alan
/
zotero
1
0
Fork 0
Code Issues Pull Requests Packages Projects Releases Wiki Activity
zotero/storage/UTT2BJ6D/.zotero-ft-cache

7744 lines
340 KiB
Plaintext
Raw Normal View History

the big one
2024-08-27 21:48:20 -05:00
PHYSICS for
SCIENTISTS & ENGINEERS
with Modern Physics
D O U G L A S C. G I A N C O L I
PEARSON
Upper Saddle River, New Jersey 07458
Library of Congress Cataloging-in-Publication Data
Giancoli, Douglas C. Physics for scientists and engineers with modern physics / Douglas C.
Giancoli.—4th ed. p. cm.
Includes bibliographical references and index. ISBN 0-13-149508-9 1. Physics—Textbooks. I. Title. QC21.3.G539 2008 530—dc22
2006039431
President, Science: Paul Corey Sponsoring Editor: Christian Botting Executive Development Editor: Karen Karlin Production Editor: Clare Romeo Senior Managing Editor: Scott Disanno Art Director and Interior & Cover Designer: John Christiana Manager, Art Production: Sean Hogan Copy Editor: Jocelyn Phillips Proofreaders: Karen Bosch, Gina Cheselka, Traci Douglas, Nancy Stevenson,
and Susan Fisher Senior Operations Specialist: Alan Fischer Art Production Editor: Connie Long Illustrators: Audrey Simonetti and Mark Landis Photo Researchers: Mary Teresa Giancoli and Truitt & Marshall Senior Administrative Coordinator: Trisha Tarricone Composition: Emilcomp/Prepare Inc.;
Pearson Education/Lissette Quinones, Clara Bartunek Photo credits appear on page A-72 which constitutes
a continuation of the copyright page.
© 2009,2000,1989,1984 by Douglas C. Giancoli Published by Pearson Education, Inc.
PEARSON Pearson Prentice Hall Pearson Education, Inc. Upper Saddle River, NJ 07458
All rights reserved. No part o f this book may be reproduced, in any form or by any means, without permission in writing from the publisher.
Pearson Prentice Hall™ is a trademark of Pearson Education, Inc.
Printed in the United States of America 10 9 8 7 6 5 4 3 2 1
ISBN-13: ISBN-1D:
Pearson Education LTD., London Pearson Education Australia PTY, Limited, Sydney Pearson Education Singapore, Pte. Ltd. Pearson Education North Asia Ltd., Hong Kong Pearson Education Canada, Ltd., Toronto Pearson Education de Mexico, S.A. de C.V. Pearson Education—Japan, Tokyo Pearson Education Malaysia, Pte. Ltd.
Contents
A p p lica tion s List P re fa c e
To S tu d e n ts
U se o f C o lo r
Xll xiv xviii xix
Volume 1
In tr o d u c t io n ,
1 M easurement, Estimating
1
l - i The Nature of Science
2
1-2 Models, Theories, and Laws
2
1-3 Measurement and Uncertainty;
Significant Figures
3
1-4 Units, Standards, and the SI System
6
1-5 Converting Units
8
1-6 Order of Magnitude: Rapid Estimating 9
*1-7 Dimensions and Dimensional Analysis 12
SUMMARY 14 PROBLEMS 14
QUESTIONS 14 GENERAL PROBLEMS 16
D escribing M otion : 2 Kinematics in O ne D im ension 18
2-1 Reference Frames and Displacement
19
2-2 Average Velocity
20
2-3 Instantaneous Velocity
22
2-4 Acceleration
24
2-5 Motion at Constant Acceleration
28
2-6 Solving Problems
30
2-7 Freely Falling Objects
34
*2-8 Variable Acceleration; Integral Calculus 39
*2-9 Graphical Analysis and
Numerical Integration
40
SUMMARY 43 PROBLEMS 44
QUESTIONS 43 GENERAL PROBLEMS 48
3
3-1 3-2 3-3
3-4 3-5 3-6 3-7 3-8 3-9
4A
4-1 4-2 4-3 4-4 4-5 4-6 4-7
4-8
D 5-1
5-2 5-3 5-4 *5-5 *5-6
KlNEMAnCS IN TWO OR
T hree D imensions; Vectors
51
Vectors and Scalars
52
Addition of Vectors-Graphical Methods 52
Subtraction of Vectors, and
Multiplication of a Vector by a Scalar 54
Adding Vectors by Components
55
Unit Vectors
59
Vector Kinematics
59
Projectile Motion
62
Solving Problems: Projectile Motion
64
Relative Velocity
71
SUMMARY 74 QUESTIONS 75
PROBLEMS 75 GENERAL PROBLEMS 80
D ynamics:
N ew ton' s Laws of M o tion
83
Force
84
Newton’s First Law of Motion
84
Mass
86
Newton’s Second Law of Motion
86
Newton’s Third Law of Motion
89
Weight—the Force of Gravity;the Normal Force 92
Solving Problems with Newton’s Laws:
Free-Body Diagrams
95
Problem Solving—A General Approach 102
SUMMARY 102 QUESTIONS 103
PROBLEMS 104 GENERAL PROBLEMS 109
U sing N ew to n 's Laws: F ric tio n , C ircu lar MonoN, D rag Forces 112
Applications of Newton’s Laws
Involving Friction
113
Uniform Circular Motion—Kinematics 119
Dynamics of Uniform Circular Motion 122
Highway Curves: Banked and Unbanked 126
Nonuniform Circular Motion
128
Velocity-Dependent Forces:
Drag and Terminal Velocity
129
su m m a r y 130 q u e s t io n s 131
PROBLEMS 132 GENERAL PROBLEMS 136
III
6
6-1 6-2
6-3
6-4 6-5 * 6-6 6-7 * 6-8
G ravitation a n d N ew ton's Synthesis
139
Newton’s Law of Universal Gravitation 140
Vector Form of Newton’s Law of
Universal Gravitation
143
Gravity Near the Earth’s Surface;
Geophysical Applications
143
Satellites and “Weightlessness”
146
Kepler’s Laws and Newton’s Synthesis 149
Gravitational Field
154
Types of Forces in Nature
155
Principle of Equivalence;
Curvature of Space; Black Holes
155
SUMMARY 157 QUESTIONS 157
PROBLEMS 158 GENERAL PROBLEMS 160
Displacement
W ork a nd Energy
163
7-1 Work Done by a Constant Force
164
7-2 Scalar Product of Two Vectors
167
7-3 Work Done by a Varying Force
168
7-4 Kinetic Energy and the
Work-Energy Principle
172
SUMMARY 176 QUESTIONS 177 PROBLEMS 177 GENERAL PROBLEMS 180
oc
C onservation o f Energy
183
8-1 Conservative and Nonconservative Forces 184
8-2 Potential Energy
186
8-3 Mechanical Energy and Its Conservation 189
8-4 Problem Solving Using
Conservation of Mechanical Energy 190
8-5 The Law of Conservation of Energy 196
8-6 Energy Conservation with Dissipative Forces: Solving Problems 197
8-7 Gravitational Potential Energy and
Escape Velocity
199
8-8 Power
201
*8-9 Potential Energy Diagrams;
Stable and Unstable Equilibrium
204
SUMMARY 205 QUESTIONS 205 PROBLEMS 207 GENERAL PROBLEMS 211
iv CONTENTS
Linear M om entum
214
9-1 Momentum and Its Relation to Force 215
9-2 Conservation of Momentum
217
9-3 Collisions and Impulse
220
9-4 Conservation of Energy and
Momentum in Collisions
222
9-5 Elastic Collisions in One Dimension 222
9-6 Inelastic Collisions
225
9-7 Collisions in Two or Three Dimensions 227
9-8 Center of Mass ( cm )
230
9-9 Center of Mass and Translational Motion 234
*9-10 Systems ofVariable Mass; Rocket Propulsion 236
SUMMARY 239 QUESTIONS 239 PROBLEMS 240 GENERAL PROBLEMS 245
10 R otational M o t io n
248
10-1 Angular Quantities
249
10-2 Vector Nature of Angular Quantities 254
10-3 Constant Angular Acceleration
255
10-4 Torque
256
10-5 Rotational Dynamics;
Torque and Rotational Inertia
258
10-6 Solving Problems in Rotational Dynamics 260
10-7 Determining Moments of Inertia
263
10-8 Rotational Kinetic Energy
265
10-9 Rotational Plus Translational Motion; Rolling 267
*10-10 Why Does a Rolling Sphere Slow Down? 273
SUMMARY 274 QUESTIONS 275 PROBLEMS 276 GENERAL PROBLEMS 281
U A n g u la r M om entum ; G eneral R otation
284
11-1 Angular Momentum—Objects
Rotating About a Fixed Axis
285
11-2 Vector Cross Product; Torque as a Vector 289
11-3 Angular Momentum of a Particle
291
11-4 Angular Momentum and Torque for a System of Particles; General Motion 292
11-5 Angular Momentum and
Torque for a Rigid Object
294
11-6 Conservation of Angular Momentum 297
*11-7 The Spinning Top and Gyroscope
299
*11-8 Rotating Frames of Reference; Inertial Forces 300
*11-9 The Coriolis Effect
301
su m m a r y 302
QUESTIONS 303
PROBLEMS 303
GENERAL
PROBLEMS 308
O scillations
369
14-1 Oscillations of a Spring
370
14-2 Simple Harmonic Motion
372
14-3 Energy in the Simple
Harmonic Oscillator
377
14-4 Simple Harmonic Motion Related
to Uniform Circular Motion
379
14-5 The Simple Pendulum
379
*14-6 The Physical Pendulum and
the Torsion Pendulum
381
14-7 Damped Harmonic Motion
382
14-8 Forced Oscillations; Resonance
385
SUMMARY 387 QUESTIONS 388 PROBLEMS 388 GENERAL PROBLEMS 392
Static Equilibrium ;
Elasticity and Fracture
311
12-1 12-2 12-3 12-4 12-5 *12-6 *12-7
The Conditions for Equilibrium
312
Solving Statics Problems
313
Stability and Balance
317
Elasticity; Stress and Strain
318
Fracture
322
Trusses and Bridges
324
Arches and Domes
327
SUMMARY 329 QUESTIONS 329 PROBLEMS 330 GENERAL PROBLEMS 334
13 Fluids
339
13-1 Phases of Matter
340
13-2 Density and Specific Gravity
340
13-3 Pressure in Fluids
341
13-4 Atmospheric Pressure and
Gauge Pressure
345
13-5 Pascal’s Principle
346
13-6 Measurement of Pressure;
Gauges and the Barometer
346
13-7 Buoyancy and Archimedes’ Principle 348
13-8 Fluids in Motion; Flow Rate
and the Equation of Continuity
352
13-9 Bernoulli’s Equation
354
13-10 Applications of Bernoulli’s Principle:
Torricelli, Airplanes, Baseballs, TIA
356
*13-11 Viscosity
358
*13-12 Flow in Tubes: Poiseuille’s
Equation, Blood Flow
358
*13-13 Surface Tension and Capillarity
359
*13-14 Pumps, and the Heart
361
SUMMARY 361 QUESTIONS 362 PROBLEMS 363 GENERAL PROBLEMS 367
15 W ave M o tio n
395
15-1 Characteristics of Wave Motion
396
15-2 Types of Waves:
Transverse and Longitudinal
398
15-3 Energy Transported by Waves
402
15-4 Mathematical Representation of a
Traveling Wave
404
*15-5 The Wave Equation
406
15-6 The Principle of Superposition
408
15-7 Reflection and Transmission
409
15-8 Interference
410
15-9 Standing Waves; Resonance
412
*15-10 Refraction
415
*15-11 Diffraction
416
SUMMARY 417 QUESTIONS 417 PROBLEMS 418 GENERAL PROBLEMS 422
16 Sound
424
16-1 Characteristics of Sound
425
16-2 Mathematical Representation
of Longitudinal Waves
426
16-3 Intensity of Sound: Decibels
427
16-4 Sources of Sound:
Vibrating Strings and Air Columns
431
*16-5 Quality of Sound, and Noise;
Superposition
436
16-6 Interference of Sound Waves; Beats
437
16-7 Doppler Effect
439
*16-8 Shock Waves and the Sonic Boom
443
*16-9 Applications: Sonar, Ultrasound,
and Medical Imaging
444
SUMMARY 446 QUESTIONS 447 PROBLEMS 448 GENERAL PROBLEMS 451
CONTENTS V
T em perature, 't f j T h erm al E xp ansion,
1 / an d th e Id eal Gas Law
454
17-1 Atomic Theory of Matter
455
17-2 Temperature and Thermometers
456
17-3 Thermal Equilibrium and the
Zeroth Law of Thermodynamics
459
17-4 Thermal Expansion
459
*17-5 Thermal Stresses
463
17-6 The Gas Laws and
Absolute Temperature
463
17-7 The Ideal Gas Law
465
17-8 Problem Solving with the
Ideal Gas Law
466
17-9 Ideal Gas Law in Terms of Molecules:
Avogadro’s Number
468
*17-10 Ideal Gas Temperature Scale—
a Standard
469
SUMMARY 470 QUESTIONS 471 PROBLEMS 471 GENERAL PROBLEMS 474
18 K in e t ic T h e o r y o f G ases
476
18-1
18-2 18-3 18-4 *18-5 *18-6 *18-7
The Ideal Gas Law and the Molecular
Interpretation of Temperature
476
Distribution of Molecular Speeds
480
Real Gases and Changes of Phase
482
Vapor Pressure and Humidity
484
Van der Waals Equation of State
486
Mean Free Path
487
Diffusion
489
SUMMARY 490 QUESTIONS 491 PROBLEMS 492 GENERAL PROBLEMS 494
vi CONTENTS
19 H eat a n d t h e First Law of T herm odynam ics
496
19-1 Heat as Energy Transfer
497
19-2 Internal Energy
498
19-3 Specific Heat
499
19-4 Calorimetry—Solving Problems
500
19-5 Latent Heat
502
19-6 The First Law of Thermodynamics
505
19-7 The First Law of Thermodynamics
Applied; Calculating the Work
507
19-8 Molar Specific Heats for Gases,
and the Equipartition of Energy
511
19-9 Adiabatic Expansion of a Gas
514
19-10 Heat Transfer: Conduction,
Convection, Radiation
515
SUMMARY 520 QUESTIONS 521 PROBLEMS 522 GENERAL PROBLEMS 526
20 S e c o n d Law o f T herm odynam ics
528
20-1 The Second Law of
Thermodynamics—Introduction
529
20-2 Heat Engines
530
20-3 Reversible and Irreversible
Processes; the Carnot Engine
533
20-4 Refrigerators, Air Conditioners, and
Heat Pumps
536
20-5 Entropy
539
20-6 Entropy and the Second Law of
Thermodynamics
541
20-7 Order to Disorder
544
20-8 Unavailability of Energy; Heat Death 545
*20-9 Statistical Interpretation of Entropy
and the Second Law
546
*20-10 Thermodynamic Temperature;
Third Law of Thermodynamics
548
*20-11 Thermal Pollution, Global Warming,
and Energy Resources
549
SUMMARY 551 QUESTIONS 552 PROBLEMS 552 GENERAL PROBLEMS 556
Volume 2
/■%-« Electric C harge and Z 1 Electric Field
559
21-1 Static Electricity; Electric
Charge and Its Conservation
560
21-2 Electric Charge in the Atom
561
21-3 Insulators and Conductors
561
21-4 Induced Charge; the Electroscope
562
21-5 Coulomb’s Law
563
21-6 The Electric Field
568
21-7 Electric Field Calculations for
Continuous Charge Distributions
572
21-8 Field Lines
575
21-9 Electric Fields and Conductors
577
21-10 Motion of a Charged Particle in
an Electric Field
578
21-11 Electric Dipoles
579
*21-12 Electric Forces in Molecular Biology;
DNA
581
*21-13 Photocopy Machines and Computer
Printers Use Electrostatics
582
SUMMARY 584 QUESTIONS 584
PROBLEMS 585 GENERAL PROBLEMS 589
G auss's Law
___ 591
22-1 Electric Flux
592
22-2 Gauss’s Law
593
22-3 Applications of Gauss’s Law
595
*22-4 Experimental Basis of Gauss’s and
Coulomb’s Laws
600
SUMMARY 601 QUESTIONS 601 PROBLEMS 601 GENERAL PROBLEMS 605
23 E lectric P otential
607
23-1
23-2
23-3 23-4 23-5 23-6 23-7 23-8
*23-9
Electric Potential Energy and
Potential Difference
607
Relation between Electric Potential
and Electric Field
610
Electric Potential Due to Point Charges 612
Potential Due to Any Charge Distribution 614
Equipotential Surfaces
616
Electric Dipole Potential
617
E Determined from V
617
Electrostatic Potential Energy; the
Electron Volt
619
Cathode Ray Tube: TV and Computer
Monitors, Oscilloscope
620
SUMMARY 622 QUESTIONS 622
PROBLEMS 623 GENERAL PROBLEMS 626
24 C apa cita nce, D ielectrics, E lectric Energ y Storage
628
24-1
24-2 24-3 24-4 24-5 *24-6
Capacitors
628
Determination of Capacitance
630
Capacitors in Series and Parallel
633
Electric Energy Storage
636
Dielectrics
638
Molecular Description of Dielectrics 640
SUMMARY 643 QUESTIONS 643 PROBLEMS 644 GENERAL PROBLEMS 648
25 E lectric C u r r en ts a n d R esistance
651
25-1 The Electric Battery
652
25-2 Electric Current
654
25-3 Ohm’s Law: Resistance and Resistors 655
25-4 Resistivity
658
25-5 Electric Power
660
25-6 Power in Household Circuits
662
25-7 Alternating Current
664
25-8 Microscopic View of Electric Current:
Current Density and Drift Velocity
666
*25-9 Superconductivity
668
*25-10 Electrical Conduction in the Nervous System 669
SUMMARY 671 QUESTIONS 671
PROBLEMS 672 GENERAL PROBLEMS 675
26 D C C ircuits
677
26-1 26-2 26-3 26-4 26-5
26-6 *26-7
EMF and Terminal Voltage
678
Resistors in Series and in Parallel
679
Kirchhoffs Rules
683
Series and Parallel EMFs; Battery Charging 686
Circuits Containing Resistor
and Capacitor (RC Circuits)
687
Electric Hazards
692
Ammeters and Voltmeters
695
SUMMARY 698 QUESTIONS 698 PROBLEMS 699 GENERAL PROBLEMS 704
CONTENTS vii
27 M ag netism
707
27-1 Magnets and Magnetic Fields
707
27-2 Electric Currents Produce Magnetic Fields 710
27-3 Force on an Electric Current in a
Magnetic Field; Definition of B
710
27-4 Force on an Electric Charge Moving
in a Magnetic Field
714
27-5 Torque on a Current Loop; Magnetic
Dipole Moment
718
*27-6 Applications: Motors, Loudspeakers,
Galvanometers
720
27-7 Discovery and Properties of the
Electron
721
27-8 The Hall Effect
723
*27-9 Mass Spectrometer
724
SUMMARY 725 QUESTIONS 726 PROBLEMS 727 GENERAL PROBLEMS 730
Sources o f M agnetic Field 7 3 3
28-1 Magnetic Field Due to a Straight Wire 734
28-2 Force between Two Parallel Wires
735
28-3 Definitions of the Ampere and the
Coulomb
736
28-4 Ampere’s Law
737
28-5 Magnetic Field of a Solenoid and
a Toroid
741
28-6 Biot-Savart Law
743
28-7 Magnetic Materials—Ferromagnetism 746
*28-8 Electromagnets and
Solenoids—Applications
747
*28-9 Magnetic Fields in Magnetic Materials;
Hysteresis
748
*28-10 Paramagnetism and Diamagnetism 749
SUMMARY 750 QUESTIONS 751
PROBLEMS 751 GENERAL PROBLEMS 755
viii CONTENTS
Electromagnetic Induction
and Faraday' s Law
758
29-1 Induced EMF
759
29-2 Faraday’s Law of Induction; Lenz’s Law 760
29-3 EMF Induced in a Moving Conductor 765
29-4 Electric Generators
766
*29-5 Back EMF and Counter Torque;
Eddy Currents
768
29-6 Transformers and Transmission of Power 770
29-7 A Changing Magnetic Flux Produces an
Electric Field
773
*29-8 Applications of Induction:
Sound Systems, Computer Memory,
Seismograph, GFCI
775
SUMMARY 111 QUESTIONS 111 PROBLEMS 778 GENERAL PROBLEMS 782
Inductance, Electromagnetic
O scillations, and AC C ircuits 7 8 5
30-1 Mutual Inductance
786
30-2 Self-Inductance
788
30-3 Energy Stored in a Magnetic Field
790
30-4 LR Circuits
790
30-5 LC Circuits and Electromagnetic
Oscillations
793
30-6 LC Oscillations with Resistance
(.LRC Circuit)
795
30-7 AC Circuits with AC Source
796
30-8 LRC Series AC Circuit
799
30-9 Resonance in AC Circuits
802
*30-10 Impedance Matching
802
*30-11 Three-Phase AC
803
SUMMARY 804 QUESTIONS 804
PROBLEMS 805 GENERAL PROBLEMS 809
M axwell' s Equations and Electromagnetic W aves
812
31-1 Changing Electric Fields Produce
Magnetic Fields; Ampere’s Law and
Displacement Current
813
31-2 Gauss’s Law for Magnetism
816
31-3 Maxwell’s Equations
817
31-4 Production of Electromagnetic Waves 817
31-5 Electromagnetic Waves, and
Their Speed, from Maxwell’s Equations 819
31-6 Light as an Electromagnetic Wave
and the Electromagnetic Spectrum
823
31-7 Measuring the Speed of Light
825
31-8 Energy in EM Waves; the Poynting Vector 826
31-9 Radiation Pressure
828
31-10 Radio and Television;
Wireless Communication
829
SUMMARY 832 QUESTIONS 832 PROBLEMS 833 GENERAL PROBLEMS 835
32 Light: R eflection and Refraction
837
32-1 The Ray Model of Light
838
32-2 Reflection; Image Formation by a
Plane Mirror
838
32-3 Formation of Images by Spherical
Mirrors
842
32-4 Index of Refraction
850
32-5 Refraction: Snell’s Law
850
32-6 Visible Spectrum and Dispersion
852
32-7 Total Internal Reflection; Fiber Optics 854
*32-8 Refraction at a Spherical Surface
856
SUMMARY 858 QUESTIONS 859
PROBLEMS 860 GENERAL PROBLEMS 864
rm
y jii'w n in H r.
th e n h ( « l K- p l4 A ‘J .11 lltL> N irti
I t u prodlM U a victual im ujw. h W h
j
f >n ^
C (
LSm .^ m
n p
irito jfjs o
n n
n f rwirt
u l p jT L { J J
IMIw c j* « ictmed.
^
ift CK K tly U I V f ^ a l p u ir L
/viewed at ihe near point wirfiSL'**vcn >™
‘w the <*jw.-i_
object subtends at the eye is much /a!k rarl(h>in
magniricttliim or mttgnifying power,
rr‘
angle subtended by an object when u s i n X ^ S j f c ^ c r,JhlJ,lf lHc
unaided eye, with the object at the nea^P^^u^itn^duMnicih.L:
normal eye):
»" ,h4: lN -
fDr *
M
(»-£i
vritcjWin[LinnL»1lilt t«*l
w here 0 and flr ars shown in Fig. 3 3 -3 3 . length by noting that 0 = h / N (Fig, 3.1-
' “ h/tl" (Hj(. 3? .l.lij, wliL’ii;
lc*#tc wialt k>rtandflrcijiml
* « .he heigh, of Ihe object and we
J - 3J
^buir suits and lungems, ff the eye is
j^t infinity and the ot>ji.>el will h>
^ / and fl' = h / f .
33 Lenses and O ptical Instruments
866
33-1 Thin Lenses; Ray Tracing
867
33-2 The Thin Lens Equation; Magnification 870
33-3 Combinations of Lenses
874
*33-4 Lensmaker’s Equation
876
33-5 Cameras: Film and Digital
878
33-6 The Human Eye; Corrective Lenses 882
33-7 Magnifying Glass
885
33-8 Telescopes
887
*33-9 Compound Microscope
890
*33-10 Aberrations of Lenses and Mirrors 891
SUMMARY 892 QUESTIONS 893 PROBLEMS 894 GENERAL PROBLEMS 897
34 T he Wave N ature of Light;
Interference
900
34-1 Waves Versus Particles; Huygens’
Principle and Diffraction
901
34-2 Huygens’ Principle and the Law of
Refraction
902
34-3 Interference—Young’s Double-Slit
Experiment
903
*34-4 Intensity in the Double-Slit
Interference Pattern
906
34-5 Interference in Thin Films
909
*34-6 Michelson Interferometer
914
*34-7 Luminous Intensity
915
SUMMARY 915 QUESTIONS 916 PROBLEMS 916 GENERAL PROBLEMS 918
35 D iffraction and P olarization
921
35-1 Diffraction by a Single Slit or Disk
922
*35-2 Intensity in Single-Slit Diffraction
Pattern
924
*35-3 Diffraction in the Double-Slit Experiment 927
35-4 Limits of Resolution; Circular Apertures 929
35-5 Resolution of Telescopes and
Microscopes; the ALimit
931
*35-6 Resolution of the Human Eye
and Useful Magnification
932
35-7 Diffraction Grating
933
35-8 The Spectrometer and Spectroscopy 935
*35-9 Peak Widths and Resolving Power for a
Diffraction Grating
937
35-10 X-Rays and X-Ray Diffraction
938
35-11 Polarization
940
*35-12 Liquid Crystal Displays (LCD)
943
*35-13 Scattering of Light by the Atmosphere 945
su m m a r y 945 q u e s t io n s 946 PROBLEMS 946 GENERAL PROBLEMS 949
CONTENTS ix
Volume 3
36 Special T heory of Relativity 951
36-1 Galilean-Newtonian Relativity
952
*36-2 The Michelson-Morley Experiment 954
36-3 Postulates of the Special Theory of Relativity 957
36-4 Simultaneity
958
36-5 Time Dilation and the Twin Paradox
960
36-6 Length Contraction
964
36-7 Four-Dimensional Space-Time
967
36-8 Galilean and Lorentz Transformations 968
36-9 Relativistic Momentum and Mass
971
36-10 The Ultimate Speed
974
36-11 E = me2;Mass and Energy
974
*36-12 Doppler Shift for Light
978
36-13 The Impact of Special Relativity
980
SUMMARY 981 QUESTIONS 981 PROBLEMS 982 GENERAL PROBLEMS 985
37 Early Q uantum T heory and
M odels of the Atom
987
37-1 Blackbody Radiation;
Planck’s Quantum Hypothesis
987
37-2 Photon Theory; Photoelectric Effect
989
37-3 Photon Energy, Mass, and Momentum 993
37-4 Compton Effect
994
37-5 Photon Interactions; Pair Production 996
37-6 Wave-Particle Duality; the Principle of
Complementarity
997
37-7 Wave Nature of Matter
997
*37-8 Electron Microscopes
1000
37-9 Early Models of the Atom
1000
37-10 Atomic Spectra: Key to Atomic Structure 1001
37-11 The Bohr Model
1003
37-12 deBroglie’s Hypothesis Applied to Atoms 1009
SUMMARY 1010 QUESTIONS 1011 PROBLEMS 1012 GENERAL PROBLEMS 1014
38 Q uantum M echanics
1017
38-1 Quantum Mechanics—A New Theory 1018
38-2 The Wave Function and Its Interpretation;
the Double-Slit Experiment
1018
38-3 The Heisenberg Uncertainty Principle 1020
38-4 Philosophic Implications; Probability
Versus Determinism
1024
38-5 The Schrodinger Equation in One
Dimension—Time-Independent Form 1025
*38-6 Time-Dependent Schrodinger Equation 1027
38-7 Free Particles; Plane Waves and Wave Packets 1028
38-8 Particle in an Infinitely Deep
Square Well Potential (a Rigid Box) 1030
38-9 Finite Potential Well
1035
38-10 Tunneling through a Barrier
1036
SUMMARY 1039 QUESTIONS 1039
PROBLEMS 1040 GENERAL PROBLEMS 1042
X CONTENTS
39 Q uantum M echanics of Atoms
1044
39-1 Quantum-Mechanical View of Atoms 1045
39-2 Hydrogen Atom: Schrodinger Equation
and Quantum Numbers
1045
39-3 Hydrogen Atom Wave Functions
1049
39-4 Complex Atoms; the Exclusion Principle 1052
39-5 Periodic Table of Elements
1053
39-6 X-Ray Spectra and Atomic Number 1054
*39-7 Magnetic Dipole Moment;
Total Angular Momentum
1057
39-8 Fluorescence and Phosphorescence 1060
39-9 Lasers
1061
*39-10 Holography
1064
SUMMARY 1066 QUESTIONS 1066 PROBLEMS 1067 GENERAL PROBLEMS 1069
40 M olecules and S olids
1071
40-1 Bonding in Molecules
1071
40-2 Potential-Energy Diagrams
for Molecules
1074
40-3 Weak (van der Waals) Bonds
1077
40-4 Molecular Spectra
1080
40-5 Bonding in Solids
1085
40-6 Free-Electron Theory of Metals;
Fermi Energy
1086
40-7 Band Theory of Solids
1090
40-8 Semiconductors and Doping
1093
40-9 Semiconductor Diodes
1094
40-10 Transistors and Integrated Circuits (Chips) 1097
SUMMARY 1098 QUESTIONS 1099
PROBLEMS 1099 GENERAL PROBLEMS 1102
41 N uclear P hysics and Radioactivity
1104
41-1 Structure and Properties of the Nucleus 1105
41-2 Binding Energy and Nuclear Forces 1108
41-3 Radioactivity
1110
41-4 Alpha Decay
1111
41-5 Beta Decay
1114
41-6 Gamma Decay
1116
41-7 Conservation of Nucleon Number
and Other Conservation Laws
1117
41-8 Half-Life and Rate of Decay
1117
41-9 Decay Series
1121
41-10 Radioactive Dating
1122
41-11 Detection of Radiation
1124
SUMMARY 1126 QUESTIONS 1126 PROBLEMS 1127 GENERAL PROBLEMS 1129
42 N uclear Energy; Effects and U ses of Radiation
1131
42-1 Nuclear Reactions and the
Transmutation of Elements
1132
42-2 Cross Section
1135
42-3 Nuclear Fission; Nuclear Reactors
1136
42-4 Nuclear Fusion
1141
42-5 Passage of Radiation Through Matter;
Radiation Damage
1146
42-6 Measurement of Radiation—Dosimetry 1147
*42-7 Radiation Therapy
1150
*42-8 Tracers in Research and Medicine
1151
*42-9 Imaging by Tomography: CAT Scans
and Emission Tomography
1153
*42-10 Nuclear Magnetic Resonance (NMR);
Magnetic Resonance Imaging (MRI) 1156
SUMMARY 1159 QUESTIONS 1159
PROBLEMS 1160 GENERAL PROBLEMS 1162
43 Elementary Particles
1164
43-1 High-Energy Particles and Accelerators 1165
43-2 Beginnings of Elementary Particle
Physics—Particle Exchange
1171
43-3 Particles and Antiparticles
1174
43-4 Particle Interactions and Conservation Laws 1175
43-5 Neutrinos—Recent Results
1177
43-6 Particle Classification
1178
43-7 Particle Stability and Resonances
1180
43-8 Strangeness? Charm? Towards a New Model 1181
43-9 Quarks
1182
43-10 The Standard Model: QCD and
Electroweak Theory
1184
43-11 Grand Unified Theories
1187
43-12 Strings and Supersymmetry
1189
SUMMARY 1189 QUESTIONS 1190
PROBLEMS 1190 GENERAL PROBLEMS 1191
Astrophysics and C osmology 1193
44-1 Stars and Galaxies
1194
44-2 Stellar Evolution: Nucleosynthesis,
and the Birth and Death of Stars
1197
44-3 Distance Measurements
1203
44-4 General Relativity: Gravity and the
Curvature of Space
1205
44-5 The Expanding Universe: Redshift and
Hubble’s Law
1209
44-6 The Big Bang and the Cosmic
Microwave Background
1213
44-7 The Standard Cosmological Model:
Early History of the Universe
1216
44-8 Inflation
1219
44-9 Dark Matter and Dark Energy
1221
44-10 Large-Scale Structure of the Universe 1224
44-11 Finally ...
1224
SUMMARY 1225 QUESTIONS 1226
PROBLEMS 1226 GENERAL PROBLEMS 1227
Appendices
A
M athematical Form ulas
A -1
B
D erivatives a n d Integrals
A -6
C
M ore o n D im ensio na l A nalysis
A -8
D
G ravitational Fo rce d u e to a
S pherical M ass D istribution
A -9
E
D ifferential Fo r m o f M axwell's Eq uations A - 1 2
F
S elected Isotopes
A -14
A nswers to O d d -N umbered P roblems
A -18
In d e x
A -47
P h o to C redits
A -72
CONTENTS xi
Applications ( s e l e c t e d )
Chapter 1
The 8000-m peaks
8
Estimating volume of a lake
10
Height by triangulation
11
Radius of the Earth
11
Heartbeats in a lifetime
12
Particulate pollution (Pr30)
15
Global positioning satellites (Pr39) 16
Lung capacity (Pr65)
17
Chapter 2
Airport runway design
29
Automobile air bags
31
Braking distances
32
CD error correction (PrlO)
44
CD playing time (Prl3)
45
Golfing uphill or down (Pr79)
48
Rapid transit (Pr83)
49
Chapter 3
Kicked football
66,69
Ball sports (Problems)
77,81,82
Extreme sports (Pr41)
77
Chapter 4
Rocket acceleration
90
What force accelerates a car?
90
How we walk
90
Elevator and counterweight
99
Mechanical advantage of pulley
100
Bear sling (Q24)
104
High-speed elevators (Prl9)
105
Mountain climbing (Pr31,82,83) 106,110
City planning, cars on hills (Pr71) 109
Bicyclists (Pr72,73)
109
“Doomsday” asteroid (Pr84)
110
Chapter 5 Push or pull a sled? Centrifugation Not skidding on a curve Banked highways Simulated gravity (Q18, Pr48) “Rotor-ride” (Pr82)
116 122 126-7 127 131,134 136
Chapter 6
Oil/mineral exploration
144,420
Artificial Earth satellites
146
Geosynchronous satellites
147
Weightlessness
148
Free fall in athletics
149
Planet discovery, extrasolar planets 152
Black holes
156
Asteroids (Pr44,78)
159,162
Navstar GPS (Pr58)
160
Black hole, galaxy center
(Pr61,64)
160,161
Tides (Pr75)
162
Chapter 7
Car stopping distance of v2
174
Lever(Pr6)
177
Spiderman (Pr54)
179
Bicycling on hills, gears (Pr85)
181
Child safety in car (Pr87)
181
Rock climber’s rope (Pr90)
182
Chapter 8 Downhill ski runs Rollercoaster Pole vault Toy dart gun
183 191,198
192-3 193
x ii APPLICATIONS
Escape velocity from Earth or Moon 201
Stair climbing power
202
Power needs of car
202-3
Cardiac treadmill (Prl04)
213
Chapter 9
Tennis serve
216
Rocket propulsion
219,236-8
Rifle recoil
220
Karate blow
221
Billiards/bowling
223,228
Nuclear collisions
225,228
Ballistic pendulum
226
Conveyor belt
237
Gravitational slingshot (Prl05)
246
Crashworthiness (Prl09)
247
Asteroids, planets (PrllO, 112,113) 247
Chapter 10
Hard drive and bit speed
253
Wrench/tire iron
256
Flywheel energy
266,281
Yo-yo
271
Car braking forces
272-3
Bicycle odometer calibration (Ql) 275
Tightrope walker (Q ll)
275
Triceps muscle and throwing
(Pr38,39)
278
CD speed (Pr84)
281
Bicycle gears (Pr89)
281
Chapter 11
Rotating skaters, divers 284,286,309
Neutron star collapse
287
Auto wheel balancing
296
Top and gyroscope
299-300
Coriolis effect
301-2
Hurricanes
302
SUV possible rollover (Pr67)
308
Triple axel jump (Pr79)
309
Bat’s “sweet spot” (Pr82)
310
Chapter 12
Tragic collapse
311,323
Lever’s mechanical advantage
313
Cantilever
315
Biceps muscle force
315
Human balance with loads
318
Trusses and bridges
324-6,335
Architecture: arches and domes 327-8
Forces on vertebrae (Pr87)
337
Chapter 13
Lifting water
345,348
Hydraulic lift, brakes
346
Pressure gauges
346-7
Hydrometer
351
Helium balloon lift
352,368
Blood flow
353,357,361
Airplane wings, lift
356
Sailing against the wind
357
Baseball curve
357
Blood to the brain, TIA
357
Blood flow and heart disease
359
Surface tension, capillarity
359-60
Walking on water
360
Pumps and the heart
361
Reynolds number (Pr69)
366
Chapter 14
Car shock absorbers
383
Resonance damage
386
Chapter 15 Echolocation by animals Earthquake waves
400 401,403,416
Chapter 16
Distance from lightning
425
Autofocus camera
426
Wide range of human hearing 427-8,431
Loudspeaker response
428
Stringed instruments
432-3
Wind instruments
433-6
Tuning with beats
439
Doppler blood flow meter
442,453
Sonar: sonic boom
AAA
Ultrasound medical imaging
445-6
Motion sensor (Pr5)
448
Chapter 17
Hot air balloon
454
Expansion joints, highways 456,460,463
Gas tank overflow
462
Life under ice
462
Cold and hot tire pressure
468
Molecules in a breath
469
Thermostat (Q10)
471
Scuba/snorkeling (Pr38,47,82,85) 473,475
Chapter 18
Chemical reactions, temperature
dependence
481
Superfluidity
483
Evaporation cools
484,505
Humidity, weather
485-6
Chromatography
490
Pressure cooker (Pr35)
493
Chapter 19
Working off the calories
498
Cold floors
516
Heat loss through windows
516
How clothes insulate
516-7
i?-values for thermal insulation
517
Convective house heating
517
Human radiative heat loss
518
Room comfort and metabolism
519
Radiation from Sun
519
Medical thermography
519
Astronomy—size of a star
520
Thermos bottle (Q30)
521
Weather, air parcel, adiabatic lapse
rate (Pr56)
525
Chapter 20
Steam engine
530
Internal combustion engine 531,535-6
Car efficiency
532
Refrigerators, air conditioners 537-8
Heat pump
538
Biological evolution, development 545
Thermal pollution, global warming 549-51
Energy resources
550
Diesel engine (Pr7)
553
Chapter 21
Static electricity
560,589 (Pr78)
Photocopiers
569,582-3
Electric shielding, safety
577
DNA structure and replication 581-2
Biological cells: electric forces
and kinetic theory
581-2,617
Laser & inkjet printers
583
Chapter 23 Breakdown voltage Lightning rods, corona CRT, oscilloscopes,
TV monitors Photocell (Pr75) Geiger counter (Pr83) Van de Graaff (Pr84)
612 612
620-1,723 626 627
627,607
Chapter 24 Capacitor uses Very high capacitance Computer key Camera flash Heart defibrillator DRAM (PrlO, 57) Electrostatic air cleaner (Pr20) CMOS circuits (Pr53)
628,631 631 631 636 638
644,647 645 647
Chapter 25
Light bulb
651,653,660
Battery construction
653
Loudspeaker wires
659
Resistance thermometer
660
Heating elements, bulb filament
660
Why bulbs burn out at turn on
661
Lightning bolt
662
Household circuits, shorts
662-3
Fuses, circuit breakers 662-3,747,776
Extension cord danger
663
Nervous system, conduction
669-70
Strain gauge (Pr 24)
673
Chapter 26
Car battery charging, jump start 686,687
RC applications: flashers, wipers
691
Heart pacemaker
692,787
Electric hazards
692-4
Proper grounding
693-4
Heart fibrillation
692
Meters, analog and digital
695-7
Potentiometers and bridges (Pr) 704,705
Chapter 27
Compass and declination
709
Aurora borealis
717
Motors, loudspeakers, galvonometers 720-1
Mass spectrometer
724-5
Electromagnetic pumping (Q14) 726
Cyclotron (Pr66)
731
Beam steering (Pr67)
731
Chapter 28 Coaxial cable Solenoid switches: car starters,
doorbell Circuit breakers, magnetic Relay (Q16) Atom trap (Pr73)
740,789
747 747,776
751 757
Chapter 29 Induction stove EM blood-flow meter Power plant generators Car alternators Motor overload Airport metal detector Eddy current damping Transformers and uses, power Car ignition, bulb ballast Microphone Read/write on disks and tape Digital coding Credit card swipe
762 765 766-7 768 769 770 770 770-3 772,773 775 775 775 776
Ground fault circuit interrupter
(GFCI)
776
Betatron (Pr55)
782
Search coil (Pr68)
783
Inductive battery charger (Pr81) 784
Chapter 30
Spark plug Pacemaker Surge protector LC oscillators, resonance Capacitors as filters Loudspeaker cross-over Impedance matching Three-phase AC 0-value (Pr86,87)
785 787 792 794,802 799 799 802-3 803 810
Chapter 31
Antennas Phone call lag time Solar sail Optical tweezers Wireless: AM/FM, TV, tuning,
cell phones, remotes
824,831 825 829 829
829-32
Chapter 32
How tall a mirror do you need 840-1
Close up and wide-view
mirrors
842,849,859
Where you can see yourself in a
concave mirror
848
Optical illusions
851,903
Apparent depth in water
852
Rainbows
853
Colors underwater
854
Prism binoculars
855
Fiber optics in
telecommunications
855-6,865
Medical endoscopes
856
Highway reflectors (Pr86)
865
Chapter 33
Where you can see a lens image
869
Cameras, digital and film
878
Camera adjustments
879-80
Pixels and resolution
881
Human eye
882-5,892
Corrective lenses
883-5
Contact lenses
885
Seeing under water
885
Magnifying glass
885-7
Telescopes
887-9,931-2
Microscopes
890-1,931,933
Chapter 34
Bubbles, reflected color
900,912-3
Mirages
903
Colors in thin soap film, details 912-3
Lens coatings
913-4
Multiple coating (Pr52)
919
Chapter 35
Lens and mirror resolution
929-30
Hubble Space Telescope
930
Eye resolution,
useful magnification
930,932-3
Radiotelescopes
931
Telescope resolution, Arule
931
Spectroscopy
935-6
X-ray diffraction in biology
939
Polarized sunglasses
942
LCDs—liquid crystal displays 943-4
Sky color
945
Chapter 36
Space travel
963
Global positioning system (GPS) 964
Chapter 37
Photocells
992
Photodiodes
992
Photosynthesis
993
Measuring bone density
995
Electron microscopes
1000
Chapter 38 Tunneling through a QM barrier Scanning tunneling electron
microscope
Chapter 39 Fluorescence analysis Fluorescent bulbs Phosphorescence, watch dials Lasers DVD and CD players Barcodes Laser surgery Holography
1038
1038-9
1060 1060 1061 1061-5 1063 1063 1064 1064-5
Chapter 40
Cell energy—activation energy, ATP 1075-7
Weak bonds in cells, DNA
1077-8
Protein synthesis
1079-80
Transparency
1092
Semiconductor diodes, transistors 1094-8
Rectifier circuits
1096
LED displays; photodiodes
1096
Integrated circuits (Chips)
1098
Chapter 41
Smoke detectors
1114
Carbon-14 dating
1122-3
Archeological, geological dating 1123-4
Oldest Earth rocks and earliest life 1124
Chapter 42
Nuclear reactors and power plants 1138^40
Manhattan Project
1141
Stellar fusion
1142-3
Fusion energy reactors
1131,1144-6
Biological radiation damage
1146-7
Radiation dosimetry
1147-9
Radon
1148,1150
Human radiation exposure
1148-9
Radiation sickness
1149
Radiation therapy
1150-1
Proton therapy
1151
Tracers in medicine and biology 1151-2
X-ray imaging
1153
CAT scans
1153-5
Emission tomography: PET
and SPET
1156
NMR and MRI
1156-9
Chapter 43 Antimatter
1174-5,1188
Chapter 44
Stars and galaxies
1194-9
Star evolution
1200-2
Supernovae
1201,1202,1203
Star distances
1194,1203^1
Black holes
1202,1208-9
Curved space
1207-8
Big Bang
1212,1213-6
Evolution of universe
1216-9
Dark matter and dark energy 1221-3
APPLICATIONS x iii
Preface
xiv PREFACE
I was motivated from the beginning to write a textbook different from others that present physics as a sequence of facts, like a Sears catalog: “here are the facts and you better learn them.” Instead of that approach in which topics are begun formally and dogmatically, I have sought to begin each topic with concrete observations and experiences students can relate to: start with specifics and only then go to the great generalizations and the more formal aspects of a topic, showing why we believe what we believe. This approach reflects how science is actually practiced.
Why a Fourth Edition?
Two recent trends in physics texbooks are disturbing: (1) their revision cycles have become short—they are being revised every 3 or 4 years; (2) the books are getting larger, some over 1500 pages. I don’t see how either trend can be of benefit to students. My response: (1) It has been 8 years since the previous edition of this book. (2) This book makes use of physics education research, although it avoids the detail a Professor may need to say in class but in a book shuts down the reader. And this book still remains among the shortest.
This new edition introduces some important new pedagogic tools. It contains new physics (such as in cosmology) and many new appealing applications (list on previous page). Pages and page breaks have been carefully formatted to make the physics easier to follow: no turning a page in the middle of a derivation or Example. Great efforts were made to make the book attractive so students will want to read it.
Some of the new features are listed below.
F, y, B
What's New
Chapter-Opening Questions: Each Chapter begins with a multiple-choice question, whose responses include common misconceptions. Students are asked to answer before starting the Chapter, to get them involved in the material and to get any preconceived notions out on the table. The issues reappear later in the Chapter, usually as Exercises, after the material has been covered. The Chapter-Opening Questions also show students the power and usefulness of Physics.
APPROACH paragraph in worked-out numerical Examples .A short introductory paragraph before the Solution, outlining an approach and the steps we can take to get started. Brief NOTES after the Solution may remark on the Solution, may give an alternate approach, or mention an application.
Step-by-Step Examples: After many Problem Solving Strategies (more than 20 in the book), the next Example is done step-by-step following precisely the steps just seen.
Exercises within the text, after an Example or derivation, give students a chance to see if they have understood enough to answer a simple question or do a simple calculation. Many are multiple choice.
Greater clarity : No topic, no paragraph in this book was overlooked in the search to improve the clarity and conciseness of the presentation. Phrases and sentences that may slow down the principal argument have been eliminated: keep to the essentials at first, give the elaborations later.
Vector notation, arrows: The symbols for vector quantities in the text and Figures now have a tiny arrow over them, so they are similar to what we write by hand. Cosmological Revolution: With generous help from top experts in the field, readers have the latest results.
Page layout: more than in the previous edition, serious attention has been paid to how each page is formatted. Examples and all important derivations and arguments are on facing pages. Students then don’t have to turn back and forth. Throughout, readers see, on two facing pages, an important slice of physics.
New Applications'. LCDs, digital cameras and electronic sensors (CCD, CMOS), electric hazards, GFCIs, photocopiers, inkjet and laser printers, metal detectors, underwater vision, curve balls, airplane wings, DNA, how we actually see images. (Turn back a page to see a longer list.)
Examples modified: more math steps are spelled out, and many new Examples added. About 10% of all Examples are Estimation Examples.
This Book is Shorter than other complete full-service books at this level. Shorter explanations are easier to understand and more likely to be read.
Content and Organizational Changes
• Rotational Motion: Chapters 10 and 11 have been reorganized. All of angular momentum is now in Chapter 11.
• First law of thermodynamics, in Chapter 19, has been rewritten and extended. The full form is given: AK + AU + AEint = Q —W, where internal energy is Ete, and U is potential energy; the form Q — W is kept so that dW = P dV.
• Kinematics and Dynamics of Circular Motion are now treated together in Chapter 5.
• Work and Energy, Chapters 7 and 8, have been carefully revised. • Work done by friction is discussed now with energy conservation (energy
terms due to friction). • Chapters on Inductance and AC Circuits have been combined into one:
Chapter 30. • Graphical Analysis and Numerical Integration is a new optional Section 2-9.
Problems requiring a computer or graphing calculator are found at the end of most Chapters. • Length of an object is a script £ rather than normal /, which looks like 1 or I (moment of inertia, current), as in F = IIB. Capital L is for angular momentum, latent heat, inductance, dimensions of length [L\. • Newton’s law of gravitation remains in Chapter 6. Why? Because the 1/r2 law is too important to relegate to a late chapter that might not be covered at all late in the semester; furthermore, it is one of the basic forces in nature. In Chapter 8 we can treat real gravitational potential energy and have a fine instance of using U = - JF •di. • New Appendices include the differential form of Maxwell’s equations and more on dimensional analysis. • Problem Solving Strategies are found on pages 30, 58, 64, 96,102,125,166, 198,229,261,314,504,551,571, 685,716,740,763,849, 871, and 913.
Organization
Some instructors may find that this book contains more material than can be covered in their courses. The text offers great flexibility. Sections marked with a star * are considered optional. These contain slightly more advanced physics material, or material not usually covered in typical courses and/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 major parts of Chapters 1, 13, 16, 26, 30, and 35, and selected parts of Chapters 9,12,19,20, 33, and the modern physics Chapters. Topics not covered in class can be a valuable resource for later study by students. Indeed, this text can serve as a useful reference for years because of its wide range of coverage.
Versions of this Book
Complete version: 44 Chapters including 9 Chapters of modern physics.
Classic version: 37 Chapters including one each on relativity and quantum theory.
3 Volume version: Available separately or packaged together (Vols. 1 & 2 or all 3 Volumes):
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 XV
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.
Mario Affatigato, Coe College Lorraine Allen, United States Coast Guard Academy Zaven Altounian, McGill University Bruce Barnett, Johns Hopkins University Michael Barnett, Lawrence Berkeley Lab Anand Batra, Howard University Cornelius Bennhold, George Washington University Bruce Birkett, University of California Berkeley Dr. Robert Boivin, Auburn University Subir Bose, University of Central Florida David Branning, Trinity College Meade Brooks, Collin County Community College Bruce Bunker, University of Notre Dame Grant Bunker, Illinois Institute of Technology Wayne Carr, Stevens Institute of Technology Charles Chiu, University of Texas Austin 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 Kathy Dimiduk, University of New Mexico John DiNardo, Drexel University
Scott Dudley, United States Air Force Academy John Essick, Reed College Cassandra Fesen, Dartmouth College Alex Filippenko, University of California Berkeley Richard Firestone, Lawrence Berkeley Lab Mike Fortner, Northern Illinois University Tom Furtak, Colorado School of Mines Edward Gibson, California State University Sacramento John Hardy, Texas A&M J. Erik Hendrickson, University of Wisconsin Eau Claire Laurent Hodges, Iowa State University David Hogg, New York University Mark Hollabaugh, Normandale Community College Andy Hollerman, University of Louisiana at Lafayette William Holzapfel, University of California Berkeley Bob Jacobsen, University of California Berkeley TerukiKamon, Texas A&M Daryao Khatri, University of the District of Columbia Jay Kunze, Idaho State University
Jim LaBelle, Dartmouth College M.A.K. Lodhi, Texas Tech Bruce Mason, University of Oklahoma Dan Mazilu, Virginia Tech Linda McDonald, North Park College Bill McNairy, Duke University Raj Mohanty, Boston University Giuseppe Molesini, Istituto Nazionale di Ottica Florence Lisa K. Morris, Washington State University Blaine Norum, University of Virginia Alexandria Oakes, Eastern Michigan University Michael Ottinger, Missouri Western State University Lyman Page, Princeton and WMAP Bruce Partridge, Haverford College R. Daryl Pedigo, University of Washington Robert Pelcovitz, Brown University Vahe Peroomian, UCLA James Rabchuk, Western Illinois University Michele Rallis, Ohio State University Paul Richards, University of California Berkeley Peter Riley, University of Texas Austin Larry Rowan, University of North Carolina Chapel Hill Cindy Schwarz, Vassar College Peter Sheldon, Randolph-Macon Woman’s College
Natalia A. Sidorovskaia, University of Louisiana at Lafayette James Siegrist, UC Berkeley, Director Physics Division LBNL George Smoot, University of California Berkeley Mark Sprague, East Carolina University Michael Strauss, University of Oklahoma Laszlo Takac, University of Maryland Baltimore Co. Franklin D.Trumpy, Des Moines Area Community College Ray Turner, Clemson University Som Tyagi, Drexel University John Vasut, Baylor University Robert Webb, Texas A&M Robert Weidman, Michigan Technological University Edward A. Whittaker, Stevens Institute of Technology John Wolbeck, Orange County Community College Stanley George Wojcicki, Stanford University Edward Wright, UCLA Todd Young, Wayne State College William Younger, College of the Albemarle Hsiao-Ling Zhou, Georgia State University
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 end of this book. Many thanks also to J. Erik Hendrickson who collaborated with Bob Davis on the solutions, and to the team they managed (Profs. Anand Batra, Meade Brooks, David Currott, Blaine Norum, Michael Ottinger, Larry Rowan, Ray Turner, John Vasut, William Younger). I am grateful to Profs. John Essick, Bruce Barnett, Robert Coakley, Biman Das, Michael Dennin, Kathy Dimiduk, John DiNardo, Scott Dudley, David Hogg, Cindy Schwarz, Ray Turner, and Som Tyagi, who inspired many of the Examples, Questions, Problems, and significant clarifications.
Crucial for rooting out errors, as well as providing excellent suggestions, were Profs. Kathy Dimiduk, Ray Turner, and Lorraine Allen. A huge thank you to them and to Prof. Giuseppe Molesini for his suggestions and his exceptional photographs for optics.
xvi PREFACE
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: George Smoot, Paul Richards, Alex Filippenko, James Siegrist, and William Holzapfel (UC Berkeley), Lyman Page (Princeton and WMAP), Edward Wright (UCLA and WMAP), and Michael Strauss (University of Oklahoma).
I especially wish to thank Profs. Howard Shugart, Chair Frances Heilman, and many others at the University of California, Berkeley, Physics Department for helpful discussions, and for hospitality. Thanks also to Prof. Tito Arecchi and others at the Istituto Nazionale di Ottica, Florence, Italy.
Finally, I am grateful to the many people at Prentice Hall with whom I worked on this project, especially Paul Corey, Karen Karlin, Christian Botting, John Christiana, and Sean Hogan.
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: Paul.Corey@Pearson.com
Post: Paul Corey One Lake Street Upper Saddle River, NJ 07458
D.C.G.
About the Author
Douglas C. Giancoli obtained his BA in physics (summa cum laude) from the University of California, Berkeley, his MS in physics at the Massachusetts Institute of Technology, and his PhD in elementary particle physics at the University of Cali­ fornia, 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 Segre and Donald Glaser.
He has taught a wide range of undergraduate courses, traditional as well as innovative ones, and continues to update his texbooks meticulously, seeking ways to better provide an understanding of physics for students.
Doug’s favorite spare-time activity is the outdoors, especially climbing peaks (here on a dolomite summit, Italy). He says climbing peaks is like learning physics: it takes effort and the rewards are great.
Online Supplements (partial list)
MasteringPhysics™ (www.masteringphysics.com) is a sophisticated online tutoring and homework system devel­ oped specially for courses using calculus-based physics. Originally developed by David Pritchard and collaborators at MIT, MasteringPhysics provides students with individualized online tutoring by responding to their wrong answers and providing hints for solving multi-step problems when they get stuck. It gives them immediate and up-to-date assessment of their progress, and shows where they need to practice more. MasteringPhysics provides instructors with a fast and effective way to assign triedand-tested online homework assignments that comprise a range of problem types. The powerful post-assignment diagnostics allow instructors to assess the progress of their class as a whole as well as individual students, and quickly identify areas of difficulty.
WebAssign (www.webassign.com)
CAPA and LON-CAPA (www.lon-capa.org)
Student Supplements (partial list)
Student Study Guide & Selected Solutions Manual (Volume I: 0-13-227324-1, Volumes U & III: 0-13-227325-X) by Frank Wolfs Student Pocket Companion (0-13-227326-8) by Biman Das Tutorials in Introductory Physics (0-13-097069-7) by Lillian C. McDermott, Peter S. Schaffer, and the Physics Education Group at the University of Washington Physlet® Physics (0-13-101969-4) by Wolfgang Christian and Mario Belloni Ranking Task Exercises in Physics, Student Edition (0-13-144851-X) by Thomas L. O’Kuma, David P. Maloney, and Curtis J. Hieggelke E&M TIPERs: Electricity & Magnetism Tasks Inspired by Physics Education Research (0-13-185499-2) by Curtis J. Hieggelke, David P. Maloney, Stephen E. Kanim, and Thomas L. O ’Kuma Mathematics for Physics with Calculus (0-13-191336-0) by Biman Das
PREFACE xvii
xviii PREFACE
To Students
HOW TO STUDY
1. Read the Chapter. Learn new vocabulary and notation. Try to respond to questions and exercises as they occur.
2. Attend all class meetings. Listen. Take notes, especially about aspects you do not remember seeing in the book. Ask questions (everyone else 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.
4. Solve 10 to 20 end of Chapter Problems (or more), especially those assigned. In doing Problems you 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 won’t cut it.
NOTES ON THE FORMAT AND PROBLEM SOLVING
1. Sections marked with a star (*) are considered optional. 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: F.
3. Few equations are valid in all situations. Where practical, the limitations 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 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, which are not arranged by Section nor ranked as to difficulty. Problems that relate to optional Sections are starred (*). Most Chapters have 1 or 2 Computer/Numerical Problems at the end, requiring a computer or graphing calculator. 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 and their solutions in the text, 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) special Problem Solving Strategies placed throughout the text to suggest a step-by-step approach to problem solving for a particular topic—but remember that 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 imme­ diately, 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 (point 4 above).
6. Conceptual Examples pose a question which hopefully starts you to think and come up with a response. Give yourself a little time to come up with your own response before reading the Response given.
7. Math review, plus some additional topics, are found in Appendices. Useful data, conversion factors, and math formulas are found inside the front and back covers.
USE OF COLOR
Vectors
A general vector resultant vector (sum) is slightly thicker components of any vector are dashed
Displacement (D, ?) Velocity (v) Acceleration (a) Force (F)
Force on second or third object in same figure Momentum (p ormv) Angular momentum (L) Angular velocity (to)
Torque ( f ) Electric field (E) Magnetic field (B)
Electricity and magnetism
Electric circuit symbols
1l TT
Electric field lines Equipotential lines Magnetic field lines Electric charge (+) Electric charge (-)
+ ) or • + Q or • -
Wire, with switch S
S
Resistor
-v w v -
Capacitor
Inductor Battery
-/n n n p -
Optics
Light rays
— *—
Object 1
Real image (dashed)
4■■■
Virtual image (dashed and paler)
4■■■
Ground
x
Other
Energy level (atom, etc.)
Measurement lines h—1.0 m—H
Path of a moving object
------------
Direction of motion -------► or current
PREFACE
Image of the Earth from a NASA satellite. The sky appears black from out in space 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
•* 4
Chapter 35.) Note the storm off the coast of Mexico.
Introduction,
Measurement, Estimating
CHAPTER-OPENING QUESTION —Guess now! 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) Give up; it is impossible using ordinary means. (b) Use an extremely long measuring tape. (c) It is only possible by flying high enough to see the actual curvature of the Earth. (d) Use a standard measuring tape, a step ladder, and a large smooth lake. (e) Use a laser and a mirror on the Moon or on a satellite.
\We start each Chapter with a Question, like the one above. Try to answer it 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 usually 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 to see the power and usefulness of physics. ]
CONTENTS
1-1 The Nature of Science 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
(b) FIGURE 1-1 (a) This Roman aqueduct was built 2000 years ago and still stands, (b) The Hartford Civic Center collapsed in 1978, just two years after it was built.
2 CHAPTER 1
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 classicalphysics which includes motion, fluids, heat, sound, light, electricity and magnetism; and modem 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 crucial 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-la). Indeed, in Chapter 12 we will see a worked-out Example of how a simple physics calculation—or even intuition based on understanding the physics of forces—would have saved hundreds of lives (Fig. 1-lb). We will see many examples in this book of how physics is useful in many fields, and in everyday life.
1—1 The Nature of Science
The principal aim of all sciences, including physics, is generally considered to be the search for order in our observations of the world around us. Many people think that science is a mechanical process of collecting facts and devising theories. But it is not so simple. Science is a creative activity that in many respects resem­ bles other creative activities of the human mind.
One important aspect of science is observation of events, which includes the design and carrying out of experiments. But observation and experiment require imagination, for scientists can never include everything in a description of what they observe. Hence, scientists must make judgments about what is relevant in their observations and experiments.
Consider, for example, how two great minds, Aristotle (384-322 b.c.) and Galileo (1564-1642), interpreted motion along a horizontal surface. Aristotle noted that objects given an initial push along the ground (or on a tabletop) always slow down and stop. Consequently, Aristotle argued that the natural state of an object is to be at rest. Galileo, in his reexamination of horizontal motion in the 1600s, 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.
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.
Although the testing of theories distinguishes science from other creative fields, it should not be assumed that a theory is “proved” by testing. First of all, no measuring instrument is perfect, so exact confirmation is not possible. Further­ more, it is not possible to test a theory in every single possible circumstance. Hence a theory cannot be absolutely verified. Indeed, the history of science tells us that long-held theories can be replaced by new ones.
1 -2 Models, Theories, and Laws
When scientists are trying to understand a particular set of phenomena, 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 picture— something to hold on to—when we cannot see what actually is happening. 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, however, not to confuse a model or a theory with the real system or the phenomena themselves.
Scientists give the title law to certain concise but general statements about how nature behaves (that energy is conserved, for example). Sometimes the state­ ment 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 principle is often used (such as Archimedes’ principle).
Scientific laws are different from political laws in that the latter 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 cases, 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.
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 an instrument 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. 1-2), 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
1%,
where ~ means “is approximately equal to.’
FIGURE 1-2 Measuring the width of a board with a centimeter ruler. The uncertainty is about ± 1 mm.
SECTION 1-3 3
(a)
(b)
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 would be 0.67. In (b), 2.5 was multiplied by 3.2. The correct result is 8.0.
p PROBLEM SOLVING
Significant figure rule: N um ber o f significant figures in final
result should be same as the least significant input value
A CAUTION
Calculators err with significantfigures
I PROBLEI VI S OL V I N G Report only the proper number o f
significant figures in the final result. Keep extra digits during the calculation
FIGURE 1 -4 Example 1-1. A protractor used to measure an angle.
4 CHAPTER 1
Often the uncertainty in a measured value is not specified explicitly. In such cases, the uncertainty is generally assumed to be one or a few units in the last digit specified. For example, if a length is given as 8.8 cm, the uncertainty is assumed to be about 0.1 cm or 0.2 cm. It is important in this case that you do not write 8.80 cm, for this implies an uncertainty on the order of 0.01 cm; it assumes that the length is probably between 8.79 cm and 8.81 cm, when actually you believe it is between 8.7 and 8.9 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 signif­ icant 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 is 80 km within an accuracy of about 1 or 2 km, and then the 80 has two significant figures. If it is precisely 80 km, to within + 0.1 km, then we write 80.0 km (three significant figures).
When making measurements, or when doing calculations, you should avoid the temptation to keep more digits in the final answer than is justified. For example, to calculate the area of a rectangle 11.3 cm by 6.8 cm, the result of multiplication would be 76.84 cm2. But this answer is clearly not accurate to 0.01 cm2, since (using the outer limits of the assumed uncertainty for each measurement) the result could be between 11.2 cm X 6.7 cm = 75.04 cm2 and 11.4 cm X 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 because they are not significant. As a rough general rule (i.e., in the absence of a detailed consideration of uncertainties), we can say that the final result o f a multiplication or division should have only as many digits as the number with the least number o f significant figures used in the calculation. 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.
When adding or subtracting numbers, the final result is no more precise than the least precise number used. For example, the result of subtracting 0.57 from 3.6 is 3.0 (and not 3.03).
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 some such thing as 0.666666666. 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 o ff only in the final result. (With a calculator, you can keep all its digits in intermediate results.) Note also that calculators sometimes give too few significant figures. For example, when you multiply 2.5 X 3.2, a calculator may give the answer as simply 8. But the answer is accurate to two significant figures, so the proper answer is 8.0. See Fig. 1-3.
CONCEPTUAL EXAMPLE 1-1 | Significant figures. Using a protractor (Fig. 1-4), 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 number like 0.866025403. However, 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 Cosine and other trigonometric functions are reviewed in Appendix A.
| EXERCISE B Do 0.00324 and 0.00056 have the same number of significant figures?
Be careful not to confuse significant figures with the number of decimal places.
EXERCISE C 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.
Scientific_Notation
We commonly write numbers in “powers of ten,” or “scientific” notation—for instance 36,900 as 3.69 X 104, or 0.0021 as 2.1 X 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 X 104, but if it is known to four, we write 3.690 X 104.
I 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 « 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. Now 92 + 1 and 97 + 1 both imply an uncertainty of about 1% (1/92 « 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 « 0.1 ~ 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 « 0.01 ~ 1%, just like the uncertainty in the original numbers 92 and 97.
SUGGESTION: Use the significant figures rule, but consider the % uncer­ tainty 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 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.” Predsion 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. 1-2 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.
SECTION 1- 3 Measurement, Uncertainty; Significant Figures 5
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. l- 5 a ] Sheet of paper
(thickness) Finger width Football field length Height of Mt. Everest
[see Fig. l- 5 b ] Earth diameter Earth to Sun Earth to nearest star Earth to nearest galaxy Earth to farthest
galaxy visible
10-15
-7
10-4 10“2
102
104 107
1011
1016 1022
1026
FIGURE 1 - 5 Som e lengths: (a) viruses (about 10-7 m long) attacking a cell; (b) Mt. E verest’s height is on the order of 104 m (8850 m, to be precise).
(a)
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 meaningless. The unit must be given; for clearly, 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 anyone needing to make a very accurate measurement can refer to the standard in the laboratory.
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 stan­ dard meter was originally chosen to be one ten-millionth of the distance from the Earth’s equator to either pole,f 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 to the side.) In 1889, the meter was defined more precisely as the distance between two finely engraved marks on a particular bar of platinum-iridium 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 lm /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.” *
British units of length (inch, foot, mile) are now defined in terms of the meter. The inch (in.) is defined as precisely 2.54 centimeters (cm; 1 cm = 0.01 m). Other conversion factors are given in the Table on the inside of the front cover of this book. Table 1-1 presents some typical lengths, from very small to very large, rounded off to the nearest power of ten. See also Fig. 1-5. [Note that the abbreviation for inches (in.) is the only one with a period, to distinguish it from the word “in”.]
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 (24h/day X 60min/h X 60s/min = 86,400 s/day). The standard second is now defined more precisely in terms of the frequency of radi­ ation emitted by cesium atoms when they pass between two particular states. [Specifically, one second is defined as the time required for 9,192,631,770 periods of this radiation.] There are, by definition, 60 s in one minute (min) and 60 minutes in one hour (h). Table 1-2 presents a range of measured time intervals, rounded off to the nearest power of ten.
Mass
The standard unit of mass is the kilogram (kg). The standard mass is a particular platinum-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 1-3. [For practical purposes, 1 kg weighs about 2.2 pounds on Earth.]
tModern measurements of the Earth’s circumference reveal that the intended length is off by about one-fiftieth of 1%. Not bad!
*The new definition of the meter has the effect of giving the speed of light the exact value of
(b)
299,792,458 m/s.
6 CHAPTER 1 Introduction, Measurement, Estimating
TABLE 1-2 Some Typical Time Intervals
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 A ge of Universe
1 0 -23 s 10~22 s to 1028 s 1(T6 s 10° s (= 1 s)
105 s
3 X 107 2 X 109
1011
1014 1017 1018
TABLE 1-3 Some Masses
Object
Kilograms (approximate)
Electron Proton, neutron D N A molecule Bacterium M osquito Plum Human Ship Earth Sun Galaxy
1(T30 kg 10-27 kg 1(T17 kg 1(T15 kg 1(T5 kg 10"1 kg 102 kg 108 kg 6 X 1024 kg 2 X 1030 kg 1041 kg
When dealing with atoms and molecules, we usually use the unified atomic mass unit (u). In terms of the kilogram,
l u = 1.6605 X 10-27kg.
The definitions of other standard units for other quantities will be given as we encounter them in later Chapters. (Precise values of this and other numbers are given inside the front cover.)
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 kilometer (km)
is 1000 m, 1 centimeter is ifem, 1 millimeter (mm) is
or ^cm , and so on.
The prefixes “centi-,” “kilo-,” and others are listed in Table 1-4 and can be applied
not only to units of length but to units of volume, mass, or any other metric unit.
For example, a centiliter (cL) is ^ liter (L)> and a kilogram (kg) is 1000 grams (g).
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 Systeme 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 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.
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
microf
V
nano
n
pico
P
femto
f
atto
a
zepto
z
yocto
y
f ju, is the Greek letter “mu.”
1024 1021 1018 1015 1012 109 106 103 102 101 KT1 1(T2 1(T3 1(T6 KT9 1 ( T 12 1 ( T 15 KT18 1(T21 KT24
Base versus Derived Quantities
Physical quantities can be divided into two categories: base quantities and derived quantities. The corresponding units for these quantities are called base units and derived units. A base quantity must be defined in terms of a standard. Scientists, in the interest of simplicity, want the smallest number of base quantities possible consistent with a full description of the physical world. This number turns out to be seven, and those used in the SI are given in Table 1-5. All other quantities can be defined in terms of these seven base quantities/ and hence are referred to as derived quantities. An example of a derived quantity is speed, which is defined as distance divided by the time it takes to travel that distance. A Table inside the front cover lists many derived quantities and their units in terms of base units. To define any quantity, whether base or derived, we can specify a rule or procedure, and this is called an operational definition.
trThe only exceptions are for angle (radians—see Chapter 8) and solid angle (steradian). No general agreement has been reached as to whether these are base or derived quantities.
TABLE 1-5 SI Base Quantities and Units
Quantity
Unit
Unit Abbreviation
Length
meter
m
Time
second
s
Mass
kilogram
kg
Electric
current
ampere
A
Temperature kelvin
K
Amount
of substance m ole
mol
Luminous
intensity candela
cd
SECTION 1- 4 Units, Standards, and the SI System 7
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 table 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 one does not change anything, the width of our table, in cm, is
21.5 inches = (21.5
X ^ 2 . 5 4 ^ ^ = 54.6 cm.
Note how the units (inches in this case) cancelled out. A Table containing many unit conversions is found inside the front cover of this book. Let’s consider some Examples.
0 PHYSICS APPLIED
The world’s tallest peaks
FIGURE 1 - 6 The w orld ’s second highest peak, K2, w hose summit is considered the m ost difficult of the “8000-ers.” K2 is seen here from the north (China).
TABLE 1-6 The 8000-m Peaks
Peak
Height (m)
Mt. Everest K2 Kangchenjunga Lhotse M akalu Cho Oyu Dhaulagiri M anaslu 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
EXAMPLE 1 -2 The 8000-m peaks. The fourteen tallest peaks in the world (Fig. 1-6 and Table 1-6) are referred to as “eight-thousanders,” meaning their summits are over 8000 m above sea level. What is the elevation, in feet, of an elevation of 8000 m?
APPROACH We need simply 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. SOLUTION One foot is 12 in., so we can write
cm 1 ft = (1 2 is.)(2 .5 4 — J = 30.48 cm = 0.3048 m,
which is exact. Note how the units cancel (colored slashes). We can rewrite this equation to find the number of feet in 1 meter:
l m = a U s = 3'28084ft We multiply this equation by 8000.0 (to have five significant figures):
8000.0m = (8000.0 la .) ^ 3 .2 8 0 8 4 ;^ = 26,247ft.
An elevation of 8000 m is 26,247 ft above sea level. NOTE We could have done the conversion all in one line:
annum -
- 26O T «.
The key is to multiply conversion factors, each equal to one (= 1.0000), and to make sure the units cancel.
EXERCISE E There are only 14 eight-thousand-m eter peaks in the world (see E xam ple 1 -2 ), and their names and elevations are given in Table 1 -6 . They are all in the H imalaya m oun­ tain range in India, Pakistan, Tibet, and China. D eterm in e the elevation o f the w orld ’s three highest peaks in feet.
8 CHAPTER 1 Introduction, Measurement, Estimating
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 lin. = 2.54cm = 0.0254m, then lft2= (12 in.)2(0.0254 m/in.)2 = 0.0929 m2. So 880 ft2 = (880ft2)(0.0929 m2/f t2) « 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 X).
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 ft in a mile and 12 inches in a foot; also, one hour contains (60min/h) X (60s/min) = 3600 s/h.
SOLUTION (a) We can write 1 mile as
1 mi = (5280ir)(
jGirr 1 m 2.54 'TRv. / \ 100 jGfTf = 1609 m.
We also know that 1 hour contains 3600 s, so
55—h =
'mi. 55
ir
m 1609
1 JT
~mLJ V3600 s
where we rounded off to two significant figures. (b) Now we use 1 mi = 1609 m = 1.609 km; then
= 25“ , s
55—h =
'm i.
55
km 1.609 'm i
_km - 88- .
NOTE Each conversion factor is equal to one. You can look up most conversion factors in the Table inside the front cover.
j PROBLEM SOLVING Conversion factors = 1
EXERCISE F Would a driver traveling at 15 m /s in a 35 m i/h zone be exceeding the speed limit?
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 1-4(a), if we had incorrectly used the factor ( n ^ ) instead of (ujoSn), the centimeter units would not have cancelled out; we would not have ended up with meters.
\ PROBLEM SOLVING Unit conversion is wrong if units do not cancel
1—6 Order of Magnitude: Rapid Estimating
We are sometimes interested only in an approximate value for a quantity. This might be because an accurate calculation would take more time than it is worth or would require additional data that are not available. In other cases, we may want to make a rough estimate in order to check an accurate calculation made on a calculator, to make sure that no blunders were made when the numbers were entered.
A rough estimate is made by rounding off all numbers to one significant figure and its power of 10, and after the calculation is made, again only one significant figure is kept. 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.
^ \ \ PROBLEM SOLVING
H ow to m ake a rough estimate
SECTION 1- 6 Order of Magnitude: Rapid Estimating 9
10 m
r = 500 in
© - PHYSICS APPLIED
Estim ating the volum e (or m ass) o f a lake; see also Fig. 1 - 7
(b)
FIGURE 1 - 7 E xam ple 1 - 5 . (a) H ow m uch water is in this lake? (Photo is of one of the R ae Lakes in the Sierra N evada of California.) (b) M odel of the lake as a cylinder. [We could go one step further and estim ate the mass or weight of this lake. We will see later that w ater has a density o f 1000 k g /m 3, so this lake has a mass of about (I0 3 k g /m 3) ( l0 7 m 3) « 1010 kg, which is about 10 billion kg or 10 m illion metric tons. (A m etric ton is 1000 kg, about 2200 lbs, slightly larger than a British ton, 2000 lbs.)]
EXAMPLE 1 -5 ESTIMATE I Volume of a lake. Estimate how much water there is in a particular lake, Fig. l-7 a , 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 bottom. 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. l-7 b ). SOLUTION The volume V of a cylinder is the product of its height h times the area of its base: V = hirr2, where r is the radius of the circular base.f The radius r is \ km = 500 m, so the volume is approximately
V = hirr2 M (10 m) X (3) X (5 X 102m)2 m 8 X 106m3 « 107m3,
where tt was rounded off to 3. So the volume is on the order of 107m3, ten million cubic meters. Because of all the estimates that went into this calculation, the order-of-magnitude estimate (l07m3) is probably better to quote than the 8 X 106m3 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 « \ gallon. Hence, the lake contains (8 X 106m3)(l gallon/4 X 10_3m3) « 2 X 109 gallons of water.
PROBLEM SOLVING Use sym m etry when possible
EXAMPLE 1 -6 ESTIMATE I Thickness of a page. Estimate the thickness of a page of this book.
APPROACH At first you might think that a special measuring device, a micrometer (Fig. 1-8), is needed to measure the thickness of one page since an ordinary ruler clearly won’t do. 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,
fFormulas like this for volume, area, etc., are found inside the back cover of this book.
10 CHAPTER 1 Introduction, Measurement, Estimating
counted front and back, is 250 separate sheets of paper. So one page must have a thickness of about
1.5 cm 250 pages
6 X 10-3 cm = 6 X 10“2mm,
or less than a tenth of a millimeter (0.1 mm).
EXAMPLE 1 -7 ESTIMATE- ! Height by triangulation. Estimate the height of the building shown in Fig. 1-9, 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. l-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. l-9 a). 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. l-9 b using these measurements. You can measure, right on the diagram, the last side of the triangle to be about x = 13 m. Alternatively, you can use similar triangles to obtain the height x :
1.5 m 2m
so 18 m
13 4 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.
FIGURE 1 - 8 Example 1-6. Micrometer used for measuring small thicknesses.
FIGURE 1 - 9 Exam ple 1 -7 . Diagrams are really useful!
EXAMPLE 1 -8 ESTIMATE I 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 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 ~ 6.1 km. Use Fig. 1-10 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.
SOLUTION For the right triangle of Fig. 1-10, the two sides are the radius of the Earth R and the distance d = 6.1 km = 6100 m. The hypotenuse is approxi­ mately the length R + h, where h = 3.0 m. By the Pythagorean theorem,
R2 + d2 « (R + h)2
« R2 + 2hR + h2.
We solve algebraically for R , after cancelling R 2 on both sides:
d2 - h2 2 h
(6100 m)2 - (3.0 m)2 = 6.2 X 106m = 6200 km.
6.0 m
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. Now you know the answer to the Chapter-Opening Question on p. 1.
18m
FIGURE 1 - 1 0 Exam ple 1 -8 , but not to scale. You can see small rocks at water level on the opposite shore o f a lake 6.1 km wide if you stand on a stepladder.
SECTION 1- 6 Order of Magnitude: Rapid Estimating 11
EXAMPLE 1-9 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 70beats/min. But during exercise it can be a lot higher. A reasonable average might be 80 beats/min.
SOLUTION One year in terms of seconds is (24h)(3600s/h)(365 d) « 3 X 107s. If an average person lives 70 years = (70yr)(3 X 107s/yr) « 2 X 109s, then the total number of heartbeats would be about
or 3 trillion.
:) ( 2 x 109 s) « 3 x 109, min / \ 60 s
j PROBLEM SOLVING Estimating h ow many piano tuners
there are in a city
Another technique for estimating, this one made famous by Enrico Fermi to his physics students, is to estimate the number of piano tuners in a city, say, 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 700,000 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 70,000 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 70,000 pianos, needs about 70 piano tuners. This is, of course, only a rough estimated It tells us that there must be many more than 10 piano tuners, and surely not as many as 1000.
1—7 Dimensions and Dimensional Analysis
When we speak of the dimensions of a quantity, we are referring to the type of base units or base quantities 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 = \bh, whereas the area of a circle of radius r is A = irr2. 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 numerical calculations, the units must also be the same on both sides of an equation.)
For example, suppose you derived the equation v = v0 + I at2, where v is the speed of an object after a time t, v0 is the object’s initial speed, and the object undergoes an acceleration a. Let’s do a dimensional check to see if this equation
tA check of the San Francisco Yellow Pages (done after this calculation) reveals about 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. *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.
12 CHAPTER 1 Introduction, Measurement, Estimating
could be correct or is surely incorrect. Note that numerical factors, like the \ here, do not affect dimensional checks. We write a dimensional equation as follows, remembering that the dimensions of speed are [L/T\ and (as we shall see in Chapter 2) the dimensions of acceleration are [L /T 2]:
.?]1M+ +
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’t tell you if it is completely right. For example, a dimensionless numerical factor (such as \ or 2t t ) could be missing.
Dimensional analysis can also be used as a quick check on an equation you are not sure about. For example, suppose that you can’t remember whether the equa­ tion for the period of a simple pendulum T (the time to make one back-and-forth swing) of length i is T = 2ttV tj g or T = 2ttV g /l, where g is the acceleration 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 14; what we are concerned about here is a person’s recalling whether it contains £/g or g/L) A dimensional check shows that the former (i/g ) is correct:
[r| - 'J\ S k ] - v W - m ,
whereas the latter (g/l) is not:
m *
'[l / t 2] = n r = 1
[l ]
\I[ r -} [t ]
Note that the constant 2tt has no dimensions and so can’t be checked using dimensions. Further uses of dimensional analysis are found in Appendix C.
[ 2 5 J 2 I 2 H H H 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 X 108m/s, the gravitational constant G = 6.67 X 10-11 m3/kg •s2, and Planck’s constant h = 6.63 X 10_34kg*m2/s. The Planck length AP (A is the Greek letter “lambda”) is given by the following combination of these three constants:
AP —
Show that the dimensions of APare length [L], and find the order of magnitude of AP. APPROACH We rewrite the above equation in terms of dimensions. The dimen­ sions of c are [L/T], of G are [L3/M T 2], and of h are [ML2/T]. SOLUTION The dimensions of AP are
|L,/”
rl- v P I- w
which is a length. The value of the Planck length is
/Gfc
(6.67 X 10-11m3A g -s2)(6.63 X 10-34kg-m2/s)
Ap = A/ —r = a / ------------------- ;-----------:----- ^
~
VC3 V
(3.0 X 10sm/s)3
4 X 10 "m ,
which is on the order of 10-34 or 10-35m. 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_35m. 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.
*SECTION 1 -7 Dimensions and Dimensional Analysis 13
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.]
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. 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.
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 partic­ ular standard or unit, and the unit used should always be stated. The commonly accepted set of units today is the Systeme 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]. Dimensional analysis can be used to check a relationship for correct form.]
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 (ib) any person’s foot. Keep in mind that it is advantagous that fundamental standards be accessible (easy to compare to), invariable (do not change), indestructible, and reproducible.
2. Why is it incorrect to think that the more digits you represent in your answer, the more accurate it is?
3. When traveling a highway in the mountains, you may see elevation signs that read “914 m (3000 ft).” Critics of the metric system claim that such numbers show the metric system is more complicated. How would you alter such signs to be more consistent with a switch to the metric system?
4. What is wrong with this road sign:
Memphis 7 mi (11.263 km)?
5. For an answer to be complete, the units need to be speci­ fied. Why?
6. Discuss how the notion of symmetry could be used to estimate the number of marbles in a 1-liter jar.
7. 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.
8. Express the sine of 30.0° with the correct number of significant figures.
9. A recipe for a souffle specifies that the measured ingredients must be exact, or the souffle will not rise. The recipe calls for 6 large eggs. The size of “large” eggs can vary by 10%, according to the USDA specifications. What does this tell you about how exactly you need to measure the other ingredients?
10. List assumptions useful to estimate the number of car mechanics in (a) San Francisco, (b) your hometown, and then make the estimates.
11. Suggest a way to measure the distance from Earth to the Sun.
*12. Can you set up a complete set of base quantities, as in Table 1-5, that does not include length as one of them?
| 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) Problems are meant mainly as a challenge for the best students, for “extra credit.” The Problems are arranged by Sections, meaning that the reader should have read up to and including that Section, but not only that Section—Problems often depend on earlier material. Each Chapter also has a group of General Problems that are 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 + 10 unless 950 is said to be “precisely” or “very nearly” 950, in which case assume 950 + 1.)
1. (I) The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this in powers of ten in (a) years, (b) seconds.
2. (I) How many significant figures do each of the following numbers have: (a) 214, (b) 81.60, (c) 7.03, (d) 0.03, (e) 0.0086, ( /) 3236, and (g) 8700?
3. (I) Write the following numbers in powers of ten notation: (a) 1.156, (b) 21.8, (c) 0.0068, (d) 328.65, (e) 0.219, and ( /) 444.
4. (I) Write out the following numbers in full with the correct number of zeros: (a) 8.69 X 104, (b) 9.1 X 103, (c) 8.8 X 10_1, (d) 4.76 X 102, and (e) 3.62 X 10“5.
5. (II) What is the percent uncertainty in the measurement 5.48 ± 0.25 m?
6. (II) Time intervals measured with a 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) 5 s, (b) 50 s, (c) 5 min?
7. (II) Add (9.2 X 103s) + (8.3 X 104s) + (0.008 X 106s).
14 CHAPTER 1 Introduction, Measurement, Estimating
8. (II) Multiply 2.079 X 102m by 0.082 X 10-1, taking into account significant figures.
9. (Ill) For small angles 6, the numerical value of sin 0 is approximately the same as the numerical value of tan0. Find the largest angle for which sine and tangent agree to within two significant figures.
10. (Ill) What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is r = 0.84 ± 0.04 m?
1-4 and 1-5 Units, Standards, SI, Converting Units
11. (I) Write the following as full (decimal) numbers with stan­ dard units: (a) 286.6 mm, (b) 85 /jlV, ( c ) 760 mg, (d) 60.0 ps, (e) 22.5 fm, ( /) 2.50 gigavolts.
12. (I) Express the following using the prefixes of Table 1-4: (a) 1 X 106volts, (b) 2 X 10-6 meters, (c) 6 X 103days, (id) 18 X 102bucks, and (e) 8 X 10-8 seconds.
13. (I) Determine your own height in meters, and your mass in kg. 14. (I) The Sun, on average, is 93 million miles from Earth. How
many meters is this? Express (a) using powers of ten, and (b) using a metric prefix. 15. (II) What is the conversion factor between (a) ft2 and yd2, (b) m2 and ft2? 16. (II) An airplane travels at 950 km /h. How long does it take to travel 1.00 km? 17. (II) A typical atom has a diameter of about 1.0 X 10-10m. (a) What is this in inches? (b) Approximately how many atoms are there along a 1.0-cm line? 18. (II) Express the following sum with the correct number of significant figures: 1.80 m + 142.5 cm + 5.34 X 105/xm. 19. (II) Determine the conversion factor between (a) km /h and m i/h, (b) m /s and ft/s, and (c) km /h and m/s. 20. (II) How much longer (percentage) is a one-mile race than a 1500-m race (“the metric mile”)? 21. (II) A light-year is the distance light travels in one year (at speed = 2.998 X 108m /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 X 108km. How many AU are there in 1.00 light-year? (c) What is the speed of light in A U /h? 22. (II) If you used only a keyboard to enter data, how many years would it take to fill up the hard drive in your computer that can store 82 gigabytes (82 X 109bytes) of data? Assume “normal” eight-hour working days, and that one byte is required to store one keyboard character, and that you can type 180 characters per minute. 23. (Ill) The diameter of the Moon is 3480 km. (a) What is the surface area of the Moon? (b) How many times larger is the surface area of the Earth?
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.) 24. (I) Estimate the order of magnitude (power of ten) of: (a) 2800,
(b) 86.30 X 102, (c) 0.0076, and (d) 15.0 X 108. 25. (II) Estimate how many books can be shelved in a college
library with 3500 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. 26. (II) Estimate how many hours it would take a runner to run (at 10 km/h) across the United States from New York to California. 27. (II) Estimate the number of liters of water a human drinks in a lifetime.
28. (II) Estimate how long it would take one person to mow a football field using an ordinary home lawn mower (Fig. 1-11). Assume the mower moves with a 1-km/h speed, and has a 0.5-m width.
FIGURE 1-11 Problem 28.
29. (II) Estimate the number of dentists (a) in San Francisco and (b) in your town or city.
30. (Ill) The rubber worn from tires mostly enters the atmos­ phere 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.
31. (Ill) You are in a hot air balloon, 200 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.
32. (Ill) I agree to hire you for 30 days and you can decide between two possible 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 to day 30. Use quick estimation to make your decision, and justify it.
33. (Ill) Many sailboats are moored 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. 1-12, where h = 1.5 m, esti­
mate the radius R of the Earth.
h-------- d
FIGURE 1-12 Problem 33. You see a sailboat across a lake (not to scale). R is the radius of the Earth. You are a distance d = 4.4 km from the sailboat when you can see only its deck and not its side. Because of the curvature of the Earth, the water “bulges out” between you and the boat.
34. (Ill) Another experiment you can do also uses the radius of the Earth. The Sun sets, fully disappearing over the horizon as you lie on the beach, your eyes 20 cm above the sand. You immediately jump up, your eyes now 150 cm above the sand, and you can again see the top of the Sun. If you count the number of seconds (= t) until the Sun fully disappears again, you can estimate the radius of the Earth. But for this Problem, use the known radius of the Earth and calculate the time t.
Problems 15
1-7 Dimensions
*35. (I) What are the dimensions of density, which is mass per volume?
*36. (II) The speed v of an object is given by the equation v = A t3 —Bt, where t refers to time, (a) What are the dimensions of A and 5 ? (b) What are the SI units for the constants A and 5?
*37. (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: (a) x = vt2 + 2at, (b) x = v0t + \a t2, and (c) x = v0t + 2at2. Which of these could possibly be correct according to a dimensional check?
1538. (II) Show that the following combination of the three funda­ mental constants of nature that we used in Example 1-10 (that is G, c, and h) forms a quantity with the dimensions of time:
h =
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.
| General Problems___________
39. Global positioning satellites (GPS) can be used to deter­ mine positions with great accuracy. If one of the satellites is at a distance of 20,000 km from you, what percent uncertainty in the distance does a 2-m uncertainty represent? How many significant figures are needed in the distance?
40. Computer chips (Fig. 1-13) etched on circular silicon wafers of thickness 0.300 mm are sliced from a solid cylindrical silicon crystal of length 25 cm. If each wafer can hold 100 chips, what is the maximum number of chips that can be produced from one entire cylinder?
48. Estimate the number of gumballs in the machine of Fig. 1-14.
F IG U R E !-13 Problem 40. The wafer held by the hand (above) is shown below, enlarged and illuminated by colored light. Visible are rows of integrated circuits (chips).
41. {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?
42. American football uses a field that is 100 yd long, whereas a regulation soccer field is 100 m long. Which field is longer, and by how much (give yards, meters, and percent)?
43. A typical adult human lung contains about 300 million tiny cavities called alveoli. Estimate the average diameter of a single alveolus.
44. One hectare is defined as 1.000 X 104m2. One acre is 4.356 X 104ft2. How many acres are in one hectare?
45. Estimate the number of gallons of gasoline consumed by the total of all automobile drivers in the United States, per year.
46. Use Table 1-3 to estimate the total number of protons or neutrons in (a) a bacterium, (b) a DNA molecule, (c) the human body, (d) our Galaxy.
47. An average family of four uses roughly 1200 L (about 300 gallons) of water per day (l L = 1000 cm3). How much depth would a lake lose per year if it uniformly covered an area of 50 km2 and supplied a local town with a population of 40,000 people? Consider only population uses, and neglect evaporation and so on.
FIGURE 1-14 Problem 48. Estimate the number of gumballs in the machine.
49. 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.
50. How big is a ton? 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 (lO3cm3) or 62 lb per cubic foot.]
51. A certain audio 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 digital information at a constant rate of 1.4 megabits per second. How many minutes does it take the player to read the entire CD?
52. Hold a pencil in front of your eye at a position where its blunt end just blocks out the Moon (Fig. 1-15). Make appro­ priate measurements to estimate the diameter of the Moon, given that the Earth-M oon distance is 3.8 X 105km.
FIGURE 1-15 Problem 52. How big is the Moon?
53. A heavy rainstorm dumps 1.0 cm of rain on a city 5 km wide and 8 km long in a 2-h period. How many metric tons (l metric ton = 103kg) 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?
16 CHAPTER 1 Introduction, Measurement, Estimating
54. 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).
55. Estimate how many days it would take to walk around the world, assuming 10 h walking per day at 4 km/h.
56. 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 mole­ cules have a diameter of 2 X 10-10 m.
57. Jean camps beside a wide river and wonders how wide it is.
She spots a large rock on the bank directly across from her.
She then walks upstream until she judges that the angle
between her and the rock, which she can still see clearly, is
now at an angle of 30° downstream (Fig. 1-16). Jean
measures her stride
to be about 1 yard
long. The distance
back to her camp is
120 strides. About
how far across, both
in yards and in
meters, is the river?
\m
v |
FIGURE 1-16 Problem 57.
120 Strides
58. A watch manufacturer claims that its watches gain or lose no more than 8 seconds in a year. How accurate is this watch, expressed as a percentage?
59. An angstrom (symbol A ) 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 21)?
60. The diameter of the Moon is 3480 km. What is the volume of the Moon? How many Moons would be needed to create a volume equal to that of Earth?
61. Determine the percent uncertainty in 6, and in sin 6, when (a) 6 = 15.0° ± 0.5°, (b) 6 = 75.0° ± 0.5°.
62. If you began walking along one of E arth’s lines of longi­ tude and walked north 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 called a “nautical mile.”
63. Make a rough estimate of the volume of your body (in m3). 64. Estimate the number of bus drivers (a) in Washington, D.C.,
and (b) in your town. 65. The American Lung Association gives the following formula
for an average person’s expected lung capacity V (in liters, where 1 L = 103cm3):
V = 4.1 H - 0.018A - 2.69,
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 2.69?
66. The density of an object is defined as its mass divided by its volume. Suppose the mass and volume of a rock are measured to be 8 g and 2.8325 cm3. To the correct number of significant figures, determine the rock’s density.
67. To the correct number of significant figures, use the infor­ mation 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.
68. One mole of atoms consists of 6.02 X 1023 individual atoms. If a mole of atoms were spread uniformly over the surface of the Earth, how many atoms would there be per square meter?
69. Recent findings in astrophysics suggest that the observable Universe can be modeled as a sphere of radius R = 13.7 X 109 light-years with an average mass density of about 1 X 10_26kg/m 3, where only about 4% of the Universe’s total mass is due to “ordinary” matter (such as protons, neutrons, and electrons). Use this information to estimate the total mass of ordinary matter in the observable Universe. (1 light-year = 9.46 X 1015m.)
Answers to Exercises
A: (d). B: No: they have 3 and 2, respectively. C: All three have three significant figures, although the
number of decimal places is (a) 2, (b) 3, (c) 4.
D: (a) 2.58 X 10“2, 3; (b) 4.23 X 104, 3 (probably); (c) 3.4450 X 102, 5. Mt. Everest, 29,035 ft; K 2,28,251 ft; Kangchenjunga, 28,169 ft.
F: No: 15 m /s ~ 34 mi/h.
General Problems 17
A high-speed car has released a parachute to reduce its sp eed quickly. The directions o f the car’s velocity and acceleration are shown by the green (v) and gold (a) arrows.
M otion is described using the concepts of velocity and acceleration. In the case shown here, the acceleration a is in the opposite direction from the velocity v, which means the object is slowing down. We exam ine in detail m otion with constant acceleration, including the vertical m otion of objects falling under gravity.
T £
Describing Motion: Kinematics in One Dimension
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 * 2 -9 Graphical Analysis and
Numerical Integration
18
CHAPTER-OPENING QUESTION— Guess now!
[D o n ’t w o r ry a b o u t getting the right an sw er n o w — y o u w ill g et another chance later in the Chapter. See also p. 1 o f C h apter 1 f o r m o re 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.
The motion of objects—baseballs, automobiles, joggers, and even the Sun and M oon—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 (1564-1642) and Isaac Newton (1642-1727).
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.
For now we only discuss objects that move without rotating (Fig. 2 - la). 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.
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 trav­ eling 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. 2-2). 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)
<h)
FIGURE 2 - 1 The pinecone in (a) undergoes pure translation as it falls, whereas in (b) it is rotating as w ell as translating.
FIGURE 2 - 2 A person walks toward the front of a train at 5 km /h. The train is moving 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 the cardinal points, north, east, south, and west, and by “up” and “down.” In physics, we often draw a set of coordinate axes, as shown in Fig. 2-3, 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. Objects positioned to the right of the origin of coordinates (0) on the x axis have an x coordinate which we usually choose to be positive; then 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 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.
+:v
+ X
-y
SECTION 2-1 Reference Frames and Displacement 19
A CAUTION
The displacement m ay not equal the total distance traveled
70 m 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 — x\ . Distances are in meters.
*1
*2
10 20 30 40 Distance (m)
FIGURE 2 - 6 For the displacem ent Ax = x 2 — x 1 = 10.0 m - 30.0 m, the displacement vector points to the left.
x2
xl
jk— Ax —
10 20 30 40 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. 2-4). The total distance traveled is 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 quan­ tities are called vectors, and are represented by arrows in diagrams. For example, in Fig. 2-4, 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 have a positive sign, whereas vectors that point in the opposite direc­ tion will have a negative sign, along with their magnitude.
Consider the motion of an object over a particular time interval. Suppose that at some initial time, call it tx, the object is on the x axis at the position x 1 in the coordinate system shown in Fig. 2-5. At some later time, t2, suppose the object has moved to position x2. The displacement of our object is x2 - x1, and is repre­ sented by the arrow pointing to the right in Fig. 2-5. It is convenient to write
Ax = x2 — x x,
where the symbol A (Greek letter delta) means “change in.” Then Ax means “the change in x,” or “change in position,” which is the displacement. Note that the “change in” any quantity means the final value of that quantity, minus the initial value.
Suppose x l = 10.0 m and x2 = 30.0 m. Then
Ax = x2 - x 1 = 30.0 m - 10.0 m = 20.0 m,
so the displacement is 20.0 m in the positive direction, Fig. 2-5. Now consider an object moving to the left as shown in Fig. 2-6. Here the
object, say, a person, starts at xx = 30.0 m and walks to the left to the point x2 = 10.0 m. In this case her displacement is
Ax = x2 - Xj = 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 has a positive sign, whereas a vector pointing to the left has a negative sign.
EXERCISE A A n ant starts at x = 20 cm on a piece of graph paper and walks along the x axis to x = - 2 0 cm. It then turns around and walks back to x = - 1 0 cm. What is the ant’s displacem ent and total distance traveled?
2 —2 Average Velocity
The most obvious 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:
distance traveled average speed = —ti:-m--e--e-l--a-p--s-e-d-—
(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
20 CHAPTER 2 Describing Motion: Kinematics in One Dimension
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.) There is a second difference between speed and velocity: namely, the average velocity is defined in terms of displacement, rather than total distance traveled:
displacement final position - initial position
average velocity = —------------- - = -------------;------------- --------------
time elapsed
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. 2-4, 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:
---d-i-s-t-a--n-c--e-- - = —100—m = 1„.4J m/s. time elapsed 70 s
The magnitude of the average velocity, on the other hand, was:
displacement 40 m ti.me e,l-a-p--s-e-d- = —70—s = 0.57 m/s.
This difference between the speed and the magnitude of the velocity can occur when we calculate average values.
To discuss one-dimensional motion of an object in general, suppose that at some moment in time, call it , the object is on the x axis at position jcxin a coor­ dinate system, and at some later time, t2, suppose it is at position x2. The elapsed time is At = t2 — t^\ during this time interval the displacement of our object is Ax = x2 ~ x 1. Then the average velocity, defined as the displacement divided by the elapsed time, can be written
/j\ CAUTION________
Average speed is not necessarily equal to the magnitude o f the average velocity
=
= A*,
t2 - tx At
where v stands for velocity and the bar (- ) over the v is a standard symbol meaning “average.”
For the usual case of the +x axis to the right, note that if x2is less than x x, the object is moving to the left, and then Ax = 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.
Note that it is always important to choose (and state) the elapsed time, or time interval, t2 — tx, the time that passes during our chosen period of observation.
\ \ PROBLEM SOLVING
+ or - sign can signify the direction fo r linear motion
Runner's average velocity. The position of a runner as a function of time 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 xx = 50.0 m to x2 = 30.5 m, as shown in Fig. 2-7. What was 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 Ax = x2 - x x = 30.5 m - 50.0 m = -19.5 m. The elapsed time, or time interval, is At = 3.00 s. The average velocity is
Ax -19.5 m
,
v = —— = ———— = -6.50 m/s.
At
3.00 s
'
The displacement and average velocity are negative, which tells us that the runner is moving to the left along the x axis, as indicated by the arrow in Fig. 2-7. Thus we can say that the runner’s average velocity is 6.50 m/s to the left.
FIGURE 2 - 7 Exam ple 2 -1 . A person runs from x i = 50.0 m to x2 = 30.5 m. The displacem ent is -1 9 .5 m.
y
Finish Start (x2) (x{)
^trA x -k — k — \— \— r« i v— u
u 10 20 30 40 50 60 Distance (m)
SECTION 2 - 2 Average Velocity 21
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, so we solve Eq. 2-2 for Ax. SOLUTION We rewrite Eq. 2-2 as Ax = v At, and find
Ax = v At = (18km/h)(2.5 h) = 45 km.
EXERCISE B A car travels at a constant 50 k m /h for 100 km. It then speeds up to 100 k m /h and is driven another 100 km. What is the car’s average speed for the 200 km trip? (a) 67 km /h; (b) 75 km /h; (c) 81 km /h; (d) 50 k m /h .
2 —3 Instantaneous Velocity
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 magni­ tude is the number, with units, indicated by a speedometer, Fig. 2-8.) More precisely, the instantaneous velocity at any moment is defined as the average velocity over an infinitesimally short time interval. That is, Eq. 2-2 is to be evalu­ ated in the limit of At becoming extremely small, approaching zero. We can write the definition of instantaneous velocity, v, for one-dimensional motion as
Ax
=
lim
Af->0
—A—t ■
(2-3)
FIGURE 2 - 8 Car speedom eter showing m i/h in white, and km /h in orange.
The notation limA^ 0 means the ratio A x/A t is to be evaluated in the limit of At approaching zero. But we do not simply set At = 0 in this definition, for then Ax would also be zero, and we would have an undefined number. Rather, we are considering the ratio A x/A t, as a whole. As we let At approach zero, Ax approaches zero as well. But the ratio A x/A t approaches some definite value, which is the instantaneous velocity at a given instant.
In Eq. 2-3, the limit as At —> 0 is written in calculus notation as dx/dt and is called the derivative of x with respect to t:
FIGURE 2 - 9 Velocity o f a car as a function o f time: (a) at constant velocity; (b) with varying velocity.
60
&40-
•3 20--
> 0 H----- 1----- 1----- h
0 0.1 0.2 0.3 0.4 0.5
(a)
Time (h)
60-
Average velocity
| 20 J
\
^ n I---- 1----- u
0 0.1 0.2 0.3 0.4 0.5
(b)
Time (h)
dx
=
lim
Af->0
^A
t
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 instan­ taneous velocity. Why? Because distance traveled and the magnitude of the displacement become the same when they become infinitesimally small.
If an object moves at a uniform (that is, constant) velocity during a particular time interval, then its instantaneous velocity at any instant is the same as its average velocity (see Fig. 2-9a). 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 destina­ tion after traveling a total of 15 km in 30 min. This trip is plotted on the graph of Fig. 2-9b. Also shown on the graph is the average velocity (dashed line), which is v = A x/A t = 15km/0.50h = 30 km/h.
22 CHAPTER 2 Describing Motion: Kinematics in One Dimension
To better understand instantaneous velocity, let us consider a graph of the position of a particular particle versus time (x vs. t), as shown in Fig. 2-10. (Note that this is different from showing the “path” of a particle on an x vs. y plot.) The particle is at position x 1 at a time tx, and at position x2 at time t2. Pi and P2 repre­ sent these two points on the graph. A straight line drawn from point Pi (x 1, to point P2(x2, t2) forms the hypotenuse of a right triangle whose sides are Ax and M. The ratio Ax/At is the slope of the straight line PiP2. But Ax/At is also the average velocity of the particle during the time interval At = t2 - tx. Therefore, we conclude that the average velocity of a particle during any time interval At = t2 — h is equal to the slope of the straight line (or chord) connecting the two points (xx, and (x2, t2) on an x vs. t graph.
Consider now a time tx, intermediate between tx and t2, at which time the particle is at x{ (Fig. 2-11). The slope of the straight line P ^ is less than the slope of PxP2in this case. Thus the average velocity during the time interval /• - tx is less than during the time interval t2 — tx.
Now let us imagine that we take the point Pj in Fig. 2-11 to be closer and closer to point Pj. That is, we let the interval tx — tx, which we now call At, to 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 . The average velocity (equal to the slope of the chord) thus approaches the slope of the tangent at point Px. The definition of the instantaneous velocity (Eq. 2-3) is the limiting value of the average velocity as At approaches zero. Thus the instantaneous velocity equals the slope o f the tangent to the 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. 2-12 (which shows the same curve as in Figs. 2-10 and 2-11), as our object moves from x 1to x2, the slope continually increases, 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. 2-12. 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 toward decreasing values of x, to the left on a standard xy plot.
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. 2-10 has no straight sections, so there are no time intervals when the velocity is constant.
X
Po
x
I
I
0 ------t1-x-------------t21---------------------- 1
FIGURE 2 - 1 0 Graph of a particle’s position x vs. time t. The slope of the straight line Pi P2 represents the average velocity of the particle during the time interval At = t2 — t\.
FIGURE 2 -1 1 Same position vs. time curve as in Fig. 2 -1 0 , but note that the average velocity over the time interval t[ — tx (which is the slope of Pi Pi) is less than the average velocity over the tim e interval t2 — 11. The slope of the thin line tangent to the curve at point Pj equals the instantaneous velocity at time t\ .
x
____ ii_____ ii ii
0] tx
l ~t 2
FIGURE 2 - 1 2 Same x vs. t curve as in Figs. 2 -1 0 and 2 -1 1 , but here showing the slope at four different points: A t P3 , the slope is zero, so v = 0. A t P4 the slope is negative, so v < 0.
t
EXERCISE C W hat is your speed at the instant you turn around to m ove in the opposite direction? {a) D epends 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 this book gives a summary in Appendix B. The derivatives of polynomial functions (which we use a lot) are:
4d-t (C tn) = nCt"-1 and ^dt = 0, where C is any constant.
SECTION 2 -3 Instantaneous Velocity 23
X (ml t) 10 20 30 4() 50 60
(a)
Tangent at P2 whose slope is v2 = 21.0 m/s
FIGURE 2 - 1 3 Exam ple 2 -3 . (a) Engine traveling on a straight track. (b) Graph of x vs. t: x = A t 2 + B.
EXAMPLE 2-3 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. 2-13a. We will treat
the engine as if it were a particle. Its position as a function of time is given by the
equation x = A t2 + B, where A = 2.10 m/s2 and B = 2.80 m, and this equa­
tion is plotted in Fig. 2-13b. (a) Determine the displacement of the engine during
the time interval from
to t2 = 5.00 s. (b) Determine the average
velocity during this time interval, (c) Determine the magnitude of the instanta­
neous velocity at t = 5.00 s.
APPROACH We substitute values for tx and t2in the given equation for x to obtain x 1 and x2. The average velocity can be found from Eq. 2-2. We take the deriva­ tive of the given x equation with respect to t to find the instantaneous velocity, using the formulas just given.
SOLUTION (a) At tx = 3.00 s, the position (point Px in Fig. 2-13b) is
X! = A t\ + B = (2.10m/s2)(3.00s)2 + 2.80 m = 21.7 m.
At t2 = 5.00 s, the position (P2 in Fig. 2-13b) is
x2 = (2.10m/s2)(5.00s)2 + 2.80 m = 55.3 m.
The displacement is thus
x2 - x 1 = 55.3 m - 21.7 m = 33.6 m.
(b) The magnitude of the average velocity can then be calculated as
Ax v =
A*
Xi to t-l
33.6 m = 16.8 m/s.
2.00 s
This equals the slope of the straight line joining points and P2 shown in Fig. 2-13b.
(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. 2-13b. 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 = A t2 + 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 bottom of previous page):
„ = ^ = l - [ A t2 + B) 2At.
dt dt
'
We are given A = 2.10 m /s2, so for t = t2 = 5.00 s,
v2 = 2A t = 2(2.10 m /s2)(5.00s) = 21.0 m/s.
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:
change of velocity average acceleration = ---- ;------------- -----
time elapsed
In symbols, the average acceleration over a time interval A£ = t2 — tx during
24 CHAPTER 2 Describing Motion: Kinematics in One Dimension
which the velocity changes by Av = v2 — v1, is defined as
v2 - V! Au a = ---------- = •
t2 - tx At
^ (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 - 4 Average acceleration. A car accelerates along a straight road from rest to 90 km /h in 5.0 s, Fig. 2-14. 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 vx = 0. The final velocity is v2 = 90 km/h = 90 X 103m/3600 s = 25 m/s. SOLUTION From Eq. 2-5, the average acceleration is
v2 ~ vx 25 m/s - Om/s ^ ^ m/s
a = ---------- = --------—--------- = 5.0------
t2 - tx
5.0 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. 2-14.
=0
v, = 0
Acceleration \n - 5.0 m/s~l
;ii t * 1.0 s r = 5.0 m/s
iil I - 2.0 s if = 10.0 m/s
at ( - t2 = 5,0 s v = th = 25 ni/s
FIGURE 2 - 1 4 Exam ple 2 -4 . The car is shown at the start with vi = 0 at t\ = 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. We assume the
acceleration is constant and equals 5.0 m /s2. The green arrows represent the velocity vectors; the length of each arrow represents the magnitude of the velocity at that moment. The acceleration vector is the orange arrow. Distances are not to scale.
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 _ m _ m
s
s •S s2
According to the calculation in Example 2-4, 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.
SECTION 2 - 4 Acceleration 25
CONCEPTUAL EXAMPLE 2 -5 I Velocity and acceleration, (a) If the velocity 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.
at f, = 0
■ SffiSE
Acceleration
tt = -2,0 m/s2
at = 5.0 s - 5.0 m/s
FIGURE 2 - 1 5 Exam ple 2 -6 , showing the position of the car at
times t\ and t2, as w ell as the car’s
velocity represented by the green arrows. The acceleration vector (orange) points to the left as the car slows down while moving to the right.
EXERCISE D A powerful car is advertised to go from zero to 60 m i/h in 6.0 s. What does this say about the car: (a) it is fast (high speed); or (b ) it accelerates well?
EXAMPLE 2 - 6 Car slowing down. An automobile is moving to the right along a straight highway, which we choose to be the positive x axis (Fig. 2-15). Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is vx = 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. 2-5 for a.
SOLUTION In Eq. 2-5, we call the initial time tx = 0, and set t2 = 5.0 s. (Note that our choice of tx = 0 doesn’t affect the calculation of a because only At = t2 — ti appears in Eq. 2-5.) Then
5.0 m/s - 15.0 m/s
5.0 s
= -2.0 m/s'
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 /s2to the left, and it is shown in Fig. 2-15 as an orange arrow.
A CAUTION
Deceleration means the magnitude o f the velocity is decreasing; a is not
necessarily negative
Deceleration
When an object is slowing down, we can say it is decelerating. But be careful: deceler­ ation does not mean that the acceleration is necessarily negative. 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. 2-15), 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. 2-16. We have a deceleration whenever the magnitude of the velocity is decreasing, and then the velocity and acceleration point in opposite directions.
FIGURE 2 - 1 6 The car of Exam ple 2 -6 ,
now moving to the left and decelerating.
The acceleration is
v2 - Vi
a = -----A--t-----
(-5 .0 m /s) - (-15.0m /s)
5.0 s
a
—5.0 m /s + 15.0 m /s
5X)s
= +2.0 m /s.
EXERCISE E A car m oves along the x axis. What is the sign o f 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 m oves in the negative direction with (c) increasing speed or (d) decreasing speed?
26 CHAPTER 2 Describing Motion: Kinematics in One Dimension
Instantaneous Acceleration
The instantaneous acceleration, a, is defined as the limiting value o f the average acceleration as we let At approach zero:
= lim —A—v = —dv ■ ►o At 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. 2-17, then the average acceleration over a time interval At = t2 - tx is represented by the slope of the straight line connecting the two points P1and P2 as shown. [Compare this to the position vs. time graph of Fig. 2-10 for which the slope of the straight line represents the average velocity.] The instantaneous acceleration at any time, say t\ , is the slope of the tangent to the v vs. t curve at that time, which is also shown in Fig. 2-17. Let us use this fact for the situation graphed in Fig. 2-17; as we go from time ti to time t2the velocity continually increases, but the acceleration (the rate at which the velocity changes) is decreasing since the slope of the curve is decreasing.
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)?2 + (2.80 m), as in Example 2-3. Calculate (a) its average acceleration during the time interval from ti = 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 tx and t2 by differentiating x: v = dx/dt. Then we use Eq. 2-5 to find the average acceleration, and Eq. 2-6 to find the instantaneous acceleration.
SOLUTION (a) The velocity at any time t is
2.80 m ] = (4.20 m/s2)t,
as we saw in Example 2-3c. Therefore, at tx = 3.00 s, v1 = (4.20m/s2)(3.00s) = 12.6 m /s and at t2 = 5.00 s, v2 = 21.0 m/s. Therefore,
_ Av _ 21.0 m /s - 12.6 m/s
= 4.20 m /s2.
At
5.00 s - 3.00 s
(b) With v = (4.20 m /s2)?, the instantaneous acceleration at any time is
U = ~^t =
= 4.20 m /s2.
The acceleration in this case is constant; it does not depend on time. Figure 2-18 shows graphs of (a) x vs. t (the same as Fig. 2 - 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 during At = t2 - ?i Slope is instantaneous acceleration
FIGURE 2 - 1 7 A graph of velocity v vs. time t. The average acceleration over a time interval A t = t2 - ti is the slope of the straight line Pi P2 : a = A v / At. The instantaneous acceleration at time t\ is the slope of the v vs. t curve at that instant. FIGURE 2 - 1 8 Exam ple 2 -7 . Graphs of (a) x vs. t, (b) v vs. t, and (c) a vs. t for the m otion x = A t2 + B. N ote that v increases linearly with t and that the acceleration a is constant. A lso, v is the slope of the x vs. t curve, whereas a is the slope o f the v vs. t curve.
■=-+—1— 1— 1— 1— b - t ( s) ' 12 3 4 5 6 (a)
--
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.” This can be expressed in equation form as follows: since a = dv/dt and v = dx/dt, then
dv _ d f d x \ _ d2x
dt d t \ d t ) dt2
Here d2x /d t2 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.
r i1 11 11 11 11 11 ' l 23456 (b)
a = 4.20 m/s2
EXERCISE F The position of a particle is given by the follow ing equation: x = (2.00 m /s3)?3 + (2.50 m /s )t.
What is the acceleration of the particle at t = 2.00 s? (a) 13.0 m /s2; (b ) 22.5 m /s2; (c) 24.0 m /s2; (d) 2.00 m /s 2.
— 1— 1— 1— 1— 1— b - t { s) 12 34 5 6
(C)
SECTION 2 - 4 Acceleration 27
v (km/li J
CONCEPTUAL EXAMPLE 2^8~1 Analyzing with graphs. Figure 2-19 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 (a) the average acceleration; (b) instantaneous acceleration; and (c) total distance traveled for the two cars.
RESPONSE («) Average acceleration is A v/A t. Both cars have the same Av (100 km/h) and the same At (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 (-M s) curve. For about the first 4 s, the top curve is steeper than the bottom curve, so car A has a greater acceleration during this interval. The bottom curve is steeper during the last 6 s, so car B has the larger acceleration for 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.
2 - 5 Motion at Constant Acceleration
We now examine the situation when the magnitude of the acceleration is constant and the motion is in a straight line. In this case, the instantaneous and average accelerations 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.
To simplify our notation, let us take the initial time in any discussion to be zero, and we call it t0: tx = t0 = 0. (This is effectively starting a stopwatch at t0.) We can then let t2 = t be the elapsed time. The initial position and the initial velocity (i^) 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 — t0will be (Eq. 2-2)
_ _ Ax _ x - x0 _ x - x0
V At t - t0
t
since we chose t0 = 0. The acceleration, assumed constant in time, is (Eq. 2-5)
v - v0
A CAUTION
Average velocity, but only if a = constant
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 to obtain:
v = v0 + at.
[constant acceleration] (2-7)
If an object starts from rest (v0 = 0) and accelerates at 4.0 m /s2, after an elapsed time t = 6.0 s its velocity will be v = 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. 2-2) is v = (x — x0)/t, which we can rewrite as
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:
_
Vq + v
v = — -----
[constant acceleration] (2-9)
(Careful: Equation 2-9 is not necessarily valid if the acceleration is not constant.) We combine the last two Equations with Eq. 2-7 and find
x = x0 + vt
x 0 . . V0
^0 + ^0 + at
28 CHAPTER 2
or
v01 + \a t2.
[constant acceleration] (2-10)
Equations 2-7, 2-9, and 2-10 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 into Eq. 2-8:
v + v0 X = x0 + v t = x0 +
Next we solve Eq. 2-7 for t, obtaining V - Vo
and substituting this into the previous equation we have
x0
v + vo) f v ~ vo 2 j\ a
V2 - Vq
x0 2 a
We solve this for v2 and obtain
v2 = vl + 2a(x — x0),
[constant acceleration] (2-11)
which is the 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 here in one place for future reference (the tan background screen emphasizes their usefulness):
v = v0 + at x = xQ+ v0t + \a t2 v2 = Vq + 2a(x - x0)
v Vo
[a = constant] (2-12a) [a = constant] (2-12b) [a = constant] (2-12c)
[a = constant] (2-12d)
Kinematic equations fo r constant acceleration (w e ’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 posi­ tion, not distance, that x — x0is the displacement, and that t is the elapsed time.
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 The plane’s acceleration is constant, so we can use the kinematic equations for constant acceleration. In (a), we want to find v, and we are given:
Known
Wanted
*o = 0
V
0 II
s?
x = 150 m a = 2.00 m /s2
SOLUTION (a) Of the above four equations, Eq. 2 - 12c will give us v when we know vQ, a, x, and x0:
v2 = vl + 2a(x — x0) = 0 + 2(2.00 m /s2)(150m) = 600m2/s2
v = \ / 600 m2/s 2 = 24.5 m/s.
This runway length is not sufficient.
(b) Now we want to find the minimum length of runway, x — x0, given v = 27.8 m/s and a = 2.00 m /s2. So we again use Eq. 2-12c, but rewritten as
(x - *o) = 2 a
(27.8 m /s)2 - 0 = 193 m.
2(2.00 m /s2)
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.
(^a j p h y s i c s a p p l i e d A irport design
\PROBLEM SOLVING Equations 2 -1 2 are valid only when the acceleration is constant, which w e assume in this Example
EXERCISE G A car starts from rest and accelerates at a constant 10 m /s2 during a \ m ile (402 m) race. H ow fast is the car going at the finish line? (a) 8090 m /s; (b) 90m /s; (c) 81 m /s; (d) 809 m /s.
SECTION 2-5 29
S.O B i
c, O L V /
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 surely not help you understand physics. A better approach is to use the following (rough) procedure, which we put in a special “Problem Solving Strategy.” (Other such Problem Solving Strategies, as an aid, will be found throughout the book.)
1. Read and reread the whole problem carefully before unknown. Sometimes several sequential calculations,
trying to solve it.
or a combination of equations, may be needed. It is
2. Decide what object (or objects) you are going to often preferable to solve algebraically for the desired
9* study, and for what time interval. You can often unknown before putting in numerical values.
choose the initial time to be t = 0.
7. Carry out the calculation if it is a numerical problem.
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
Keep one or two extra digits during the calculations, but round off the final answer(s) to the correct number of significant figures (Section 1-3).
like to make your calculations easier.]
8. Think carefully about the result you obtain: Is it
4. Write down what quantities are “known” or “given,” and then what you want to know. Consider quanti­ ties both at the beginning and at the end of the chosen time interval.
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 ten, as discussed in Section 1-6. Often it is preferable to do a rough estimate at the start of a numerical problem
5. Think about which principles of physics apply in this because it can help you focus your attention on
problem. Use common sense and your own experi­ finding a path toward a solution.
ences. Then plan an approach.
9. A very important aspect of doing problems is
6. Consider which equations (and/or definitions) relate keeping track of units. An equals sign implies the
the quantities involved. Before using them, be sure units on each side must be the same, just as the
their range of validity includes your problem (for numbers must. If the units do not balance, a mistake
example, Eqs. 2-12 are valid only when the accelera­ has no doubt been made. This can serve as a check
tion is constant). If you find an applicable equation on your solution (but it only tells you if you’re
that involves only known quantities and one desired wrong, not if you’re right). Always use a consistent
unknown, solve the equation algebraically for the set of units.
FIGURE 2-20 Exam ple 2 -1 0 .
a - 2,00 m/s2
a = 2.00 m/a2
Vo = 0
30.0 m
EXAMPLE 2 -1 0 Acceleration of a car. How long does it take a car to cross a 30.0-m-wide intersection after the light turns green, if the car accelerates from rest at a constant 2.00 m/s2?
APPROACH We follow the Problem Solving Strategy above, step by step.
SOLUTION
1. Reread the problem. Be sure you understand what it asks for (here, a time interval).
2. The object under study is the car. We choose the time interval: t = 0, the initial time, is 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 width of the intersection.
3. Draw a diagram: the situation is shown in Fig. 2-20, 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.
30 CHAPTER 2 Describing Motion: Kinematics in One Dimension
4. The “knowns” and the “wanted” are shown in the Table in the margin, and we choose x0 = 0. Note that “starting from rest” means v = 0 at t = 0; that is, v0 = 0.
5. The physics: the motion takes place at constant acceleration, so we can use the kinematic equations, Eqs. 2-12.
6. Equations: we want to find the time, given the distance and acceleration; Eq. 2-12b is perfect since the only unknown quantity is t. Setting v0 = 0 and jc0 = 0 in Eq. 2-12b (x = xQ+ v0t + \a t2), we can solve for t:
2'
2x tl = —a so
t = A—
7. The calculation:
2(30.0 m)
t = A— =
= 5.48 s. 2.00 m /s2
This is our answer. Note that the units come out correctly.
8. We can check the reasonableness of the answer by calculating the final velocity v = at = (2.00m/s2)(5.48 s) = 10.96 m/s, and then finding x = x0 + vt = 0 + \ (10.96 m /s + 0)(5.48s) = 30.0 m, which is our given distance.
9. We checked the units, and they came out perfectly (seconds).
NOTE In steps 6 and 7, when we took the square root, we should have written t = ± \^ 2 x /a = ± 5.48 s. Mathematically there are two solutions. But the second solution, 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.
o II
5*
Known
*o = 0 x = 30.0 m a = 2.00 m /s2
Wanted t
We explicitly followed the steps of the Problem Solving Strategy for Example 2-10. In upcoming Examples, we will use our usual “Approach” and “Solution” to avoid being wordy.
EXAMPLE 2-11 ESTIMATE"! Air bags. Suppose you want to design an airbag 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. 2-21) to effec­ tively 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. 2-12. Both Eqs. 2-12a and 2-12b contain t, our desired unknown. They both contain a, so we must first find a, which we can do using Eq. 2-12c 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 X 103m/3600 s = 28 m/s. We then find the acceleration from Eq. 2-12c:
(28 m /s)2
a = - — = ----- — ------ = -390 m /s.
2x
2.0 m
This enormous acceleration takes place in a time given by (Eq. 2-12a):
v 3 0 - 28 m /s
t =
a
-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 against the expanding air bag.
PHYSICS APPLIED Car safety— air bags
FIGURE 2-21 Exam ple 2 -1 1 . A n air bag deploying on impact.
SECTION 2 - 6 Solving Problems 31
FIGURE 2-22 Exam ple 2 -1 2 : stopping distance for a braking car.
Tmvet during —
Tnivel during
.
reaction litne
^
FOTmJ
'
v = constant = 14 m/s t = 0.50 s a =0
v decreases from 14 m/s to zero a = - 6.0 m/s2
0 PHYSICS APPLIED Braking distances
Part 1: Reaction tim e
Known
t = 0.50 s Vq = 14 m /s v = 14 m /s a = 0 *o = 0
Wanted
X
Part 2: Braking
Known
X q = 7.0 m Vq = 14 m /s v = 0 a = -6 .0 m /s2
Wanted
X
FIGURE 2-23 E xam ple 2 -1 2 . Graphs o f (a) v vs. t and (b ) x vs. t.
32 CHAPTER 2
EXAMPLE 2 -1 2 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 acceleration 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 « 31m i/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 normal 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. 2-12.
SOLUTION Part (1). We take x 0 = 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. 2-22 and the Table in the margin. To find x, the position of the car at t = 0.50 s (when the brakes are applied), we cannot use Eq. 2 - 12c because x is multiplied by a, which is zero. But Eq. 2 - 12b works:
x = v0t + 0 = (14m /s)(0.50s) = 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 x 0 = 7.0 m (result of part (1)), and other variables are shown in the second Table in the margin. Equation 2 - 12a doesn’t contain x; Eq. 2-12b contains x but also the unknown t. Equation 2-12c, v2 — vl = 2a(x — jc0), is what we want; after setting x0 = 7.0 m, we solve for x, the final position of the car (when it stops):
x0
2 a
0 - (14 m /s)2
-196 m /s
= 7.0 m H----- ----------- ——- = 7.0 m H------ —— —r~
2( - 6.0 m /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 2-23 shows graphs of (a) v vs. t 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 - x 0) 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.
EXAMPLE 2-13 ESTIMATE"! Two Moving Objects: Police and Speeder. A car speeding at 150 km /h 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 assump­ tions 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
100 km/h
km/h 11000 m l h
cip — 5.0 s = 20
= 5.6 m /s2. 1 km j \ 3600 s
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. 2-12b for each car:
xs = vost + ^ast2 = (150 km/h)? = (42 m/s )?
xP = v0Pt + \a Yt2 = ^(5.6 m /s2)?2,
where we have set t>0P = 0 and as = 0 (speeder assumed to move at constant speed). We want the time when the cars meet, so we set xs = xF and solve for ?:
(42 m/s)? = (2.8 m /s2)?2.
The solutions are
42 m/s
0 and ? =
= 15 s.
2.8 m /s2
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 ? = 15 s is
Vp = -%> + aFt = 0 + (5.6m/s2)(15 s) = 84 m/s
or 300 km /h (« 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 those speeds. Also, the speeder, if a reasonable person, would slow down upon hearing the police siren. Figure 2-24 shows (a) x vs. ? and (b) v vs. ? graphs, based on the original assumption of = constant, whereas (c) shows v vs. ? for more reasonable assumptions.
A CAUTION
Initial assumptions need to be checked out for reasonableness
FIGURE 2-24 Exam ple 2 -1 3 .
(a)
(b)
(c)
SECTION 2 - 6 Solving Problems 33
FIGURE 2-25 G alileo Galilei (1564-1642).
/j\ CAUTION_______
A freely falling object increases in speed, but not in proportion
to its mass or weight FIGURE 2-26 Multiflash photograph of a falling apple, at equal time intervals. The apple falls farther during each successive interval, which means it is accelerating.
Acceleration due to gravity
2 - 7 Freely Falling Objects
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 accel­ erating may not be obvious at first. And beware of thinking, as was widely believed before the time of Galileo (Fig. 2-25), that heavier objects fall faster than lighter objects and that the speed of fall is proportional to how heavy the object is.
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. 2-26); that is, d oc t2. We can see this from Eq. 2-12b; but Galileo was the first to derive 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 will the same stone 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 horizontally in one hand and a heavier object—say, a baseball—in the other, and release them at the same time as in Fig. 2-27a, the heavier object will reach the ground first. But if you repeat the experiment, this time crumpling the paper into a small wad (see Fig. 2-27b), you will find 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 negli­ gible. 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. 2-28). 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 approach to science (idealization and simplification, mathematization of theory, theories that have testable consequences, experiments to test theoretical predictions).
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.
We call this acceleration the acceleration due to gravity on the surface of the Earth, and we give it the symbol g. Its magnitude is approximately
g = 9.80 m /s2.
[at surface of Earth]
In British units g is about 32 ft/s2. Actually, g varies slightly according to latitude and elevation, but these variations are so small that we will ignore them for most
FIGURE 2-27 (a) A ball and a light piece of paper are dropped at the same time, (b) Repeated, with the paper wadded up.
<
M
M
(a)
<b>
FIGURE 2-28 A rock and a feather are dropped sim ultaneously (a) in air, (b) in a vacuum.
34 CHAPTER 2 Describing Motion: Kinematics in One Dimension
Air-filled lube
(a)
<
Evacuated lube
<b)
purposes. The effects of air resistance are often small, and we will neglect them for the most part. However, air resistance will be noticeable even on a reasonably heavy object if the velocity becomes large.1Acceleration 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. 2-12, 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 jc0. 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.
EXERCISE H Return to the Chapter-Opening Q uestion, page 18, and answer it again now. Try to explain why you may have answered differently the first time.
U 2 E H H H E B Falling from a tower. Suppose that a ball is dropped (v0 = 0) from a tower 70.0 m high. How far will it have fallen after a time ti = 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 = +9.80m/s2. We set v0 = 0 and = 0. We want to find the position y of the ball after three different time intervals. Equation 2 - 12b, with x replaced by y, relates the given quantities (t, a, and v0) to the unknown y. SOLUTION We set t = tx = 1.00 s inE q.2-12b:
yi = v0t-i + \a t\ = 0 + \a t\ = ^(9.80m/s2)(1.00s)2 = 4.90 m.
The ball has fallen a distance of 4.90 m during the time interval t = 0 to tx = 1.00 s. Similarly, after 2.00 s (= t2), the ball’s position is
y2 = \a t\ = |(9.80 m /s2)(2.00 s)2 = 19.6 m.
Finally, after 3.00 s (= t3), the ball’s position is (see Fig. 2-29)
y3 = \a t\ = |(9.80 m /s2)(3.00 s)2 = 44.1m.
■
Thrown down from a tower. Suppose the ball in Example 2-14
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. 2-12b, but now v0is not zero, it is v0 = 3.00 m/s. SOLUTION (a) At t = 1.00 s, the position of the ball as given by Eq. 2-12b is
y = v0t + \a t2 = (3.00 m/s) (1.00 s) + |(9.80m /s2)(1.00s)2 = 7.90 m.
At t = 2.00 s, (time interval t = 0 to t = 2.00 s), the position is
y = v0t + \a t2 = (3.00 m/s) (2.00 s) + |(9.80m /s2)(2.00s)2 = 25.6 m.
As expected, the ball falls farther each second than if it were dropped with vQ= 0. (b) The velocity is obtained from Eq. 2 - 12a:
v = v0 + at = 3.00 m /s + (9.80 m /s2)(1.00s) = 12.8 m/s [at tx = 1.00 s] = 3.00 m /s + (9.80m/s2)(2.00s) = 22.6 m/s. [at ?2 = 2.00 s]
In Example 2-14, 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 ^ = 1.00 s]
= (9.80m/s2)(2.00s) = 19.6 m/s.
[at ?2 = 2.00 s]
NOTE For both Examples 2-14 and 2-15, 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) higher than that of a dropped ball.
j PROBLEM SOLVING You can choose y to be positive either up or dow n
FIGURE 2-29 Exam ple 2 -1 4 . (a) A n object dropped from a tower falls with progressively greater speed and covers greater distance with each successive second. (See also Fig. 2 -2 6 .) (b) Graph of y vs. t. Acceleration due to
y\~ ^
(After 1.00 s) 'y2 = 19.6 m (After 2.00 s)
_y3 = 44.1 m (After 3.00 s)
(a)
trThe 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 2 - 7 Freely Falling Objects 35
B(u = 0)
ft
FIGURE 2-30 A n 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. Exam ples 2 -1 6 ,2 -1 7 ,2 -1 8 , and 2-19.
EXAMPLE 2-16 Ball thrown upward, I. A person throws a ball upward into the air with an initial velocity of 15.0 m/s. Calculate (a) how high it goes, and (b) how long the ball is in the air before it comes back to the hand. 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. 2-30) and until it comes back to the hand again. Let us choose y to be positive in the upward direc­ tion and negative in the downward direction. (This is a different convention from that used in Examples 2-14 and 2-15, 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. 2-30), where its speed is zero for an instant; then it descends, with increasing speed.
SOLUTION (a) We consider the time interval from when the ball leaves the thrower’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. 2-30) we have
= 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. 2-12c, replacing x with y: v2 = Vq + 2ay. We solve this equation for y:
y =
2
2
V L ~ Vq
2 a
0 - (15.0 m /s)2 = 11.5 m.
2(-9.80 m /s2)
The ball reaches a height of 11.5 m above the hand.
(b) Now we need to choose a different 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. 2-30) in one step and use Eq. 2 - 12b. We can do this because y represents position or displacement, and not the total distance traveled. Thus, at both points A and C, y = 0. We use Eq. 2-12b with a = -9.80 m /s2 and find
y = y0 + v0t + \a t2
0 = 0 + (15.0 m/s)^ + !(-9.80 m/s2)t2.
This equation is readily factored (we factor out one t):
(15.0 m/s - 4.90 m /s2t)t = 0.
There are two solutions:
t = 0 and t
15.0 m/s = 3.06 s.
4.90 m /s2
The first solution (t = 0) corresponds to the initial point (A) in Fig. 2-30, when the ball was first thrown from y = 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, which could be significant, so our result is only an approximation to a real, practical situation.
A CAUTION
Quadratic equations have two solutions. Sometimes only one
corresponds to reality, sometimes both
We did not consider the throwing action in this Example. 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 in Example 2-10, in which case we ignore the “unphysical” solution. But in Example 2-16, both solutions to our equation in t2 are physically meaningful: t = 0 and t = 3.06 s.
36 CHAPTER 2 Describing Motion: Kinematics in One Dimension
CONCEPTUAL EXAMPLE 2 -1 7 1 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. 2-30).
RESPONSE Both are wrong. (1) Velocity and acceleration are not necessarily in the same direction. When the ball in Example 2-16 is moving upward, its velocity is positive (upward), whereas the acceleration is negative (downward). (2) At the highest point (B in Fig. 2-30), 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 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. In sum, the acceleration of gravity always points down toward the Earth, even when the object is moving up.
/j\ CAUTION___________
(1) Velocity and acceleration are not always in the sam e direction; the acceleration (o f gravity) always points down (2) a 0 even at the highest point o f a trajectory
EXAMPLE 2 -1 8 Ball thrown upward, II. Let us consider again the ball thrown upward of Example 2-16, and make more calculations. Calculate (a) how much time it takes for the ball to reach the maximum height (point B in Fig. 2-30), 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. 2-12. We have the height of 11.5 m from Example 2-16. Again we take y as positive upward.
SOLUTION (a) We consider the time interval between the throw (t = 0, = 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. 2-12a and 2-12b contain the time t with other quantities known. Let us use Eq. 2-12a with a = -9.80 m /s2, v0 = 15.0 m/s, and v = 0:
v = v0 + at;
setting v = 0 and solving for t gives
v0
15.0 m /s
t = ----- = ------- ——----—r = 1.53 s.
a
-9.80 m/s2
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 part (b) of Example 2-16]. Thus it takes the same time to reach the maximum height as to fall back to the starting point.
( b ) Now we consider the time interval from the throw (t = 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 2-16), and we want to find v when t = 3.06 s:
v = vQ + at = 15.0 m/s - (9.80m/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, a plane pulling out of a dive and under­ going 3.00 g’s would have an acceleration of (3.00)(9.80m/s2) = 29.4 m/s2.
| EXERCISE I If a car is said to accelerate at 0.50 g, what is its acceleration in m /s2?
SECTION 2 - 7 Freely Falling Objects 37
B(v = 0)
A
ii
c
FIGURE 2-30 (Repeated for Example 2-19)
EXAMPLE 2-19 Ball thrown upward. III; the quadratic formula. For the ball in Example 2-18, calculate at what time t the ball passes a point 8.00 m above the person’s hand. (See repeated Fig. 2-30 here).
APPROACH We choose the time interval from the throw (t = 0, = 15.0 m/s) until the time t (to be determined) when the ball is at position y = 8.00 m, using Eq. 2-12b.
SOLUTION We want to find t, given y = 8.00 m, y0 = 0, vQ= 15.0 m/s, and a = -9.80 m /s2. We use Eq. 2-12b:
y = y0 + v0t + \a t2
8.00 m = 0 + (15.0m/s)f + \ (-9.80 m /s2)t2.
To solve any quadratic equation of the form at2 + bt + c = 0, where 0, b, and c are constants (a is not acceleration here), we use the quadratic formula:
b ± \ f b 2 4ac t =
2 a 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
15.0 m /s +
m /s)2 - 4(4.90 m /s2)(8.00m)
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 2-31 shows graphs of (a) y vs. t and (b) v vs. t for the ball thrown upward in Fig. 2-30, incorporating the results of Examples 2-16,2-18, and 2-19.
FIGURE 2-31 Graphs of (a) y vs. t, (b) v vs. t for a ball thrown upward, Examples 2 -1 6 ,2 -1 8 , and 2-19.
EXAMPLE 2 -2 0 Ball thrown upward at edge of cliff. Suppose that the person of Examples 2-16, 2-18, and 2-19 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 in Fig. 2-32. (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. 2-12b, 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).
38 CHAPTER 2 Describing Motion: Kinematics in One Dimension
SOLUTION (a) We use Eq. 2-12b with a = -9.80 m /s2, v0 = 15.0 m/s, y0 = 0, and y = -50.0 m:
y = yo + v0t + \a t2
-50.0m = 0 + (15.0m /s)? - |(9.80m /s2)t2.
Rewriting in the standard form we have
(4.90m /s2)?2 - (15.0m/s)? - (50.0m) = 0.
Using the quadratic formula, we find as solutions ? = 5.07 s and ? = -2.01 s. The first solution, ? = 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 2-16); so it took an additional 2.01 s to fall to the base. But what is the meaning of the other solution, ? = -2.01 s? This is a time before the throw, when our calculation begins, so it isn’t relevant here.* (b) From Example 2-16, the ball moves up 11.5 m, falls 11.5 m back down to the top of the cliff, and then down another 50.0 m to the base of the cliff, for a total distance traveled of 73.0 m. Note that the displacement, however, was -50.0 m. Figure 2-33 shows the y vs. ? graph for this situation.
EXERCISE J Two balls are thrown from a cliff. O ne 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 VariableAcceleration; Integral Calculus
In this brief optional Section we use integral calculus to derive the kinematic equa­ tions for constant acceleration, Eqs. 2 - 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 7-3, where we begin to use it in the physics.
First we derive Eq. 2-12a, assuming as we did in Section 2-5 that an object has velocity at ? = 0 and a constant acceleration a. We start with the definition of instantaneous acceleration, a = dv/dt, which we rewrite as
dv = adt.
We take the definite integral of both sides of this equation, using the same nota­ tion we did in Section 2-5:
{Vdv = I adt
Jv=Vn
Jt =-o
which gives, since a = constant,
v - v0 = at.
This is Eq. 2-12a, v = v0 + at. Next we derive Eq. 2-12b starting with the definition of instantaneous
velocity, Eq. 2-4, v = dx/dt. We rewrite this as
dx = v dt or
dx = (v0 + at)dt
where we substituted in Eq. 2 - 12a.
.. I
y =0
AM
■y = - 5 0 m FIGURE 2-32 Exam ple 2 -2 0 . The person in Fig. 2 -3 0 stands on the edge of a cliff. The ball falls to the base of the cliff, 50.0 m below. FIGURE 2-33 Exam ple 2 -2 0 , the y vs. ? graph.
of cliff
trThe solution t = -2.01 s 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 him 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.
*SECTION 2 - 8 Variable Acceleration; Integral Calculus 39
Now we integrate:
rx
\ dx = J'x = x 0n
rt
(v0 + a t ) d t Jt=0
{ v0dt + \a t dt
Jt=o
J t =o
X - x0 = v0t + \a t2
since v0 and a are constants. This result is just Eq. 2-12b, x = x0 + v0t + I at2. Finally let us use calculus to find velocity and displacement, given an accelera­
tion that is not constant but varies in time.
■frfJ u U
■ Integrating a time-varying acceleration. An experimental
vehicle starts from rest (v0 = 0) at t = 0 and accelerates 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. 2-12 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 arbi­ trary time t:
f d v = Ca dt
Jo
Jo
f (7.00 m /s3)t dt
Jo
(7.00 m/s3) ( - = (7.00m/s3) ^ y - o j = (3.50m /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 jc0 = 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:
Jro * ■ Jro v dt
r•2Z.U0U0 Ss
fj-.3 2.00 s
x = (3.50 m/s3)t2dt = (3.50 m/s3) -
= 9.33 m.
Jo
In sum, at t = 2.00 s, v = 14.0 m /s and x = 9.33 m.
*2—9 Graphical Analysis and Numerical Integration
This Section is optional. It discusses how to solve certain Problems numerically, often needing a computer to do the sums. Some of this material is also covered in Chapter 7, Section 7-3.
If we are given the velocity v of an object as a function of time t, we can obtain the displacement, x. Suppose the velocity as a function of time, v(t), is given as a graph (rather than as an equation that could be integrated as discussed in Section 2-8), as shown in Fig 2-34a. If we are interested in the time interval from txto t2, as shown, we divide the time axis into many small subintervals, , A.t2, k t3, ..., which are indicated by the dashed vertical lines. For each subinterval, a horizontal dashed line is drawn to indicate the average velocity during that time interval. The displacement during any subinterval is given by A.xt, where the subscript i represents the particular subinterval
40 CHAPTER 2 Describing Motion: Kinematics in One Dimension
(i = 1 ,2, 3 , . . .). From the definition of average velocity (Eq. 2-2) we have
Axt = vt ^ .
Thus the displacement during each subinterval equals the product of vt and Att,
and equals the area of the dark rectangle in Fig. 2-34a for that subinterval. The
total displacement between times tx and t2 is the sum of the displacements over all
the subintervals:
h
x2 X\ = 2 V i^ti,
(2-13a)
h
where x 1is the position at tx and x2 is the position at t2. This sum equals the area
of all the rectangles shown.
It is often difficult to estimate vt with precision for each subinterval from the
graph. We can get greater accuracy in our calculation of x2 — x1 by breaking the
interval t2 — tx into more, but narrower, subintervals. Ideally, we can let each Aff
approach zero, so we approach (in principle) an infinite number of subintervals. In
this limit the area of all these infinitesimally thin rectangles becomes exactly equal
to the area under the curve (Fig. 2-34b). Thus the total displacement between any
two times is equal to the area between the velocity curve and the t axis between the
two times tx and t2. This limit can be written
h
x2 — x-i = lim ViLAf,
z
1
A t—*0 "h 1 l
or, using standard calculus notation,
x2 - x1 = [ v(t)dt.
Jt,
(2-13b)
We have let At —> 0 and renamed it dt to indicate that it is now infinitesimally small. The average velocity, v, over an infinitesimal time dt is the instantaneous velocity at that instant, which we have written v(t) to remind us that v is a function of t. The symbol J is an elongated S and indicates a sum over an infinite number of infinitesimal subintervals. We say that we are taking the integral of v(t) over dt from time tx to time t2, and this is equal to the area between the v(t) curve and the t axis between the times tx and t2(Fig. 2-34b). The integral in Eq. 2 -13b is a definite integral, since the limits tx and t2 are specified.
Similarly, if we know the acceleration as a function of time, we can obtain the velocity by the same process. We use the definition of average acceleration (Eq. 2-5) and solve for Av:
Av = a At.
If a is known as a function of t over some time interval t\ to t2, we can subdivide
this time interval into many subintervals, Att, just as we did in Fig. 2-34a. The
change in velocity during each subinterval is Avt = at Att . The total change in
velocity from time tx until time t2is
h _ v2 ~ v1 = ^ a t A t i ,
(2-14a)
where v2represents the velocity at t2and vxthe velocity at tl .This relation can be written as an integral by letting A£ —> 0 (the number of intervals then approaches infinity)
h
v
0 ----- 1---------------------- 1------ 1-----1 0 (b)
FIGURE 2-34 Graph of v vs. t for
the motion of a particle. In (a), the time axis is broken into subintervals of width Ati, the average velocity during each A i s V(, and the area of all the rectangles, 2 ^ ; A ^, is numerically equal to the total displacem ent (x2 - jci) during the total time (t2 - 11). In (b), A^ —» 0 and the area under the curve is equal to (x2 — xi).
or
v2 - vx = a{t)dt.
(2-14b)
Jt,
Equations 2-14 will allow us to determine the velocity v2 at some time t2 if the
velocity is known at tx and a is known as a function of time.
If the acceleration or velocity is known at discrete intervals of time, we can use the
summation forms of the above equations, Eqs. 2-13a and 2-14a, to estimate velocity
or displacement. This technique is known as numerical integration. We now take an
Example that can also be evaluated analytically, so we can compare the results.
*SECTION 2 - 9 Graphical Analysis and Numerical Integration 41
FIGURE 2-35 Exam ple 2 -2 2 . 42 CHAPTER 2
EXAMPLE 2-22 Numerical integration. An object starts from rest at t = 0 and accelerates at a rate a(t) = (8.00m /s4)?2. Determine its velocity after 2.00s using numerical methods.
APPROACH Let us first divide up the interval t = 0.00 s to ? = 2.00 s into four subintervals each of duration Att = 0.50 s (Fig. 2-35). We use Eq. 2-14a with v2 = v, v1 = 0, t2 = 2.00 s, and tx = 0. For each of the subintervals we need to estimate at . There are various ways to do this and we use the simple method of choosing at to be the acceleration a{t) at the midpoint of each interval (an even simpler but usually less accurate procedure would be to use the value of a at the start of the subinterval). That is, we evaluate a(t) = (8.00 m /s4)?2 at ? = 0.25 s (which is midway between 0.00 s and 0.50 s), 0.75 s, 1.25 s, and 1.75 s.
SOLUTION The results are as follows:
i
1
2
3
4
M m /s2)
0.50
4.50
12.50
24.50
Now we use Eq. 2-14a, and note that all A?; equal 0.50 s (so they can be factored out): r=2.00s
v(t = 2.00 s) = 2
t=0
= (0.50 m /s2 + 4.50 m /s2 + 12.50 m /s2 + 24.50 m /s2)(0.50s)
= 21.0 m/s.
We can compare this result to the analytic solution given by Eq. 2-14b since the
functional form for a is integrable analytically:
r 2.00 s
(8.00 m /s4) ?2dt
Jo
8.00 m /s4 2.00s 3
8.00 m /s4 [(2.00 s)3 -
(0) 31
_
21.33 m/s
or 21.3 m/s to the proper number of significant figures. This analytic solution is precise, and we see that our numerical estimate is not far off even though we only used four A? intervals. It may not be close enough for purposes requiring high accu­ racy. If we use more and smaller subintervals, we will get a more accurate result. If we use 10 subintervals, each with A? = 2.00 s/10 = 0.20 s, we have to evaluate a(t) at ? = 0.10 s, 0.30 s,..., 1.90 s to get the at, and these are as follows:
i
12
3
4
5
6
7
8
9
10
fl/(m /s2) 0.08 0.72 2.00 3.92 6.48 9.68 13.52 18.00 23.12 28.88
Then, from Eq. 2-14a we obtain
v(t = 2.oos) =
= (2 ^ )(° -200s)
= (106.4 m /s2)(0.200s) = 21.28 m/s,
where we have kept an extra significant figure to show that this result is much closer to the (precise) analytic one but still is not quite identical to it. The percentage difference has dropped from 1.4% (0.3 m /s2/21.3 m/s2) for the foursubinterval computation to only 0.2% (0.05/21.3) for the 10-subinterval one.
In the Example above we were given an analytic function that was integrable, so we could compare the accuracy of the numerical calculation to the known precise one. But what do we do if the function is not integrable, so we can’t compare our numerical result to an analytic one? That is, how do we know if we’ve taken enough subintervals so that we can trust our calculated estimate to be accurate to within some desired uncer­ tainty, say 1 percent? What we can do is compare two successive numerical calculations: the first done with n subintervals and the second with, say, twice as many subintervals (2n). If the two results are within the desired uncertainty (say 1 percent), we can usually assume that the calculation with more subintervals is within the desired uncertainty of the true value. If the two calculations are not that close, then a third calculation, with more subintervals (maybe double, maybe 10 times as many, depending on how good the previous approximation was) must be done, and compared to the previous one.
The procedure is easy to automate using a computer spreadsheet application.
If we wanted to also obtain the displacement x at some time, we would have to do a second numerical integration over v, which means we would first need to calculate v for many different times. Programmable calculators and computers are very helpful for doing the long sums.
Problems that use these numerical techniques are found at the end of many Chapters of this book; they are labeled N um erical/Com puter and are given an asterisk to indicate that they are optional.
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, At, the time period over which we choose to make our observations. An object’s average velocity over a particular time interval At is its displacement Ax during that time interval, divided by At:
v=§ •
(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 (At —» 0):
Ax dx v = Alifm^o —a—t = —dt »
(2-4)
where dx/dt is the derivative of x with respect to t.
Questions
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 At is
_ Av a =
(2-5)
where Av is the change of velocity during the time interval At. Instantaneous acceleration is the average acceleration
taken over an infinitesimally short time interval:
Av dv a = lim —— = — •
M dt
(2- 6)
If an object moves in a straight line with constant acceleration, the
velocity v and position x are related to the acceleration a, the elapsed
time t, the initial position x0, and the initial velocity v0by Eqs. 2-12:
v = v0 + at,
x = x0 + v0t + \a t2,
v2L = v%2+ 2aw(x - x0),^ v- = —v -+----
( 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.
[*The kinematic Equations 2-12 can be derived using inte­ gral calculus.]
1. Does a car speedometer measure speed, velocity, or both? 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? 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 accelera­ tion is positive? What about vice versa? 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 trav­ eling with a speed of 60 km /h and has an acceleration of 40km/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. In the absence of air resistance, how fast would the ball be traveling when the catcher catches it?
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 for the entire trip from A to C 80 km/h? 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?
17. 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. The sounds will not occur at equal time intervals. Why? Will the time between clinks increase or decrease near the end of the fall? How could the bolts be tied so that the clinks occur at equal intervals?
Questions 43
18. Describe in words the motion plotted in Fig. 2-36 in terms of v, a, etc. [Hint: First try to duplicate the motion plotted by walking or moving your hand.]
19. Describe in words the motion of the object graphed in Fig. 2-37.
0 10 20 30 40 50 60 70 80 90 100 110 120 t( s)
FIGURE 2-37 Question 19, Problem 23.
FIGURE 2-36 Question 18, Problems 9 and 86.
|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 Prob­ lems 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. Finally, there is a set of unranked “General Problems” not arranged by Section number.]
2-1 to 2-3 Speed and Velocity
1. (I) If you are driving 110 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 3.25 h?
3. (I) A particle at t\ = -2.0 s is at x\ = 4.3 cm and at t2 = 4.5 s is at x2 = 8.5 cm. What is its average velocity? Can you calculate its average speed from these data?
4. (I) A rolling ball moves from x 1 = 3.4 cm to x2 = —4.2 cm during the time from t\ = 3.0 s to t2 = 5.1 s. What is its average velocity?
5. (II) According to a rule-of-thumb, every five seconds between a lightning flash and the following thunder gives the distance to the flash in miles. Assuming that the flash of light arrives in essentially no time at all, estimate the speed of sound in m/s from this rule. What would be the rule for kilometers?
6. (II) You are driving home from school steadily at 95 km/h for 130 km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 3 hours and 20 minutes. (a) How far is your hometown from school? (b) What was your average speed?
7. (II) A horse canters away from its trainer in a straight line, moving 116 m away in 14.0 s. It then turns abruptly and gallops halfway back in 4.8 s. Calculate (a) its average speed and (b) its average velocity for the entire trip, using “away from the trainer” as the positive direction.
8. (II) T x = 34 + lOt — 213, where t is in seconds and x in meters. {a) Plot jc as a function of t from t = 0 to f = 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?
9. (II) The position of a rabbit along a straight tunnel as a function of time is plotted in Fig. 2-36. What is its instanta­ neous 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?
10. (II) On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about 0.28 /xm. A CD player’s readout laser scans along the spiral’s sequence of bits at a constant speed 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 one would (at first glance) think the required bit rate for a CD player is
samplings ^
bits
Nn = 2 44,100 second )(m
second
where the 2 is for the 2 loudspeakers (the 2 stereo channels). Note that No 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?
11. (II) A car traveling 95 km/h is 110 m behind a truck trav­ eling 75 km/h. How long will it take the car to reach the truck?
12. (II) Two locomotives approach each other on parallel tracks. Each has a speed of 95 km /h with respect to the ground. If they are initially 8.5 km apart, how long will it be before they reach each other? (See Fig. 2-38).
-8,5 km—-
v = 95 km/h
FIGURE 2-38 Problem 12.
44 CHAPTER 2 Describing Motion: Kinematics in One Dimension
13. (II) Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spiraling path that starts at radius R i = 2.5 cm and finishes at radius R 2 = 5.8 cm. The distance between the centers of neighboring spiralwindings is 1.6/xm(= 1.6 X 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 ^m , and note that the original spiral and the straight path both occupy the same area.] (b) To read information, a CD player adjusts the rotation of the CD so that the player’s readout laser moves along the spiral path at a constant speed of 1.25 m/s. Estimate the maximum playing time of such a CD.
14. (II) An airplane travels 3100 km at a speed of 720 km /h, and then encounters a tailwind that boosts its speed to 990 km /h for the next 2800 km. What was the total time for the trip? What was the average speed of the plane for this trip? [Hint: Does Eq. 2-12d apply, or not?]
15. (II) Calculate the average speed and average velocity of a complete round trip in which the outgoing 250 km is covered at 95 km /h, followed by a 1.0-h lunch break, and the return 250 km is covered at 55 km /h.
16. (II) The position of a ball rolling in a straight line is given by x = 2.0 - 3.61 + 1.112, where x is in meters and t in seconds, (a) Determine the position of the ball at t = 1.0 s, 2.0 s, and 3.0 s. (b) What is the average velocity over the interval t = 1.0 s to t = 3.0 s? (c) What is its instanta­ neous velocity at t = 2.0 s and at t = 3.0 s?
17. (II) A dog runs 120 m away from its master in a straight line in 8.4 s, and then runs halfway back in one-third the time. Calculate (a) its average speed and (b) its average velocity.
18. (Ill) An automobile traveling 95 km /h overtakes a 1.10-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. 2-39. What are the results if the car and train are traveling in opposite directions?
---------------------------- IJO k m ----------------------------v - 75 km/h
j ' * t- = 95 km/h
« 4 _ * — 4 i-
FIGURE 2-3 9 Problem 18.
19. (Ill) 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 2.50 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?
2 -4 Acceleration 20. (I) A sports car accelerates from rest to 95 km /h in 4.5 s.
What is its average acceleration in m /s2? 21. (I) 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 80 km /h to 110 km /h? 22. (I) A sprinter accelerates from rest to 9.00 m /s in 1.28 s. What is her acceleration in (a) m /s2; (b) km /h2?
23. (I) Figure 2-37 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?
24. (II) A sports car moving at constant speed travels 110 m in 5.0 s. If it then brakes and comes to a stop in 4.0 s, what is the magnitude of its acceleration in m /s2, and in g’s (g = 9.80 m /s2)?
25. (II) A car moving in a straight line starts at x = 0 at t = 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 particular automobile can accelerate approximately as shown in the velocity vs. time graph of Fig. 2-40. (The short flat spots in the curve represent shifting of the gears.) Estimate the average acceleration of the car in (a) second gear; and (b) fourth gear, (c) What is its average accelera­ tion through the first four gears?
50
40
^ 20
10
0
0
10 20 30 40
FIGURE 2 -4 0 Problem 26. The velocity of a high-performance automobile as a function of time, starting from a dead stop. The flat spots in the curve represent gear shifts.
27. (II) A particle moves along the x axis. Its position as a func­
tion of time is given by x = 6.8 1 + 8.5 t2, where t is in
seconds and x is in meters. What is the acceleration as a function of time?
28. (II) The position of a racing 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) jt(m)
f(s) x(m)
0 0.25 0.50 0.75 1.00 1.50 2.00 2.50 0 0.11 0.46 1.06 1.94 4.62 8.55 13.79
3.00 3.50 4.00 4.50 5.00 5.50 6.00 20.36 28.31 37.65 48.37 60.30 73.26 87.16
29. (II) The position of an object is given by x = A t + B t2, 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 = 5.0 s? (d) What is the velocity as a function of time if x = A t + Bt~3l
Problems 45
2 -5 and 2 -6 Motion at Constant Acceleration 30. (I) A car slows down from 25 m /s to rest in a distance of
85 m. What was its acceleration, assumed constant? 31. (I) A car accelerates from 12 m/s to 21 m/s in 6.0 s. What
was its acceleration? How far did it travel in this time? Assume constant acceleration. 32. (I) A light plane must reach a speed of 32 m/s for takeoff. How long a runway is needed if the (constant) acceleration is 3.0 m/s2? 33. (II) A baseball pitcher throws a baseball with a speed of 41 m/s. Estimate the average acceleration of the ball during the throwing motion. In throwing the baseball, the pitcher accelerates the ball through a displacement of about 3.5 m, from behind the body to the point where it is released (Fig. 2-41).
FIGURE 2-41 Problem 33.
34. (II) Show that v = (v + v0)/2 (see Eq. 2-12d) is not valid when the acceleration a = A + Bt, where A and B are constants.
35. (II) A world-class sprinter can reach a top speed (of about 11.5 m/s) in the first 15.0 m of a race. What is the average acceleration of this sprinter and how long does it take her to reach that speed?
36. (II) An inattentive driver is traveling 18.0 m /s when he notices a red light ahead. His car is capable of decelerating at a rate of 3.65 m /s2. If it takes him 0.200 s to get the brakes on and he is 20.0 m from the intersection when he sees the light, will he be able to stop in time?
37. (II) A car slows down uniformly from a speed of 18.0 m /s to rest in 5.00 s. How far did it travel in that time?
38. (II) In coming to a stop, a car leaves skid marks 85 m long on the highway. Assuming a deceleration of 4.00 m /s2, esti­ mate the speed of the car just before braking.
39. (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.
40. (II) A car traveling at 105 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) Determine the stopping distances for an automobile with an initial speed of 95 km /h and human reaction time of 1.0 s: (a) for an acceleration a = -5.0 m /s2; (b) for a = —7.0 m /s2.
42. (II) A space vehicle accelerates uniformly from 65 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 = 6.0 s?
43. (II) A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 23 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. 2-42.)
v = 23 m/s
t'
f
FIGURE 2-42 Problem 43.
44. (II) 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 2.00 m /s2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?
45. (Ill) Assume in Problem 44 that the speeder’s speed is not known. If the police car accelerates uniformly as given above and overtakes the speeder after accelerating for 7.00 s, what was the speeder’s speed?
46. (Ill) 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 1100 m to go. The runner must then accelerate at 0.20 m /s2 for how many seconds in order to achieve the desired time?
47. (Ill) Mary and Sally are in a foot race (Fig. 2-43). 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, decelerates at a constant rate of 0.50 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
Sully
. 4.0 mA , 5.0 m/s
k-
I
FIGURE 2-43 Problem 47.
2-7 Freely Falling Objects [Neglect air resistance.] 48. (I) A stone is dropped from the top of a cliff. It is seen to hit
the ground below after 3.75 s. How high is the cliff? 49. (I) If a car rolls gently (v0 = 0) off a vertical cliff, how long
does it take it to reach 55 km/h? 50. (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.” 51. (II) A baseball is hit almost straight up into the air with a speed of about 20 m/s. (a) How high does it go? (b) How long is it in the air?
46 CHAPTER 2 Describing Motion: Kinematics in One Dimension
52. (II) A ball player catches a ball 3.2 s after throwing it verti­ cally upward. With what speed did he throw it, and what height did it reach?
53. (II) A kangaroo jumps to a vertical height of 1.65 m. How long was it in the air before returning to Earth?
54. (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?
55. (II) A helicopter is ascending vertically with a speed of 5.10 m/s. At a height of 105 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [Hint: v0 for the package equals the speed of the helicopter.]
56. (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. 2-26 and 2-29.
57. (II) A baseball is seen to pass upward by a window 23 m above the street with a vertical speed of 14 m/s. If the ball was thrown from the street, (a) what was its initial speed, (ib) what altitude does it reach, (c) when was it thrown, and ((d) when does it reach the street again?
58. (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 950 m. After this point, its acceleration is that of gravity, down­ ward. (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? (/) How long (total) is it in the air?
59. (II) Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 0.83 s after passing his window. Roger’s room is on the third floor, 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 floor are they being released? Each floor of the dorm is 5.0 m high.
60. (II) A stone is thrown vertically upward with a speed of 24.0m/s. (a) How fast is it moving when it reaches a height of 13.0 m? (b) How much time is required to reach this height? (c) Why are there two answers to (b)l
61. (II) A falling stone takes 0.33 s to travel past a window 2.2 m tall (Fig. 2-44). From what height above the top of the window did the stone fall?
2.2 m
To travel this Y distance took 0.33 s
&
62. (II) Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of 1.5 m above the ground (Fig. 2-45). When you quickly turn off the nozzle, you hear the water striking the ground next to you for another 2.0 s. What is the water speed as it leaves the nozzle?
1.5 m
FIGURE 2 -4 5 Problem 62.
63. (Ill) A toy rocket moving vertically upward passes by a 2.0-m-high window whose sill 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.
64. (Ill) A ball is dropped from the top of a 50.0-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?
65. (Ill) A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 3.4 s later. If the speed of sound is 340 m/s, how high is the cliff?
66. (Ill) A rock is thrown vertically upward with a speed of 12.0m/s. Exactly 1.00 s later, a ball is thrown up vertically along the same path with a speed of 18.0 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.
* 2 - 8 Variable Acceleration; Calculus *67. (II) Given v(t) = 25 + 181, where v is in m/s and t is in s,
use calculus to determine the total displacement from ti = 1.5 s to t2 = 3.1 s. *68. (Ill) The acceleration of a particle is given by a = A \ / i where A = 2.0 m /s5/2. At t = 0, v = 7.5 m/s and x = 0. (a) What is the speed as a function of time? (b) What is the displacement as a function of time? (c) What are the accel­ eration, speed and displacement at t = 5.0 s? *69. (Ill) Air resistance acting on a falling body can be taken into account by the approximate relation for the acceleration:
dv a = - = g -kv,
where A: 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 — k v .] (b) Determine an expression for the terminal velocity, which is the maximum value the velocity reaches.
FIGURE 2-44 Problem 61.
* 2 - 9 Graphical A nalysis and Num erical Integration [See Problems 95-97 at the end of this Chapter.]
Problems 47
|General Problems
70. 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 = 1.2 m /s2 to his maximum speed of 6.0 m/s. (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?
71. The acceleration due to gravity on the Moon is about onesixth 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?
72. A person jumps from a fourth-story window 15.0 m above a firefighter’s safety net. The survivor stretches the net 1.0 m before coming to rest, Fig. 2-46. (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.
78. Consider the street pattern shown in Fig. 2-47. Each inter­ section has a traffic signal, and the speed limit is 50 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?
FIGURE 2-47 Problem 78.
15.0 m
1.0 m
FIGURE 2-46 Problem 72.
73. A person who is properly restrained by an over-theshoulder seat belt has a good chance of surviving a car colli­ sion if the deceleration does not exceed 30 “g’s” (l.OOg = 9.80 m /s2). Assuming uniform deceleration of thi$ value, calculate the distance over which the front end of the car must be designed to collapse if a crash brings the car tq rest from 100 km/h.
74. Pelicans tuck their wings and free-fall straight down whei} diving for fish. Suppose a pelican starts its dive from q 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.
75. Suppose a car manufacturer tested its cars for front-en4 collisions by hauling them up on a crane and dropping then; 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 \ / 2 g H . What height corresponds tq a collision at (b) 50 km/h? (c) 100 km/h?
76. 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?
77. 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 75 km/h. What was his average acceleration (in m /s2) while riding down the mountain?
79. 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. 2-48) 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 2.8 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?
*
Downhill He
-Jjato
Uphill lie
FIGURE 2-48 Problem 79.
80. A robot used in a pharmacy picks up a medicine bottle at t = 0. It accelerates at 0.20 m /s2 for 5.0 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 pharma­ cist will take the medicine from the robot. From how far away did the robot fetch the medicine?
48 CHAPTER 2 Describing Motion: Kinematics in One Dimension
81. A stone is thrown vertically upward with a speed of 12.5 m /s
from the edge of a cliff
V
75.0 m high (Fig. 2-49).
||
(a) How much later does
it reach the bottom of
_= {I
the cliff? (b) What is its
speed just before hitting?
(c) What total distance did
it travel?
85. 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?
86. Sketch the v vs. t graph for the object whose displacement as a function of time is given by Fig. 2-36.
87. A person driving her car at 45 km /h approaches an intersec­
tion 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. 2-51). 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.
y = - 1 5 m FIGURE 2-49 Problem 81.
82. Figure 2-50 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 direc­
tion? (b) Is the object speeding up or slowing down? (c) Is
the acceleration of the object positive or negative? Next,
consider the time interval from D to E. (<d) Is the object
moving in the positive or negative direction? (e) Is the
object speeding up or slowing down? ( /) Is the acceleration
30
A 25
of the object posi­ tive or negative? (g) Finally, answer these same three
questions for the
20
time interval from
E C to D.
15
\
10
C 00 1
c D
FIGURE 2-50 *(s) Problem 82.
83. In the design of a rapid transit system, it is necessary to
balance the average speed of a train against the distance between stops. The more stops there are, the slower the train’s average speed. To get an idea of this problem, calcu­ late the time it takes a train to make a 9.0-km trip in two situations: (a) the stations at which the trains must stop are 1.8 km apart (a total of 6 stations, including those at the ends); and (b) the stations are 3.0 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.
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 2.0 m below the surface of the water. Estimate the average deceleration of the person while under the water.
FIGURE 2-51 Problem 87.
88. A car is behind a truck going 25 m /s on the highway. The driver looks for an opportunity to pass, guessing that his car can accelerate at 1.0 m /s2, and he gauges that he has to cover the 20-m length of the truck, plus 10-m clear room 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 also traveling at 25 m/s. He estimates that the car is about 400 m away. Should he attempt the pass? Give details.
89. Agent Bond is standing on a bridge, 13 m above the road
below, and his pursuers are getting too close for comfort. He spots a flatbed truck approaching at 25m /s, which he measures by knowing that the telephone poles the truck is passing are 25 m apart in this country. The bed of the truck is 1.5 m above the road, and Bond quickly calculates how many poles away the truck should be when he jumps down from the bridge onto the truck, making his getaway. How many poles is it?
90. A police car at rest, passed by a speeder traveling at a
constant 130 km /h, takes off in hot pursuit. The police officer catches up to the speeder in 750 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 accelera­ tion, and (d) the speed of the police car at the overtaking point.
91. A fast-food restaurant uses a conveyor belt to send the
burgers through a grilling machine. If the grilling machine is 1.1 m long and the burgers require 2.5 min to cook, how fast must the conveyor belt travel? If the burgers are spaced 15 cm apart, what is the rate of burger production (in burgers/min)?
92. 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 49
93. Figure 2-52 shows the position vs. time graph for two bicycles, A and B. (a) Is there 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 highest instantaneous velocity? (e) Which bicycle has the higher average velocity?
FIGURE 2-52 Problem 93.
*96. (Ill) The acceleration of an object (in m/s2) is measured at 1.00-s intervals starting at t = 0 to be as follows: 1.25, 1.58, 1.96, 2.40, 2.66, 2.70, 2.74, 2.72, 2.60, 2.30, 2.04, 1.76, 1.41,1.09, 0.86, 0.51, 0.28, 0.10. Use numerical integration (see Section 2-9) to estimate (a) the velocity (assume that v = 0 at t = 0) and (b) the displacement at t = 17.00 s.
*97. (Ill) A lifeguard standing at the side of a swimming pool spots a child in distress, Fig. 2-53. The lifeguard runs with average speed vR along the pool’s edge for a distance x, then jumps into the pool and swims with average speed v$ on a straight path to the child, (a) Show that the total time t it takes the lifeguard to get to the child is given by
X_ V o 2 + (d - J )2
1
+
vs
(b) Assume vR = 4.0 m /s and vs = 1.5 m/s. Use a graphing calculator or computer to plot t vs. x in part (a), and from this plot determine the optimal distance x the life­ guard should run before jumping into the pool (that is, find the value of x that minimizes the time t to get to the child).
94. You are traveling at a constant speed vM, and there is a car
in front of you traveling with a speed vA. You notice that
> va->so you start slowing down with a constant acceler­
m
ation a when the distance between you and the other car
is x. What relationship between a and x determines whether
or not you run into the car in front of you?
*Numerical/Computer
*95. (II) The Table below gives the speed of a particular drag racer as a function of time, (a) Calculate the average acceleration (m/s2) during each time interval. (b) Using numerical integration (see Section 2-9) estimate the total distance traveled (m) as a function of time. [Hint, for v in each interval sum the velocities at the beginning and end of the interval and divide by 2; for example, in the second interval use v = (6.0 + 13.2)/2 = 9.6] (c) Graph each of these.
7(s)
0 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00
v(km/h) 0.0 6.0 13.2 22.3 32.2 43.0 53.5 62.6 70.6 78.4 85.1
d = 10.0 m x
D = 8.0 m FIGURE 2-53 Problem 97.
A nsw ers to Exercises
A: -3 0 cm; 50 cm. B: (a). C: (b). D: (b). E: (fl) + ; ( £ , ) - ; ( C) - ; ( d ) + .
F: (c). G: 0b). H: (*). I: 4.9 m/s2 J: (c).
50 CHAPTER 2 Describing Motion: Kinematics in One Dimension
This snowboarder flying through the air shows an exam ple of m otion 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, g. Galileo analyzed the
motion of objects in 2 dimensions under the action of gravity near the Earth’s surface (now called “projectile m otion”) into its horizontal and vertical components.
We will discuss how to manipulate vectors and how to add them. Besides analyzing projectile motion, we will also see how to work with relative velocity.
T£ ^
Kinematics in Two or Three Dimensions; Vectors
CHAPTER-OPENING QUESTION—Guess now!
[D o n ’t w o rry abou t getting the right answ er n o w —yo u w ill get another chance later in the Chapter. See also p. 1 o f Chapter 1 fo r m ore explanation.]
A small heavy box of emergency supplies is dropped from a moving helicopter at point A as it flies along in a horizontal direction. Which path in the drawing below best describes the path of the box (neglecting air resistance) as seen by a person standing on the ground?
I n Chapter 2 we dealt with motion along a straight line. We now consider the description of the motion of objects that move in paths in two (or three) dimensions. To do so, we first need to discuss vectors and how they are added. We will examine the description of motion in general, followed by an interesting special case, the motion of projectiles near the Earth’s surface. We also discuss how to determine the relative velocity of an object as measured in different reference frames.
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
51
f 9
A
Scale for velocity: 1 cm = 90 km/h
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)
I I I * I I I > I I-----x (km)
8 km
6 km
East
(a)
Resultant = 2 km (east) 6 km
8 km (b)
x (km) East
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 spec­ ified 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. 3-1, 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. 3-1 by measuring the length of the corresponding arrow and using the scale shown (1cm = 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 v. 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 D , and velocity vectors, v. But the results will apply for other vectors we 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. 3-2a). 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. 3-2b), 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. 3-3. On this graph, we draw an arrow, labeled Dx, to represent the 10.0-km displacement to the east. Then we draw a second arrow, D2, to represent the 5.0-km displacement to the north. Both vectors are drawn to scale, as in Fig. 3-3.
FIGURE 3-3 A person walks 10.0 km east and then 5.0 km north. These two displacem ents are represented by the vectors £>! and D 2, which are shown as arrows. The resultant displacem ent vector, D R , which is the vector sum of D x and D 2, is also shown. M easurem ent on the graph with ruler and protractor shows that D R has a magnitude of 11.2 km and points at an angle 6 = 27° north of east.
y (km) North
West
x (km) East
South
52 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
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 DR in Fig. 3-3. Using a ruler and a protractor, you can measure on this diagram that the person is 11.2 km from the origin at an angle 6 = 27° north of east. In other words, the resultant displacement vector has a magnitude of 11.2 km and makes an angle 6 = 27° with the positive x axis. The magnitude (length) of DRcan also be obtained using the theorem of Pythagoras in this case, since D1, D2, and DR form a right triangle with DR as the hypotenuse. Thus
Dr = \ / D \ + D\ = ^(lO -O km )2 + (5.0 km)2
= y j 125 km2 = 11.2 km.
You can use the Pythagorean theorem, of course, only when the vectors are perpendicular to each other.
The resultant displacement vector, DR, is the sum of the vectors Di and D2. That is,
D r = Dx + D2.
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 - A + A >
where the equals sign applies only if the two vectors point in the same direction. In our example (Fig. 3-3), Dr = 11.2 km, whereas D1 + D2 equals 15 km, which is the total distance traveled. Note also that we cannot set DR 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: Dr = Dj + D2 = (11.2 km, 27° N of E).
EXERCISE A Under what conditions can the magnitude of the resultant vector above be D r = £>1 + Z>2?
Figure 3-3 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 £>!—to scale.
2. Next draw the second vector, D2, 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 translated parallel to themselves (maintaining the same length and angle) to accomplish these manipulations. The length of the resultant can be 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. 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 6 = 27° (see Fig. 3-4), the same as when they were added in reverse order (Fig. 3-3). That is, now using V to represent any type of vector,
Vx + V2 = V2 + V i,
[commutative property] (3-la)
which is known as the commutative property of vector addition.
FIGURE 3-4 If the vectors are added in reverse order, the resultant is the same. (Compare to Fig. 3 -3 .)
y (km)
SECTION 3 -2 Addition of Vectors - Graphical Methods 53
FIGURE 3-5 The resultant of three vectors:
VR = Vi + v 2 + v 3.
The tail-to-tip method of adding vectors can be extended to three or more vectors. The resultant is drawn from the tail of the first vector to the tip of the last one added. An example is shown in Fig. 3-5; the three vectors could repre­ sent 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,
(Vi + V2) + V3 = Vi + (V2 + V3), [associative property] (3-lb)
which is known as the associative property of vector addition. A second way to add two vectors is the parallelogram method. It is fully equiv­
alent 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. 3-6b. The resultant is the diagonal drawn from the common origin. In Fig. 3-6a, the tail-to-tip method is shown, and it is clear that both methods yield the same result.
A
=
FIGURE 3-6 Vector addition by two different methods, (a) and (b). Part (c) is incorrect.
y g g ? (c) Wrong
A CAUTION
Be sure to use the correct diagonal on parallelogram to get the resultant
It is a common error to draw the sum vector as the diagonal running between the tips of the two vectors, as in Fig. 3-6c. This is incorrect: it does not represent the sum of the two vectors. (In fact, it represents their difference, V2 - V j, as we will see in the next Section.)
CONCEPTUAL EXAMPLE 3^i~l Range of vector lengths. Suppose two vectors each have length 3.0 units. What is the range of possible lengths for the vector repre­ senting 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.
EXERCISE B If the two vectors of Exam ple 3 -1 are perpendicular to each other, what is | the resultant vector length?
FIGURE 3-7 The negative of a vector is a vector having the same length but opposite direction.
//•
3 - 3 Subtraction ofVectors, and Multiplication of a Vector by a Scalar
Given a vector V, we define the negative of this vector (—v) to be a vector with the same magnitude as V but opposite in direction, Fig. 3-7. 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.
54 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
y-
= y
+j —
-V, = v ,-K ^ 7
FIGURE 3-8 Subtracting two vectors: V? — V i .
We can now define the subtraction of one vector from another: the difference between two vectors V2 - Vi is defined as
v 2 - Vi = V2 + (-V ,).
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. 3-8 using the tail-to-tip method.
A vector V can be multiplied by a scalar c. We define their product so that cV has the same direction as V 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 cY is still \c\V (where \c\ means the magnitude of c), but the direction is precisely opposite to that of V. See Fig. 3-9.
EXERCISEC W hat d oes the “incorrect” vector in Fig. 3 - 6 c represent? (a) V2 - V i, (b) Vi — V2, (c) som ething else (specify).
FIGURE 3-9 Multiplying a vector V by a scalar c gives a vector whose magnitude is c times greater and in the same direction as V (or opposite direction if c is negative).
/
= -2 .0 V
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 V that lies in a particular plane. It can be expressed as the sum of two other vectors, called the components of the original vector. The compo­ nents are usually chosen to be along two perpendicular directions, such as 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. 3-10; the vector V could be a displacement vector that points at an angle 6 = 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 V is resolved into its x and y compo­ nents 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 OB and OC represent the x and y components of V, respectively, as shown in Fig. 3-10b. These vector components are written V* and \ y. We generally 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 positive or negative sign depending on whether they point along the positive or negative x or y axis. As can be seen in Fig. 3-10, V* + \ y = V by the parallelogram method of adding vectors.
FIGURE 3-10 Resolving a vector V into its com ponents along an arbitrarily chosen set of x and y axes. The components, once found, themselves represent the vector. That is, the com ponents contain as much information as the vector itself.
(a)
(b)
SECTION 3 - 4 Adding Vectors by Components 55
A VX COS ^ = y
Vv
ta n 0 = v v 2= y2 v 2
vy
FIGURE 3-11 Finding the components of a vector using trigonometric functions.
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 V*, \ y , and \ z . Resolution of a vector in three dimensions is merely an extension of the above technique.
The use of trigonometric functions for finding the components of a vector is illustrated in Fig. 3-11, 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. 3-11. If we multiply the definition of sin 0 = Vy/V by V on both sides, we get
Vy = V sind.
(3-2a)
Similarly, from the definition of cos 0, we obtain
Vx = VcosO.
(3-2b)
Note that 0 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 0 it makes with the positive x axis.
We can shift from one description to the other using Eqs. 3-2, and, for the reverse, by using the theorem of Pythagoras* and the definition of tangent:
V = V v } + Vy2
(3-3a)
vtayn* = -
(3-3b)
as can be seen in Fig. 3-11. We can now discuss how to add vectors using components. The first step is to
resolve each vector into its components. Next we can see, using Fig. 3-12, that the addition of any two vectors Yxand V2to give a resultant, V = % + V2, implies that
Vr = V1r + V7 v y = iy v2y.
(3-4)
That is, the sum of the x components equals the x component of the resultant, and
the sum of the y components equals the y component of the resultant, as can be
verified by a careful examination of Fig. 3-12. Note that we do not add x components
to y components.
tIn three dimensions, the theorem of Pythagoras becomes V = \ / v £ + Vy + V?, where Vz is the component along the third, or z, axis.
56 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
If the magnitude and direction of the resultant vector are desired, they can be obtained using Eqs. 3-3.
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 direc­ tion 60.0° south of east for 47.0 km (Fig. 3-13a). 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. 3-13b. Since 6 1 has magnitude 22.0 km and points north, it has only a y component:
= 0, Dly = 22.0 km.
D 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 nega­ tive y axis. The resultant vector, D, has components:
Dlx + D2x
0 km + 23.5 km = +23.5 km
Dy = Diy D2y 22.0 km + (-40.7 km) = -18.7 km.
This specifies the resultant vector completely:
Dx = 23.5 km, Dy = -18.7 km.
We can also specify the resultant vector by giving its magnitude and angle using Eqs. 3-3:
D = V d! A
tan 6 = —- = Dy
UDy2 = y j (23.5 km)2 + (-18.7 km)2 = 30.0 km 18.7 km
= -0.796. 23.5 km
A calculator with an in v t a n , an a r c t a n , or a t a n -1 key gives 6 = tan-1(-0.796) = -38.5°. The negative sign means 0 = 38.5° below the x axis, Fig. 3-13c. 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.
1North Dr
x
Post 0 \
East
office
(a) y
Dr
d 2x 0 ^60°
D 2y (b)
0
fir*
(c)
FIGURE 3-13 Exam ple 3 -2 . (a) The two displacement vectors, D i and D 2 . (b) D 2 is resolved into its components, (c) Dx and D 2 are added graphically to obtain the resultant D. The component method of adding the vectors is explained in the Example.
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. The best way to keep track of angles, and to check any vector result, is always to draw a vector diagram. 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.
^P R O B L E M S O L V I N G
Identify the correct quadrant by drawing a careful diagram
SECTION 3 - 4 Adding Vectors by Components 57
s O LV/ Ar.
v
fiQ Adding Vectors
Pay careful attention to signs: any component that
o
points along the negative x or y axis gets a minus
& Here is a brief summary of how to add two or more
sign.
vectors using components:
5. Add the x components together to get the x compo­
1. Draw a diagram, adding the vectors graphically by nent of the resultant. Ditto for y:
either the parallelogram or tail-to-tip method.
Vx = Vlx + V2x + any others
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 so that vector 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.
Vy = Viy + 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. 3-3:
4. Calculate each component (when not given) using sines and cosines. If is the angle that vector
/-----------
Vy
V = V v f + V } , tanfl = - f -
makes with the positive x axis, then:
The vector diagram you already drew helps to obtain
Vix = V1cosd1, Vly = V ^sin^.
the correct position (quadrant) of the angle 0.
(a)
(b) FIGURE 3 - 1 4 Exam ple 3 -3 .
Vector
Di D2 d3 Dr
Components
x (km)
y (km)
620 311 -331
0 -311 -439
600
-750
58 CHAPTER 3
EXAMPLE 3 -3 Three short trips. An airplane trip involves three legs, with two stopovers, as shown in Fig. 3-14a. 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. 3-14a, where , D2, and D3represent the three
legs of the trip, and DRis the plane’s total displacement. 2. Choose axes: Axes are also shown in Fig. 3-14a: x is east, y north. 3. Resolve components: It is imperative to draw a good diagram. The components
are drawn in Fig. 3-14b. Instead of drawing all the vectors starting from a common origin, as we did in Fig. 3-13b, here we draw them “tail-to-tip” style, which is just as valid and may make it easier to see. 4. Calculate the components:
D i : AIXY — + A COS 0° A y = +DXsin 0°
Di = 620 km 0 km
D2: D2x = + A cos 45c Ay = - D 2sin 45°
+ (440 km) (0.707) -(4 4 0 km) (0.707)
+311 km -311 km
D3: D3x = - D 3cos 53c -(550 km) (0.602) -331 km
Ay = - D 3sin 53° -(550 km) (0.799) -439 km. We have given a minus sign to each component that in Fig. 3-14b 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:
Dx = DXx + D2x + D3x = 620 km + 311 km —331 km = 600 km
Dv = Dly + A*2vy + A'3vy = 0 km - 311 km - 439 km = -750 km. The x and y components 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 tan 6
\ j D l + Dyl = \/(6 0 0 ) 2 + (-7 5 0 )2km = 960 km
A
-750 km
1.25,
so e = -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. 3-14a.
3 —5 Unit Vectors
Vectors can be conveniently written in terms of unit vectors. A unit vector is defined to have a magnitude exactly equal to one (1). 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 i, j, and k. They point, respectively, along the positive x, y, and z axes as shown in Fig. 3-15. Like other vectors, i, j, and k 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”: i, j, k (and we will do so in this book) as a reminder that each is a unit vector.
Because of the definition of multiplication of a vector by a scalar (Section 3-3), the components of a vector V can be written \ x = Vxi, = Vyj, and \ z = Vz k. Hence any vector V can be written in terms of its components as
V = Vx\ + Vyi + VZL
(3-5)
Unit vectors are helpful when adding vectors analytically by components. For example, Eq. 3-4 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):
v = ( v x ) i + (v y) i = Vi + V2
= { v j + vlyj) + (v2J + v2yj) = {vlx + v2x)\ + (ivly + vly) i
Comparing the first line to the third line, we get Eq. 3-4.
EXAMPLE 3 -4 Using unit vectors. Write the vectors of Example 3-2 in unit vector notation, and perform the addition.
APPROACH We use the components we found in Example 3-2,
Dlx = 0, Dly = 22.0 km, and Dlx = 23.5 km, D2y = -40.7 km,
and we now write them in the form of Eq. 3-5.
SOLUTION We have
Dx = Oi + 22.0 km j
Then
£>2 = 23.5 km i - 40.7 km j. D = £>! + D2 = (0 + 23.5) km i + (22.0 - 40.7) km j
= 23.5 km i - 18.7 km j.
The components of the resultant displacement, D, are Dx = 23.5 km and Dy = -18.7 km. The magnitude of D is D = V(23.5km)2 + (18.7 km)2 = 30.0 km, just as in Example 3-2.
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. 3-16. At time tx, the particle is at point Px, and at time t2, it is at point P2. The vector rl 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 r2 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 Ar,f where
Ar = r2 — ?i.
This represents the displacement during the time interval At = t2 - tx.
fWe used D for the displacement vector earlier in the Chapter for illustrating vector addition. The new notation here, A?, emphasizes that it is the difference between two position vectors.
y
z FIGURE 3-15 U nit vectors i, j, and k along the x, y, and z axes.
FIGURE 3-16 Path o f a particle in the xy plane. A t time t\ the particle is at point Pi given by the position vector ?!; at t2 the particle is at point P2 given by the position vector r2 . The displacement vector for the time interval t2 ~ h is A? = r2 — ? i .
y
0 SECTION 3 -6 59
y Af
In unit vector notation, we can write
?! = xxi + y j + zik,
(3-6a)
where x l ,y l , and Z\ are the coordinates of point . Similarly,
Hence
r2 = x2i + y2j + z2k. Ar = (x2 - x j i + (y2 - y j j + (z2 - Zi)k.
(3-6b)
If the motion is along the x axis only, then y2 — yx = 0, z2 — Z\ = 0, and the magnitude of the displacement is Ar = x2 — x x, which is consistent with our earlier one-dimensional equation (Section 2-1). Even in one dimension, displace­ ment is a vector, as are velocity and acceleration.
The average velocity vector over the time interval At = t2 — tx is defined as
Ar average velocity = — •
(3-7)
Now let us consider shorter and shorter time intervals—that is, we let At approach zero so that the distance between points P2 and also approaches zero, Fig. 3-17. We define the instantaneous velocity vector as the limit of the average velocity as At approaches zero:
FIGURE 3 - 1 7 (a) A s we take At and Ar smaller and smaller [compare to Fig. 3-16] we see that the direction of Ar and of the instantaneous velocity ( A r /A t, where At —> 0) is (b) tangent to the curve at P j.
FIGURE 3 - 1 8 (a) Velocity vectors \ i and v2 at instants fj and t2 for a particle
at points Pi and P2, as in Fig. 3-16.
(b) The direction of the average acceleration is in the direction of
Av = v2 —\ 1.
y
V = lim 4 ? = f . A^O At dt
(3-8)
The direction of v at any moment is along the line tangent to the path at that moment (Fig. 3-17).
Note that the magnitude of the average velocity in Fig. 3-16 is not equal to the average speed, which is the actual distance traveled along the path, A£, divided by At. In some special cases, the average speed and average velocity are equal (such as motion along a straight line in one direction), but in general they are not. However, in the limit At —» 0, Ar always approaches A£, so the instantaneous speed always equals the magnitude of the instantaneous velocity at any time.
The instantaneous velocity (Eq. 3-8) is equal to the derivative of the position vector with respect to time. Equation 3-8 can be written in terms of components starting with Eq. 3-6a as:
dr dx - dy ~ dz -
■? ? -
v = —dt = —dt 1 + —d t jJ + —dt k = vxx \ + vyvJj + t>7zk,
v(3-9)'
where vx = dx/dt, vy = dy/dt, vz = dz/dt are the x, y, and z components of the velocity. Note that di/dt = dj/dt = d i/d t = 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 At = t2 - tx is defined as
average acceleration =
= —---- —>
At t2 t-^
(3-10)
where Av is the change in the instantaneous velocity vector during that time interval: Av = v2 - vx. Note that v2 in many cases, such as in Fig. 3-18a, may not be in the same direction as \ 1. Hence the average acceleration vector may be in a different direction from either \ 1or v2 (Fig. 3-18b). Furthermore, v2and \ 1may 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 At is allowed to approach zero:
Av dy a = Ali—m>o—A—t = dt
v(3-11)7
(b)
and is thus the derivative of v with respect to t.
60 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
We can write a using components:
d\ dvx - dvy * dvz *
a = — = —— 1 + —— l + —— k
dt dt
dt J dt
= ax i + ayj + az k,
(3-12)
where 0* = dvx/dt, etc. Because vx = dx/dt, then ax = d v j d t = d2x /d t2, as we saw in Section 2-4. Thus we can also write the acceleration as
a = --d-r-2x-r ii- H+---t-d-2ty—Jj-i H+---Td--2-z-ru~t
(3-12c)
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 instan­ taneous values. If we want to discuss average values, we will use the word “average.”
■
Position given as a function of time. The position of a
particle as a function of time is given by
r = [(5.0m/s)^ + (6.0m/s2)£2]i + [(7.0m) - (3.0 m /s3)^3]j,
where r is in meters and t is in seconds, {a) What is the particle’s displacement between tx = 2.0 s and t2 = 3.0 s? (b) Determine the particle’s instantaneous velocity and acceleration as a function of time, (c) Evaluate v and a at t = 3.0 s.
APPROACH For (a), we find Ar = r2 - ? i, inserting tx = 2.0 s for finding ^ , and t2 = 3.0 s for ?2. For (b), we take derivatives (Eqs. 3-9 and 3-11), and for (c) we substitute t = 3.0 s into our results in (b).
SOLUTION (a) At tx = 2.0 s,
?! = [(5.0m/s)(2.0s) + (6.0m/s2)(2.0s)2]i + [(7.0m) - (3.0m/s3)(2.0s)3]j
= (34 m )i - (17 m) j.
Similarly, at t2 = 3.0 s,
Thus
f2 = (15m + 54m )i + (7.0m - 81m )j = (69m)i - (74m)j.
Ar = r2 - Tj = (69 m - 34 m) i + (-7 4 m + 17 m) j = (35 m) i - (57 m) j.
That is, Ax = 35 m, and Ay = -5 7 m.
(b) To find velocity, we take the derivative of the given ? with respect to time, noting (Appendix B-2) that d(t2)/dt = 21, and d (f)/d t = 312:
v = f = [5.0 m/s + (l2 m /s2)f]i + [0 - (9.0m/s3)?2]j.
The acceleration is (keeping only two significant figures):
a = — = (l2 m /s2)i - (l8 m /s3)fj.
Thus ax = 12m/s2 is constant; but ay = - ( l8 m /s 3)? 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 v and a:
v = (5.0 m/s + 36 m/s) i - (81 m /s)j = (41 m/s) i - (81 m /s)j
a = (l2 m /s2)i - (54 m/s2)j.
Their magnitudes at t = 3.0 s are v = (41 m /s)2 + (81 m /s)2 = 91 m/s, and a = V (l2 m /s 2)2 + (54 m/s2)2 = 55 m/s2.
SECTION 3 - 6 Vector Kinematics 61
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, a, 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 - 12a, b, and c, apply separately to each perpendicular component of two- or three-dimensional motion. In two dimensions we let v0 = + vyoi be the initial velocity, and we apply Eqs. 3-6a, 3-9, and 3-12b for the position vector, r, velocity, v, and acceleration, a. We can then write Eqs. 2 -12a, b, and c, for two dimensions as shown in Table 3-1.
TABLE 3-1 Kinematic Equations for Constant Acceleration in 2 Dimensions
x Component (horizontal)
y Component (vertical)
v x = vx0 + ax t x = * 0 + v x01 + \ax t2 v \ = v\o + 2ax (x ~ * 0)
+ o
II 55s
(Eq. 2-12a) (Eq. 2-12b) (Eq. 2-12c)
y = yo + vy0t + \ay t2 v 2y = v 2y0 + 2ay (y - y^)
The first two of the equations in Table 3-1 can be written more formally in vector notation.
v = v0 + a£
fa = constant] (3-13a)
r = f 0 + %t + t2.
[a = constant] (3-13b)
Here, r is the position vector at any time, and r0 is the position vector at t = 0. These equations are the vector equivalent of Eqs. 2 - 12a and b. In practical situa­ tions, we usually use the component form given in Table 3-1.
FIGURE 3-19 This strobe photograph of a ball making a series of bounces shows the characteristic “parabolic” path o f projectile motion.
3 —7 Projectile Motion
In Chapter 2, we studied one-dimensional motion of an object in terms of displace­ ment, velocity, and acceleration, including purely vertical motion of a falling object undergoing acceleration due to gravity. Now we examine the more general transla­ tional 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. 3-19), which we can describe as taking place in two dimensions.
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.1
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).
Let us look at a (tiny) ball rolling off the end of a horizontal table with an initial velocity in the horizontal (x) direction, vxQ. See Fig. 3-20, where an object falling vertically is also shown for comparison. The velocity vector v at each instant points in the direction of the ball’s motion at that instant and is always tangent to the path. Following Galileo’s ideas, we treat the horizontal and vertical compo­ nents of the velocity, vx and vy , separately, and we can apply the kinematic equations (Eqs. 2-12a through 2-12c) 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
62 CHAPTER 3
trThis restricts us to objects whose distance traveled and maximum height above the Earth are small compared to the Earth’s radius (6400 km).
FIGURE 3-20 Projectile m otion of a small ball projected horizontally. The dashed black line represents the path of the object. The velocity vector v at each point is in the direction of m otion and thus is tangent to the path. The velocity vectors are green arrows, and velocity components are dashed. (A vertically falling object starting at the same point is shown at the left for comparison; v y is the same for the falling object and the projectile.)
ball leaves the table (at t = 0), it experiences a vertically downward acceleration g, the acceleration due to gravity. Thus vy is initially zero {vyQ = 0) but increases continually in the downward direction (until the ball hits the ground). Let us take y to be positive upward. Then ay = —g, and from Eq. 2-12a we can write vy = —gt since we set vyQ = 0. The vertical displacement is given by y = —\g t2.
In the horizontal direction, on the other hand, the acceleration is zero (we are 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 is then given by x = vx0t. The two vector components, \ x and \ y, can be added vectorially at any instant to obtain the velocity v at that time (that is, for each point on the path), as shown in Fig. 3-20.
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 vertically. This is because the vertical motions are the same in both cases, as shown in Fig. 3-20. Figure 3-21 is a multiple-exposure photograph of an experi­ ment that confirms this.
EXERCISE D Return to the Chapter-Opening Q uestion, page 51, and answer it again now. Try to explain why you may have answered differently the first time.
If an object is projected at an upward angle, as in Fig. 3-22, the analysis is similar, except that now there is an initial vertical component of velocity, vy{). 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. Subsequently the object moves downward (Fig. 3-22) and vy increases in the downward direction, as shown (that is, becoming more negative). As before, vx remains constant.
FIGURE 3-21 M ultiple -exposure photograph showing positions of two balls at equal time intervals. One 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.
FIGURE 3-22 Path of a projectile fired with initial velocity v0 at angle 0Oto the horizontal. Path is shown dashed in black, the velocity vectors are green arrows, and velocity components are
dashed. The acceleration a = d \ / d t is downward. That is, a = g = - g j where j is the unit vector in
the positive y direction.
SECTION 3 - 7 Projectile Motion 63
PROBLEM SOLVING Choice o f time interval
3 —8 Solving Problems Involving Projectile Motion
We now work through several Examples of projectile motion quantitatively. We can simplify Eqs. 2-12 (Table 3-1) for the case of projectile motion
because we can set ax = 0. See Table 3-2, which assumes y is positive upward, so ay = —g = -9.80 m /s2. Note that if 0 is chosen relative to the +x axis, as in Fig. 3-22, then
V x0 = Vq COS 0o,
V y o = Vosin 0O.
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 a = g.
TABLE 3-2 Kinematic Equations for Projectile Motion (y positive upward; ax = 0, ay = - g = -9 .8 0 m/s2)
Horizontal Motion (ax = 0,vx = constant)
Vertical Motion^ [ay = —g = constant)
X = x0 + vx0t
(Eq. 2-12a) (Eq. 2-12b) (Eq. 2-12c)
s'
II II
s'
o s' +
&0 I o
Vy = Vyo ~ 2g ( y - y0)
vHlrvi 1
*If y is taken positive downward, the minus (—) signs in front of g become plus (+) signs.
CN
^9-° B z
£ II £O
Projectile Motion
Our approach to solving problems in Section 2-6 also applies here. Solving problems involving projec­ tile motion can require creativity, and cannot be done just by following some rules. Certainly 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 to the object.
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.
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, and using the + or - sign, depending on whether you choose y positive down or up. Remember that vx never changes throughout the trajectory, and that 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. A little planning goes a long way. Apply the relevant equations (Table 3-2), combining equations if neces­ sary. You may need to combine components of a vector to get magnitude and direction (Eqs. 3-3).
64 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
EXAMPLE 3 -6 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 cliff top 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 above.
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. 3-23.
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 cliff top at position x0 = 0, ;y0 = 0; our time interval ends just before the motorcycle hits 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 accelera­ tion 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 motorcycle 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 hits 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. 2-12b (Table 3-2) for the vertical (y) direction with y0 = 0 and vy0 0:
VyQt + \ a yt2
Xo
II II
oo
T
= g
50.0 m
8
y = -50.0
h----------------90.0 m ----------------- H
FIGURE 3 - 2 3 Exam ple 3 -6 .
j?
II
£
Known
x = 90.0 m y = -50.0 m
II
o
Unknown
Vxo t
ay = - g = -9.80 m /s2
or
y = - i s * 2-
We solve for t and set y = -50.0 m:
2z
2(-50.0 m) = 3.19 s.
-9.80 m/s2
To calculate the initial velocity, vx0, we again use Eq. 2-12b, but this time for the horizontal (x) direction, with ax = 0 and x0 = 0:
X = XQ + vx01 + \ a x t2 = 0 + vx0 t + 0
Then
X = vx0t.
x
vx0 = t
90.0 m = 28.2 m/s, 3.19 s
which is about 100 km /h (roughly 60mi/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.
SECTION 3 -8 Solving Problems Involving Projectile Motion 65
FIGURE 3-24 Example 3-7.
® -PH YS I CS A P P L I E D Sports
EXAMPLE 3 -7 A kicked football. A football is kicked at an angle 0O= 37.0° with a velocity of 20.0 m/s, as shown in Fig. 3-24. Calculate (a) the maximum height, (b) the time of travel before the football hits the ground, (c) how far away it hits the ground, (d) the velocity vector at the maximum height, and (e) the acceleration vector at maximum height. Assume the ball leaves the foot at ground level, and ignore air resistance 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. 3-24):
vxo = cos 37.0° = (20.0 m/s) (0.799) = 16.0 m/s
= ^osin 37.0° = (20.0 m/s) (0.602) = 12.0 m/s.
(a) We consider a time interval that begins just after the football loses contact with the foot until it reaches its maximum height. During this time interval, the acceleration is g downward. At the maximum height, the velocity is horizontal (Fig. 3-24), so vy = 0; and this occurs at a time given by vy = vy0 - gt with )y = 0 (see Eq. 2 - 12a in Table 3-2). Thus
vyo t =
§
(12.0 m/s) = 1.224 s « 1.22 s.
(9.80 m /s2)
From Eq. 2-12b, with yQ = 0, we have
y = Vyot - \g t2 = (12.0m/s)(1.224s) —j(9.80m /s2)(1.224s)2 = 7.35m.
Alternatively, we could have used Eq. 2-12c, solved for y, and found
Vyo - V y
y = 2g
(12.0 m /s)2 - (Om/s)2 = 7.35 m.
2(9.80 m /s2)
The maximum height is 7.35 m.
(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. 2-12b with y0 = 0 and also set y = 0 (ground level):
y = y» + vy0t - \g t2
o = o + vy0t - \g t2.
This equation can be easily factored:
t(kgt - Vyo) = 0.
There are two solutions, t = 0 (which corresponds to the initial point, y0), and
2vy0 t =
g
2(12.0 m/s) = 2.45 s,
(9.80 m /s2)
which is the total travel time of the football
66 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
NOTE 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). (c) The total distance traveled in the x direction is found by applying Eq. 2 - 12b with Xq = 0, ax = 0, vxq = 16.0 m/s:
x = vxot = (16.0 m /s) (2.45 s) = 39.2 m.
(d) At the highest point, there is no vertical component to the velocity. There is only the horizontal component (which remains constant throughout the flight), so v = vx0 = v0cos 37.0° = 16.0 m/s. (e) The acceleration vector is the same at the highest point as it is throughout the flight, which is 9.80 m /s2 downward. NOTE 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.
EXERCISE E Two balls are thrown in the air at different angles, but each reaches the same height. W hich ball remains in the air longer: the one thrown at the steeper angle or the one thrown at a shallower angle?
CONCEPTUAL EXAMPLE 5 -8 I Where does the apple land? A child sits upright in a wagon which is moving to the right at constant speed as shown in Fig. 3-25. The child extends her hand and throws an apple straight upward (from her own point of view, Fig. 3-25a), while the wagon continues to travel forward at constant speed. If air resistance is neglected, 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 from her own reference frame with initial velocity \ y0 (Fig. 3-25a). But when viewed by someone on the ground, the apple also has an initial horizontal component of velocity equal to the speed of the wagon, v ^ . Thus, to a person on the ground, the apple will follow the path of a projectile as shown in Fig. 3-25b. The apple experiences no horizontal acceleration, so v*o will stay constant and 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).
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. 3-26. 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 he made the wrong move. (He hadn’t studied physics yet.) Ignore air resistance. 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 = \g t2, much like Fig. 3-21. In the time it takes the water balloon to travel the horizontal 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 humiliation.
V,<>
t L
(a) Wagon reference frame
A
(h) Ground reference frame FIGURE 3 - 2 5 Exam ple 3 -8 .
y = 0 FIGURE 3 - 2 6 Exam ple 3 -9 .
SECTION 3 -8 Solving Problems Involving Projectile Motion 67
(b)
FIGURE 3-27 Exam ple 3-1 0 . (a) The range R of a projectile; (b) there are generally two angles 0O that will give the same range. Can you show that if one angle is 0O1 > th e other is 0O2 = 90° - 601 ?
EXAMPLE 3-10 Level horizontal range. (a) Derive a formula for the hori­ zontal range R of a projectile in terms of its initial speed vQand angle 0O. The horizontal range is defined as the horizontal distance the projectile travels before returning to its original height (which is typically the ground); that is, y (final) = yQ. See Fig. 3-27a. (b) 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 The situation is the same as in Example 3-7, except we are now not given numbers in (a). We will algebraically manipulate equations to obtain our result.
SOLUTION (a) We set x0 = 0 and y0 = 0 at t = 0. After the projectile travels a horizontal distance R, it returns to the same level, y = 0, the final point. We choose our time interval to start (t = 0) just after the projectile is fired and to end when it returns to the same vertical height. To find a general expression for R, we set both y = 0 and y0 = 0 ill Eq. 2-12b for the vertical motion, and obtain
y = yo + vy0t + \a y t2 so
0 = 0 + VyO t - \g t2.
We solve for t, which gives two solutions: t = 0 and t = 2vyJ g . The first solu­ tion corresponds to the initial instant of projection and the second is the time when the projectile returns to y = 0. Then the range, R, will be equal to x at the moment t has this value, which we put into Eq. 2 - 12b for the horizontal motion (x = vx0t, with x0 = 0).Thus we have:
(2vy0\ 2vx0vy0 2vq sin 0Ocos 0O
r
n
r = vx0t = vxoy— J = — - — = -------- ---------- b = yoI
where we have written vx0 = v0cos 0O and vy0 = v0sin 0O. This is the result we sought. It can be rewritten, using the trigonometric identity 2 sin 0 cos 0 = sin 20 (Appendix A or inside the rear cover):
Vo sin 20o R ---------------
8
r , ,e „ [only if y (final) = y0]
We see that the maximum range, for a given initial velocity , is obtained when sin 20 takes on its maximum value of 1.0, which occurs for 20o = 90°; so
0O = 45° for maximum range, and Rmax = vl/g.
[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°.] NOTE The maximum range increases by the square of v0, so doubling the muzzle velocity of a cannon increases its maximum range by a factor of 4.
(b) We put R = 320 m into the equation we just derived, and (assuming, unrealistically, no air resistance) we solve it to find
. _
Rg (320 m)(9.80 m/s2) _
sin 20o =
= -----—— — —;----- = 0.871.
0 v20
(60.0 m /s)2
We want to solve for an angle 0Othat is between 0° and 90°, which means 20O in this equation can be as large as 180°. Thus, 20O= 60.6° is a solution, but 20o = 180° - 60.6° = 119.4° is also a solution (see Appendix A-9). In general we will have two solutions (see Fig. 3-27b), which in the present case are given by
0O = 30.3° or 59.7°.
Either angle gives the same range. Only when sin 20o = 1 (so 0O= 45°) is there a single solution (that is, both solutions are the same).
68 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
EXERCISE F The maximum range of a projectile is found to be 100 m. If the projectile strikes the ground a distance of 82 m away, what was the angle o f launch? (a) 35° or 55°; (b) 30° or 60°; (c) 27.5° or 72.5°; (d) 13.75° or 76.25°.
The level range formula derived in Example 3-10 applies only if takeoff and landing are at the same height (y = y0)- Example 3-11 below considers a case where they are not equal heights (y ^ y0)-
EXAMPLE 3-11 A punt. Suppose the football in Example 3-7 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 from Example 3-10 because it 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. 3-28). 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. 2-12b, x = vx0t, since we know that vx0 = 16.0 m /s from Example 3-7. But first we must find t, the time at which the ball hits the ground, which we obtain from the y motion.
0 PHYSICS APPLIED Sports
33,
jf| P R O B L E M S O L V I N G D o not use any form ula unless you are sure its range o f validity fits the problem ; the range form ula does n ot apply here because y ^ yo
FIGURE 3 - 2 8 Exam ple 3-11: the football leaves the punter’s foot at y = 0, and reaches the ground where y = —1.00 m.
Ground
SOLUTION With y = -1.00 m and vy0 = 12.0 m /s (see Example 3-7), we use the equation
y = yo + vy01 - \g t2,
and obtain
-1 .0 0 m = 0 + (12.0m/s)f - (4.90m/s2)f2.
We rearrange this equation into standard form (ax2 + bx + c = 0) so we can use the quadratic formula:
(4.90m /s2)t2 — (12.0m /s)t - (1.00m) = 0.
The quadratic formula (Appendix A - l) gives
12.0m/s + \ / ( - Y 2 t i m / s ) 2 - 4(4.90m/s2)(-1.00m )
1 ~
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, 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 Example 3-7):
x = vxot = (16.0 m /s) (2.53 s) = 40.5 m.
Our assumption in Example 3-7 that the ball leaves the foot at ground level would result in an underestimate of about 1.3 m in the distance our punt traveled.
SECTION 3 -8 Solving Problems Involving Projectile Motion 69
200 m
“Dropped”
v ( > = 0)
200 m
J- h
Thrown downward?
(><o)
400 m
FIGURE 3 - 2 9 Exam ple 3 -1 2 .
PHYSICS APPLIED Reaching a target
from a m oving 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. 3-29a)? (b) Suppose, instead, that the heli­ copter 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. 3-29b)? (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 3-2).
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, vyo = 0. Then, since y = - \g t2, we have
t = -2y 8
-2(-200m) = 6.39 s.
9.80 m /s2
The horizontal motion of the falling package is at constant speed of 70 m/s. So
x = vxot = (70 m/s) (6.39 s) = 447 m ~ 450 m,
assuming the given numbers were good to two significant figures.
(p) We are given x = 400 m, vx0 = 70 m/s, y = -200 m, and we want to find vyQ (see Fig. 3-29b). 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 = vx0t, so
t= x
400 m __
Vxo 70 m/s
Now let’s try to use the vertical motion to get uyQ, y = y0 + vy0t - \g t2. Since >>o = 0 and y = -200m , we can solve for vy0:
y + \g t2 -200 m + \ (9.80 m /s2)(5.71 s)2 Vyo = ------t----- - -----------------5--.7--1--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 = V x o = 70 m/s V y = V yo ~ gt = -7.0 m/s - (9.80 m /s2)(5.71 s) = -6 3 m/s.
So v = x /(1 0 m /s )2 + (-6 3 m /s)2 = 94m/s. (Better not to release the package from such an altitude, or use a parachute.)
70 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
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 g is 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. 2-12b in Table 3-2), and for simplicity we set x0 = y0 = 0:
* = vx0t 2 y = V y 0 t - J g t From the first equation, we have t = x /v x0, and we substitute this into the second one to obtain
We see that y as a function of x has the form y = A x - B x2,
where A and B are constants for any specific projectile motion. This is the well-known equation for a parabola. See Figs. 3-19 and 3-30.
The idea that projectile motion is parabolic was, in Galileo’s day, at the forefront of physics research. Today we discuss it in Chapter 3 of introductory physics!
FIGURE 3 - 3 0 Exam ples of projectile m otion— sparks (small hot glowing pieces of m etal), water, and fireworks. The parabolic path characteristic of projectile m otion is affected by air resistance.
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 /hr for the speed of each of the trains. Observers on either one of the trains (a different frame of reference) will measure a speed of 160 km /h for the 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 2-1, that when specifying a velocity, it is important to specify what the reference frame is.
SECTION 3 - 9 Relative Velocity 71
River current
FIGURE 3 -3 1 To m ove directly across the river, the boat must head upstream at an angle 0. Velocity vectors are shown as green arrows:
vBs = velocity o f Boat with respect to the Shore,
yBw = velocity o f Boat with respect to the Water,
vWs = velocity o f the Water with respect to the Shore (river current).
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 tw o su b scrip ts: the first
refers to the object, the seco n d to the reference fra m e in w h ich it has this velocity.
For example, suppose a boat is to cross a river to the opposite side, as shown in Fig. 3-31. We let vBWbe 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, vBS is the velocity of the Boat with respect to the Shore, and vws is the velocity of the Water with respect to the Shore (this is the river current). Note that vBWis what the boat’s motor produces (against the water), whereas vBS is equal to vBWplus the effect of the current, vws. Therefore, the velocity of the boat relative to the shore is (see vector diagram, Fig. 3-31)
VBS — VBW + VWS*
(3-15)
By writing the subscripts using this convention, we see that the inner subscripts (the two W’s) on the right-hand side of Eq. 3-15 are the same, whereas the outer subscripts on the right of Eq. 3-15 (the B and the S) are the same as the two subscripts for the sum vector on the left, vBS. 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.f Figure 3-32 gives a derivation of Eq. 3-15.
Equation 3-15 is valid in general and can be extended to three or more veloc­ ities. For example, if a fisherman on the boat walks with a velocity vFB relative to the boat, his velocity relative to the shore is vFS = vFB + vBW+ vws. 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:
VBA = - V a b
(3-16)
For example, if a train is traveling 100 km /h relative to the Earth in a certain direc­ tion, objects on the Earth (such as trees) appear to an observer on the train to be traveling 100 km /h in the opposite direction.
fWe thus would know by inspection that (for example) the equation VBW = VBS + Vws is wrong.
FIGURE 3 - 3 2 Derivation of relative velocity equation (Eq. 3 -1 5 ), 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:
rPT = position vector of person (P) relative to train (T ), rPE = position vector of person (P) relative to Earth (E ), ?te = position vector of train’s coordinate system (T) relative to Earth (E ). From the diagram we see that
?p e = ? p t + ?TE-
We take the derivative with respect to time to obtain
f f e ) = J t ( i pt) + |( ? t e ) .
or, since d r /d t = v,
VpE = VPT + VTE.
This is the equivalent of Eq. 3 -1 5 for the present situation (check the subscripts!).
72 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
CONCEPTUAL EXAMPLE 3^151 Crossing a river. A woman in a small motor boat is trying to cross a river that flows due west with a strong current. The woman starts on the south bank and is trying to reach the north bank directly north from her starting point. Should she (a) head due north, (b) head due west, (c) head in a north­ westerly direction, (d) head in a northeasterly direction?
RESPONSE If the woman heads straight across the river, the current will drag the boat downstream (westward). To overcome the river’s westward current, the boat must acquire an eastward component of velocity as well as a northward compo­ nent. Thus the boat must (d) head in a northeasterly direction (see Fig. 3-33). The actual angle depends on the strength of the current and how fast the boat moves relative to the water. If the current is weak and the motor is strong, then the boat can head almost, but not quite, due north.
Heading upstream. A boat’s speed in still water is vQW = 1.85 m/s. If the boat is to travel directly across a river whose current has speed vws = 1.20 m/s, at what upstream angle must the boat head? (See Fig. 3-33.)
APPROACH We reason as in Example 3-13, and use subscripts as in Eq. 3-15. Figure 3-33 has been drawn with vBS, the velocity of the Boat relative to the Shore, pointing directly across the river because this is how the boat is supposed to move. (Note that vBS = vBW+ vws.) To accomplish this, the boat needs to head upstream to offset the current pulling it downstream.
SOLUTION Vector vBWpoints upstream at an angle 0 as shown. From the diagram,
sin0 =
VBW
1.20 m/s = 0.6486.
1.85 m/s
Thus 0 = 40.4°, so the boat must head upstream at a 40.4° angle.
River current
N
y E
. ^ws ,i
/ S
vBs <9 ^ VBW
J 'J
FIGURE 3-33 Exam ples 3 -1 3 and 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 down­ stream by the current, as shown in Fig. 3-34. The boat’s velocity with respect to the shore, vBS, is the sum of its velocity with respect to the water, vBW, plus the velocity of the water with respect to the shore, vws:
VBS = VBW + vws, just as before. SOLUTION (a) Since vBW is perpendicular to vws, we can get vBS using the theorem of Pythagoras:
vBS = V ^bw + vws = \/(1 .8 5 in /s)2 + (1.20 m /s)2 = 2.21 m/s. We can obtain the angle (note how 0 is defined in the diagram) from:
tan0 = v w s/ v b w = (1.20 m/s)/(1.85 m/s) = 0.6486.
Thus 0 = tan1(0.6486) = 33.0°. Note that this angle is not equal to the angle calculated in Example 3-14. (b) The travel time for the boat is determined by the time it takes to cross the river. Given the river’s width D = 110 m, we can use the velocity component in the direction of D, vBW = D /t. Solving for t, we get t = 110 m/1.85 m /s = 59.5 s. The boat will have been carried downstream, in this time, a distance
d = Vwst = (1.20m /s)(59.5 s) = 71.4m « 71m.
NOTE There is no acceleration in this Example, so the motion involves only constant velocities (of the boat or of the river).
FIGURE 3-34 Exam ple 3 -1 5 . A boat heading directly across a river whose current m oves at 1.20 m /s.
■punw m m m
River current
SECTION 3 - 9 Relative Velocity 73
FIGURE 3 - 3 5 Example 3-16.
i
(a)
(b)
EXAMPLE 3 -1 6 Car velocities at 90°. Two automobiles approach a street corner at right angles to each other with the same speed of 40.0 km /h (= 11.11 m /s), as shown in Fig. 3-35a. What is the relative velocity of one car with respect to the other? That is, determine the velocity of car 1 as seen by car 2.
APPROACH Figure 3-35a 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. 3-35b. In this reference frame (the world as seen by the driver of car 2), the Earth moves toward car 2 with velocity vE2 (speed of 40.0 km /h), which is of course equal and opposite to v2E, the velocity of car 2 with respect to the Earth (Eq. 3-16):
V2E = -V E2Then the velocity of car 1 as seen by car 2 is (see Eq. 3-15)
Vl2 = V1E + VE2 SOLUTION Because vE2 = - v 2E, then
= Vie - v2E. That is, the velocity of car 1 as seen by car 2 is the difference of their velocities, Vie “ v2E, both measured relative to the Earth (see Fig. 3-35c). Since the magnitudes of v1E, v2E, and vE2 are equal (40.0km /h = 11.11 m /s), we see (Fig. 3-35b) that v12 points at a 45° angle toward car 2; the speed is
v12 = 'x /(11.11 m /s)2 + (11.11 m /s)2 = 15.7 m /s (= 56.6 km /h).
Summary
A quantity that has both a magnitude and a direction is called a vector. A quantity that has only a magnitude is called a scalar.
Addition of vectors can be done graphically by placing the tail of each successive arrow (representing each vector) at the tip of the previous one. The sum, or resultant vector, is the arrow drawn from the tail of the first to the tip of the last. Two vectors can also be added using the parallelogram method.
Vectors can be added more accurately using the analytical method of adding their components along chosen axes with the aid of trigonometric functions. A vector of magnitude V making an angle 6 with the x axis has components
Vx = V cosd
Vy = Vsind. (3-2)
Given the components, we can find the magnitude and direction from
,_______ v = V v 2x + V2y,
Vv tan 6 = - f -
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 i, j, and k.
The general definitions for the instantaneous velocity, v, and acceleration, a, of a particle (in one, two, or three dimen­ sions) are
where r is the position vector of the particle. The kinematic equations 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. 2-12). Or they can be written in the more general vector form:
v = y0 + a?
r = rQ+ yQt + jat2
(3-13)
Projectile motion of an object moving in the air near the Earth’s surface can be analyzed as two separate motions if air
74 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
resistance can be ignored. The horizontal component of the motion is at constant velocity, whereas the vertical component is at constant acceleration, g, just as for an object falling vertically under the action of gravity.
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 speedometer indicates a steady 60 km/h?
3. Can you 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 ever be longer than the length of path traveled by the particle over the same time interval? Can it ever be less? Discuss.
5. During baseball practice, a batter 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?
6. If V = Vi + V2, is V necessarily greater than V\ and/or V2? Discuss.
7. Two vectors have length V\ = 3.5 km and V2 = 4.0 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. Can a particle with constant speed be accelerating? What if it has constant velocity?
11. Does the odometer of a car measure a scalar or a vector quantity? What about the speedometer?
12. A child wishes to determine the speed a slingshot imparts to a rock. How can this be done using only a meter stick, a rock, and the slingshot?
13. In archery, should the arrow be aimed directly at the target? How should your angle of aim depend on the distance to the target?
The velocity of an object relative to one frame of reference 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.
14. 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?
15. A projectile has the least speed at what point in its path? 16. It was reported in World War I that a pilot flying at an
altitude of 2 km caught in his bare hands a bullet fired at the plane! Using the fact that a bullet slows down consid­ erably due to air resistance, explain how this incident occurred. 17. Two cannonballs, A and B, are fired from the ground with identical initial speeds, but with 0A larger than 0B. (a) Which cannonball reaches a higher elevation? (b) Which stays longer in the air? (c) Which travels farther? 18. A person sitting in an enclosed train car, moving at constant velocity, throws a ball straight up into the air in her refer­ ence 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? 19. 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? 20. 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? 21. 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?
Problems
3-2 to 3-5 Vector Addition; Unit Vectors
1. (I) A car is driven 225 km west and then 78 km southwest (45°). What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram.
2. (I) A delivery truck travels 28 blocks north, 16 blocks east, and 26 blocks south. What is its final displacement from the origin? Assume the blocks are equal length.
3. (I) If Vx = 7.80 units and Vy = —6.40 units, determine the magnitude and direction of V.
4. (II) Graphically determine the resultant of the following three vector displacements: (1) 24 m, 36° north of east; (2) 18 m, 37° east of north; and (3) 26 m, 33° west of south.
5. (II) V is a vector 24.8 units in magnitude and points at an angle of 23.4° above the negative x axis, (a) Sketch this vector. (b) Calculate Vx and Vy . (c) Use Vx and Vy to obtain (again) the magnitude and direction of V. [Note: Part (c) is a good way to check if you’ve resolved your vector correctly.]
6. (II) Figure 3-36 shows two vectors, A and B, whose magni­ tudes are A = 6.8 units and B = 5.5 units. Determine C if (a) C = A + B, (b) C = A - B, (c) C = B - A. Give the magnitude and direction for each. y
FIGURE 3-36 Problem 6. Problems 75
7. (II) An airplane is traveling 835 km /h in a direction 41.5° west
of north (Fig. 3-37). (a) Find
the components of the
N
velocity vector in
the northerly and westerly directions. (b) How far north
v v 41.5° (835 km/h)
and how far west
has the plane trav­ W -
eled after 2.50 h?
FIGURE 3-37 Problem 7.
8. (II) Let Vl = —6.0i + 8.0j and % = 4.51 - 5.0j. D eter­ mine the magnitude and direction of (a) Y1? (b) V2, (c) Yj + V2 and (d) V2 - Vi.
9. (II) (a) Determ ine the magnitude and direction of the sum of the three vectors Y i = 4.0i - 8.0j, V2 = i + j, and V3 = —2.0i + 4.0j. (b) Determine Yj - V2 + V3.
10. (II) Three vectors are shown in Fig. 3-38. Their magnitudes are given in arbitrary units. D etermine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with x axis.
3 -6 Vector Kinematics
17. (I) The position of a particular particle as a function of time is given by ? = (9.601i + 8.85j - 1.00£2k)m . Determine the particle’s velocity and acceleration as a function of time.
18. (I) What was the average velocity of the particle in Problem 17 between t = 1.00 s and t = 3.00 s? What is the magnitude of the instantaneous velocity at t = 2.00 s?
19. (II) W hat is the shape of the path of the particle of Problem 17?
20. (II) A car is moving with speed 18.0 m /s due south at one moment and 27.5 m /s due east 8.00 s later. Over this time interval, determine the magnitude and direction of (a) its average velocity, (b) its average acceleration, (c) What is its average speed. [Hint: Can you determine all these from the information given?]
21. (II) A t t = 0, a particle starts from rest at x = 0, y = 0,
and moves in the xy plane with an acceleration a = (4.0i + 3.0j) m /s2. Determine (a) the x and y compo­ nents of velocity, (b) the speed of the particle, and (c) the position of the particle, all as a function of time. (d) Eval­ uate all the above at t = 2.0 s.
22. (II) (a) A skier is accelerating down a 30.0° hill at 1.80 m /s2
(Fig. 3-39). What is the vertical component of her accelera­ tion? (b) How long will it take her to reach the bottom of the hill, assuming she starts from rest and accelerates uniformly, if the elevation change is 325 m?
a = 1.80 m/s2
FIGURE 3-38 Problems 10,11,12,13, and 14. Vector magnitudes are given in arbitrary units.
11. (II) (a) Given the vectors A and B shown in Fig. 3-38, determine B - A. (b) Determine A - B without using your answer in (a). Then compare your results and see if they are opposite.
12. (II) Determine the vector A —C, given the vectors A and C in Fig. 3-38.
13. (II) For the vectors shown in Fig. 3-38, determine (a) B - 2A, (b) 2A —3B + 2C.
14. (II) For the vectors given in Fig. 3-38, determine (a) A - B + C, (b) A + B - C, and (c) C - A - B.
15. (II) The summit of a mountain, 2450 m above base camp, is measured on a map to be 4580 m horizontally from the camp in a direction 32.4° west of north. What are the components of the displacement vector from camp to summit? What is its magnitude? Choose the x axis east, y axis north, and z axis up.
16. (Ill) You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of —55.0 units. (a) What are the two possibilities for its x component? (b) Assuming the x component is known to be positive, specify the vector which, if you add it to the original one, would give a resultant vector that is 80.0 units long and points entirely in the —x direction.
FIGURE 3-39 Problem 22.
23. (II) An ant walks on a piece of graph paper straight along the x axis a distance of 10.0 cm in 2.00 s. It then turns left 30.0° and walks in a straight line another 10.0 cm in 1.80 s. Finally, it turns another 70.0° to the left and walks another 10.0 cm in 1.55 s. Determine (a) the x and y components of the ant’s average velocity, and (b) its magnitude and direction.
24. (II) A particle starts from the origin at t = 0 with an initial velocity of 5.0 m /s along the positive x axis. If the accelera­ tion is (—3.0i + 4.5j)m /s2, determine the velocity and posi­ tion of the particle at the moment it reaches its maximum x coordinate.
25. (II) Suppose the position of an object is given by r = (3.012\ — 6.0 £3j)m . (a) Determine its velocity v and acceleration a, as a function of time, (b) Determine r and v at time t = 2.5 s.
26. (II) A n object, which is at the origin at time t — 0, has initial velocity v0 = (—14.0i - 7.0j)m /s and constant acceleration a = (6.0i + 3.0j)m /s2. Find the position r where the object comes to rest (momentarily).
76 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
27. (II) A particle’s position as a function of time t is given by r = (5.01 + 6.011) m i + (7.0 - 3.0£3)m j. A t t = 5.0 s, find the magnitude and direction of the particle’s displace­ ment vector Ar relative to the point r0 = (O.Oi + 7.0j) m.
3 -7 and 3 -8 Projectile Motion (neglect air resistance)
28. (I) A tiger leaps horizontally from a 7.5-m-high rock with a speed of 3.2 m/s. How far from the base of the rock will she land?
29. (I) A diver running 2.3 m /s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. How high was the cliff and how far from its base did the diver hit the water?
30. (II) Estimate how much farther a person can jump on the Moon as compared to the Earth if the takeoff speed and angle are the same. The acceleration due to gravity on the Moon is one-sixth what it is on Earth.
31. (II) A fire hose held near the ground shoots water at a speed of 6.5 m/s. A t what angle(s) should the nozzle point in order that the water land 2.5 m away (Fig. 3-40)? Why are there two different angles? Sketch the two trajectories.
39. (II) In Example 3-11 we chose the x axis to the right and y axis up. Redo this problem by defining the x axis to the left and y axis down, and show that the conclusion remains the same—the football lands on the ground 40.5 m to the right of where it departed the punter’s foot.
40. (II) A grasshopper hops down a level road. On each hop, the grasshopper launches itself at angle 60 = 45° and achieves a range R = 1.0 m. What is the average hori­ zontal speed of the grasshopper as it progresses down the road? Assume that the time spent on the ground between hops is negligible.
41. (II) Extreme-sports enthusiasts have been known to jump off the top of El Capitan, a sheer granite cliff of height 910 m in Yosemite National Park. Assume a jumper runs horizontally off the top of El Capitan with speed 5.0 m /s and enjoys a freefall until she is 150 m above the valley floor, at which time she opens her parachute (Fig. 3-41). (a) How long is the jumper in freefall? Ignore air resistance. (b) It is important to be as far away from the cliff as possible before opening the parachute. How far from the cliff is this jumper when she opens her chute?
5.0 m/s
\
910 m
FIGURE 3-40 J # Problem 31.
-2.5 m-
32. (II) A ball is thrown horizontally from the roof of a building 9.0 m tall and lands 9.5 m from the base. What was the ball’s initial speed?
33. (II) A football is kicked at ground level with a speed of 18.0 m /s at an angle of 38.0° to the horizontal. How much later does it hit the ground?
34. (II) A ball thrown horizontally at 23.7 m /s from the roof of a building lands 31.0 m from the base of the building. How high is the building?
35. (II) A shot-putter throws the shot (mass = 7.3 kg) with an initial speed of 14.4 m /s at a 34.0° angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete’s hand at a height of 2.10 m above the ground.
36. (II) Show that the time required for a projectile to reach its highest point is equal to the time for it to return to its orig­ inal height if air resistance is neglible.
37. (II) You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to find its maximum horizontal range. You shoot the gun straight up, and it takes 4.0 s for the dart to land back at the barrel. What is the maximum horizontal range of your gun?
38. (II) A baseball is hit with a speed of 27.0 m /s at an angle of
45.0°. It lands on the flat roof of a 13.0-m-tall nearby building. If the ball was hit when it was 1.0 m above the ground, what horizontal distance does it travel before it lands on the building?
FIGURE 3-41 Problem 41.
,1
I
150 m
42. (II) Here is something to try at a sporting event. Show that the maximum height h attained by an object projected into the air, such as a baseball, football, or soccer ball, is approx­ imately given by
h ~ 1.212m,
where t is the total time of flight for the object in seconds. Assume that the object returns to the same level as that from which it was launched, as in Fig. 3-42. For example, if you count to find that a baseball was in the air for t = 5.0 s, the maximum height attained was h = 1.2 X (5.0)2 = 30 m. The beauty of this relation is that h can be determined without knowledge of the launch speed vq or launch angle 0O.
FIGURE 3-4 2 Problem 42.
Problems 77
43. (II) The pilot of an airplane traveling 170 km /h wants to drop supplies to flood victims isolated on a patch of land 150 m below. The supplies should be dropped how many seconds before the plane is directly overhead?
44. (II) (a) A long jumper leaves the ground at 45° above the horizontal and lands 8.0 m away. What is her “takeoff” speed vq ? (b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0m away horizontally and 2.5 m, vertically below. If she long jumps from the edge of the left bank at 45° with the speed calculated in (a), how long, or short, of the opposite bank will she land (Fig. 3-43)?
47 (II) Suppose the kick in Example 3-7 is attempted 36.0 m from the goalposts, whose crossbar is 3.00 m above the ground. If the football is directed perfectly between the goalposts, will it pass over the bar and be a field goal? Show why or why not. If not, from what horizontal distance must this kick be made if it is to score?
48. (II) Exactly 3.0 s after a projectile is fired into the air from the ground, it is observed to have a velocity v = (8.6i + 4.8j) m/s, where the x axis is horizontal and the y axis is positive upward. Determine {a) the horizontal range of the projectile, (b) its maximum height above the ground, and (c) its speed and angle of motion just before it strikes the ground.
49. (II) Revisit Example 3-9, and assume that the boy with the slingshot is below the boy in the tree (Fig. 3-45) and so aims upward, directly at the boy in the tree. Show that again the boy in the tree makes the wrong move by letting go at the moment the water balloon is shot.
2,5 m
10.0 m
FIGURE 3-43 Problem 44.
45. (II) A high diver leaves the end of a 5.0-m-high diving board and strikes the water 1.3 s later, 3.0 m beyond the end of the board. Considering the diver as a particle, determine {a) her initial velocity, v0; (b) the maximum height reached; and (c) the velocity Vf with which she enters the water.
46. (II) A projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of 65.0 m/s at an angle of 35.0° with the horizontal, as shown in Fig. 3-44. (a) Deter­ mine the time taken by the projectile to hit point P at ground level. (b) Determine the distance X of point P from the base of the vertical cliff. At the instant just before the projectile hits point P, find (c) the horizontal and the vertical components of its velocity, (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal. ( /) Find the maximum height above the cliff top reached by the projectile.
= 65.0 m/s
FIGURE 3-45 Problem 49.
50. (II) A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3-46). (a) With what minimum speed must he drive off the hori­ zontal ramp? The vertical height of the ramp is 1.5 m above the cars and the horizontal distance he must clear is 22 m. (b) If the ramp is now tilted upward, so that “takeoff angle” is 7.0° above the horizontal, what is the new minimum speed?
22 m
Musi dear this print!
FIGURE 3-46 Problem 50. 51. (II) A ball is thrown horizontally from the top of a cliff
with initial speed v0 (at t = 0). At any moment, its direction of motion makes an angle 6 to the horizontal (Fig. 3-47). Derive a formula for 6 as a function of time, t, as the ball follows a projectile’s path.
FIGURE 3-44 Problem 46.
78 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
FIGURE 3-47 Problem 51.
52. (II) A t what projection angle will the range of a projectile equal its maximum height?
53. (II) A projectile is fired with an initial speed of 46.6 m /s at an angle of 42.2° above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total hori­ zontal distance covered (that is, the range), and (d) the velocity of the projectile 1.50 s after firing.
54. (II) A n athlete executing a long jump leaves the ground at a 27.0° angle and lands 7.80 m away, (a) What was the takeoff speed? (b) If this speed were increased by just 5.0%, how much longer would the jump be?
55. (Ill) A person stands at the base of a hill that is a straight incline making an angle <f>with the horizontal (Fig. 3-48). For a given initial speed vQ, at what angle 0 (to the hori­ zontal) should objects be thrown so that the distance d they land up the hill is as large as possible?
61. (II) A child, who is 45 m from the bank of a river, is being carried helplessly downstream by the river’s swift current of 1.0 m/s. As the child passes a lifeguard on the river’s bank, the lifeguard starts swimming in a straight line until she reaches the child at a point downstream (Fig. 3-50). If the lifeguard can swim at a speed of 2.0 m /s relative to the water, how long does it take her to reach the child? How far downstream does the lifeguard intercept the child?
11 .0 m/s
1
1
2.0 m/s
—»
i
FIGURE 3-4 8 Problem 55. Given (f>and v0, determine 0 to make d maximum.
56. (Ill) Derive a formula for the horizontal range R, of a projectile when it lands at a height h above its initial point. (For h < 0, it lands a distance —h below the starting point.) Assume it is projected at an angle 0Owith initial speed v0.
3 - 9 Relative Velocity
57. (I) A person going for a morning jog on the deck of a cruise ship is running toward the bow (front) of the ship at 2.0 m /s while the ship is moving ahead at 8.5 m/s. What is the velocity of the jogger relative to the water? Later, the jogger is moving toward the stern (rear) of the ship. What is the jogger’s velocity relative to the water now?
58. (I) Huck Finn walks at a speed of 0.70 m /s across his raft
(that is, he walks perpendicular to the raft’s motion relative
to the shore). The raft is traveling down the Mississippi
River at a speed of
1.50 m /s relative to the
river bank (Fig. 3-49).
What is Huck’s velocity
(speed and direction)
relative to the river
bank?
, 0.70 m/s
River CLUTCnL
FIGURE 3-49 Problem 58.
59. (II) Determine the speed of the boat with respect to the shore in Example 3-14.
60. (II) Two planes approach each other head-on. Each has a speed of 780 km /h, and they spot each other when they are initially 12.0 km apart. How much time do the pilots have to take evasive action?
45 m
FIGURE 3-50 Problem 61.
62. (II) A passenger on a boat moving at 1.70 m /s on a still lake walks up a flight of stairs at a speed of 0.60 m/s, Fig. 3-51. The stairs are angled at 45° pointing in the direction of motion as shown. Write the vector velocity of the passenger relative to the water.
FIGURE 3-51 Problem 62.
63. (II) A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with speed 10.0 m /s (Fig. 3-52). What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground (a) if the hot-air balloon is rising at 5.0 m /s rela­ tive to the ground during this throw, (b) if the hot-air balloon is descending at 5.0 m /s relative to the ground.
FIGURE 3-52 Problem 63.
Problems 79
64. (II) An airplane is heading due south at a speed of 580 km/h. If a wind begins blowing from the southwest at a speed of 90.0 km /h (average), calculate (a) the velocity (magnitude and direction) of the plane, relative to the ground, and (b) how far from its intended position it will be after 11.0 min if the pilot takes no corrective action. [Hint. First draw a diagram.]
65. (II) In what direction should the pilot aim the plane in Problem 64 so that it will fly due south?
66. (II) Two cars approach a street corner at right angles to each other (see Fig. 3-35). Car 1 travels at 35 km /h and car 2 at 45 km /h. What is the relative velocity of car 1 as seen by car 2? What is the velocity of car 2 relative to car 1?
67. (II) A swimmer is capable of swimming 0.60 m /s in still water, (a) If she aims her body directly across a 55-m-wide river whose current is 0.50 m/s, how far downstream (from a point opposite her starting point) will she land? (b) How long will it take her to reach the other side?
68. (II) (a) A t what upstream angle must the swimmer in Problem 67 aim, if she is to arrive at a point directly across the stream? (b) How long will it take her?
69. (II) A motorboat whose speed in still water is 3.40 m/s must aim
upstream at an angle of 19.5° (with respect to a line perpendic­
ular to the shore) in order to travel directly across the stream.
(a) What is the speed of the current? (b) What is the resultant
speed of the boat with respect to the shore? (See Fig. 3-31.)
70. (II) A boat, whose speed in still water is 2.70 m/s, must cross
a 280-m-wide river and arrive at a
120 m
point 120 m upstream
Hinish
from where it starts
(Fig. 3-53). To do so, the pilot must head the boat at a 45.0° upstream angle. What
280 m
^ River cunvnl
is the speed of the
river’s current?
FIGURE 3-53 Problem 70.
1 m P Stan
71. (Ill) An airplane, whose air speed is 580 km /h, is supposed to fly in a straight path 38.0° N of E. But a steady 72 km /h wind is blowing from the north. In what direction should the plane head?
| General Problems__________
72. Two vectors, Vi and V2, add to a resultant V = Vi + V2. Describe % and V2 if (a) V = V1 + V2, (b) V 2 = V? + V (c) Vi + V2 = Vi - V2.
73. A plumber steps out of his truck, walks 66 m east and 35 m south, and then takes an elevator 12 m into the subbasement of a building where a bad leak is occurring. What is the displacement of the plumber relative to his truck? Give your answer in components; also give the magnitude and angles, with respect to the x axis, in the vertical and horizontal plane. Assume x is east, y is north, and z is up.
74. On mountainous downhill roads, escape routes are sometimes placed to the side of the road for trucks whose brakes might fail. Assuming a constant upward slope of 26°, calculate the horizontal and vertical components of the acceleration of a truck that slowed from 110 km /h to rest in 7.0 s. See Fig. 3-54.
75. A light plane is headed due south with a speed relative to still air of 185 km /h. After 1.00 h, the pilot notices that they have covered only 135 km and their direction is not south but southeast (45.0°). What is the wind velocity?
76. An Olympic long jumper is capable of jumping 8.0 m. Assuming his horizontal speed is 9.1 m /s as he leaves the ground, how long is he in the air and how high does he go? Assume that he lands standing upright—that is, the same way he left the ground.
77. Romeo is chucking pebbles gently up to Juliet’s window,
and he wants the pebbles
to hit the window with
only a horizontal compo­
nent of velocity. He is standing at the edge of a
.- - - a
rose garden 8.0 m below
her window and 9.0 m 8.0 m from the base of the wall
(Fig. 3-55). How fast are
the pebbles going when
they hit her window?
FIGURE 3-55 Problem 77.
78. Raindrops make an angle 6 with the vertical when viewed through a moving train window (Fig. 3-56). If the speed of the train is vT, what is the speed of the raindrops in the reference frame * of the Earth in which they are assumed to fall vertically?
FIGURE 3-56 Problem 78.
79. Apollo astronauts took a “nine iron” to the Moon and hit a golf ball about 180 m. Assuming that the swing, launch angle, and so on, were the same as on Earth where the same astronaut could hit it only 32 m, estimate the acceleration due to gravity on the surface of the Moon. (We neglect air resistance in both cases, but on the Moon there is none.)
80 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors
80. A hunter aims directly at a target (on the same level) 68.0 m
away, (a) If the bullet leaves the gun at a speed of 175 m/s,
by how much will it miss the target? (b) A t what angle
should the gun be aimed so the target will be hit?
81. The cliff divers of Acapulco push off horizontally from rock
platforms about 35 m above the water, but
they must clear rocky outcrops at water level
that extend out into the water 5.0 m from
f
the base of the cliff directly under their
I
launch point. See Fig. 3-57. What
\
minimum pushoff speed is necessary to clear the rocks? How long are they
I t t
in the air?
I I ^
I >V1
15 . 0 mi
FIGURE 3-57
Problem 81.
82. When Babe Ruth hit a homer over the 8.0-m-high rightfield fence 98 m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.0 m above the ground and its path initially made a 36° angle with the ground.
83. The speed of a boat in still water is v. The boat is to make a round trip in a river whose current travels at speed u. Derive a formula for the time needed to make a round trip of total distance D if the boat makes the round trip by moving (a) upstream and back downstream, and (b) directly across the river and back. We must assume u < v; why?
84. A t serve, a tennis player aims to hit the ball horizontally. What minimum speed is required for the ball to clear the 0.90-m-high net about 15.0 m from the server if the ball is “launched” from a height of 2.50 m? Where will the ball land if it just clears the net (and will it be “good” in the sense that it lands within 7.0 m of the net)? How long will it be in the air? See Fig. 3-58.
T
2.50 m‘ n
15.0 m
h- 7.0 m -
FIGURE 3 -5 8 Problem 84.
85. Spymaster Chris, flying a constant 208 km /h horizontally in a low-flying helicopter, wants to drop secret documents into her contact’s open car which is traveling 156 km /h on a level highway 78.0 m below. A t what angle (with the hori­ zontal) should the car be in her sights when the packet is released (Fig. 3-59)?
208 km/h
86. A basketball leaves a player’s hands at a height of 2.10 m above the floor. The basket is 3.05 m above the floor. The player likes to shoot the ball at a 38.0° angle. If the shot is made from a horizontal distance of 11.00 m and must be accurate to + 0.22 m (horizontally), what is the range of initial speeds allowed to make the basket?
87. A particle has a velocity of v = (—2.0i + 3.5£j)m/s. The particle starts at r = (l.5i —3.1j)m at t = 0. Give the position and acceleration as a function of time. What is the shape of the resulting path?
88. A projectile is launched from ground level to the top of a cliff which is 195 m away and 135 m high (see Fig. 3-60). If the projectile lands on top of the cliff 6.6 s after it is fired, find the initial velocity of the projectile (magnitude and direction). Neglect air resistance.
Landing point
135 m
FIGURE 3-60
A
Problem 88.
195 m -
89. In hot pursuit, Agent Logan of the FBI must get directly across a 1200-m-wide river in minimum time. The river’s current is 0.80 m/s, he can row a boat at 1.60 m/s, and he can run 3.00 m/s. Describe the path he should take (rowing plus running along the shore) for the minimum crossing time, and determine the minimum time.
90. A boat can travel 2.20 m /s in still water, (a) If the boat points its prow directly across a stream whose current is 1.30 m/s, what is the velocity (magnitude and direction) of the boat relative to the shore? (b) What will be the position of the boat, relative to its point of origin, after 3.00 s?
91. A boat is traveling where there is a current of 0.20 m /s east (Fig. 3-61). To avoid some offshore rocks, the boat must clear a buoy that is NNE (22.5°) and 3.0 km away. The boat’s speed through still water is 2.1 m/s. If the boat wants to pass the buoy 0.15 km on its right, at what angle should the boat head?
a
PI
Buoy
Current
FIGURE 3-59 Problem 85.
FIGURE 3-61 Problem 91.
92. A child runs down a 12° hill and then suddenly jumps upward at a 15° angle above horizontal and lands 1.4 m down the hill as measured along the hill. What was the child’s initial speed?
General Problems 81