner? (a) The faster one.
(b) The slower one. (c) Both the same.
10. You are riding in an enclosed train car moving at 90 km>h. If you throw a baseball straight up, where will the baseball land? (a) In front of you. (b) Behind you. (c) In your hand. (d) Can’t decide from the given information.
11. Which of the three kicks in Fig. 3 9 38 is in the air for the longest time? They all reach the same maximum height h. Ignore air resistance. (a), (b), (c), or (d) all the same time.
h
(a)
(b)
(c)
Figure 3 – 38   MisConceptual Question 11.
12. A baseball is hit high and far. Which of the following ­statements is true? At the highest point, (a) the magnitude of the acceleration is zero. (b) the magnitude of the velocity is zero. (c) the magnitude of the velocity is the slowest. (d) more than one of the above is true. (e) none of the above are true.
13. A hunter is aiming horizontally at a monkey who is sitting in a tree. The monkey is so terrified when it sees the gun that it falls off the tree. At that very instant, the hunter pulls the trigger. What will happen? (a) The bullet will miss the monkey because the monkey falls down while the bullet speeds straight forward. (b) The bullet will hit the monkey because both the monkey and the bullet are falling downward at the same rate due to gravity. (c) The bullet will miss the monkey because although both the monkey and the bullet are falling downward due to gravity, the monkey is falling faster. (d) It depends on how far the hunter is from the monkey.
14. Which statements are not valid for a projectile? Take up as positive and ignore air resistance. (a) The projectile has the same x velocity at any point on its path. (b) The acceleration of the projectile is positive and decreasing when the projectile is moving upwards, zero at the top, and increasingly negative as the projectile descends. (c) The acceleration of the projectile has a constant negative value. (d) The y component of the velocity of the projectile is zero at the highest point of the projectile’s path. (e) The velocity at the highest point is zero.
15. A car travels 10 m>s east. Another car travels 10 m>s north. The relative speed of the first car with respect to the second is (a) less than 20 m>s.   (b) exactly 20 m>s.    (c) more than 20 m>s. MisConceptual Questions  77

GIAN_PSE5_CH03_054-084_ca.indd 77
01/07/20 16:23
Reference in New Issue View Git Blame Copy Permalink