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) 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) 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 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, ( 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 = 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 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