Solution Manual for Physics: for Scientists and Engineers with Modern Physics, 3rd Edition

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I N S T R U C T O RS O L U T I O N SM A N U A LT H I R DE D I T I O NphysicsF O RS C I E N T I S T SA N DE N G I N E E R Sa strategic approachrandall d. knightLarry SmithSnow CollegeBrett KraabelPhD-Physics, University of Santa BarbaraBostonColumbusIndianapolisNew YorkSan FranciscoUpper Saddle RiverAmsterdamCape TownDubaiLondonMadridMilanMunichParisMontréalTorontoDelhiMexicoCitySão PauloSydneyHong KongSeoulSingaporeTaipeiTokyo

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Publisher: James SmithSenior Development Editor: Alice Houston, Ph.D.Senior Project Editor: Martha SteeleAssistant Editor: Peter AlstonMedia Producer: Kelly ReedSenior Administrative Assistant: Cathy GlennDirector of Marketing: Christy LeskoExecutive Marketing Manager: Kerry McGinnisManaging Editor: Corinne BensonProduction Project Manager: Beth CollinsProduction Management, Illustration, and Composition: PreMediaGlobal, Inc.Copyright ©2013, 2008, 2004Pearson Education, Inc. All rights reserved. Manufactured in the United Statesof America. This publication is protected by Copyright, and permission should be obtained from the publisherprior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means,electronic, mechanical, photocopying, recording, or likewise. To obtain permission(s) to use material from thiswork, please submit a written request to Pearson Education, Inc., Permissions Department, 1900 E. Lake Ave.,Glenview, IL 60025. For information regarding permissions, call (847) 486-2635.Many of the designations used by manufacturers and sellers to distinguish their products are claimed astrademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim,the designations have been printed in initial caps or all caps.MasteringPhysicsis a trademark, in the U.S. and/or other countries, of Pearson Education, Inc. or its affiliates.ISBN 13:978-0-321-76940-4ISBN 10:0-321-76940-6www.pearsonhighered.com

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ContentsPreface.......................................................................................................................................vPART INewton’s LawsChapter 1Concepts of Motion..............................................................................................1-1Chapter 2Kinematics in One Dimension.............................................................................2-1Chapter 3Vectors and Coordinate Systems.........................................................................3-1Chapter 4Kinematics in Two Dimensions...........................................................................4-1Chapter 5Force and Motion.................................................................................................5-1Chapter 6Dynamics I: Motion Along a Line.......................................................................6-1Chapter 7Newton’s Third Law............................................................................................7-1Chapter 8Dynamics II: Motion in a Plane...........................................................................8-1PART IIConservation LawsChapter 9Impulse and Momentum......................................................................................9-1Chapter 10Energy................................................................................................................10-1Chapter 11Work..................................................................................................................11-1PART IIIApplications of Newtonian MechanicsChapter 12Rotation of a Rigid Body...................................................................................12-1Chapter 13Newton’s Theory of Gravity..............................................................................13-1Chapter 14Oscillations........................................................................................................14-1Chapter 15Fluids and Elasticity...........................................................................................15-1PART IVThermodynamicsChapter 16A Macroscopic Description of Matter...............................................................16-1Chapter 17Work, Heat, and the First Law of Thermodynamics..........................................17-1Chapter 18The Micro/Macro Connection............................................................................18-1Chapter 19Heat Engines and Refrigerators.........................................................................19-1

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ivC O N T E N T SChapter 20Traveling Waves................................................................................................20-1Chapter 21Superposition.....................................................................................................21-1Chapter 22Wave Optics.......................................................................................................22-1Chapter 23Ray Optics..........................................................................................................23-1Chapter 24Optical Instruments............................................................................................24-1PART VIElectricity and MagnetismChapter 25Electric Charges and Forces...............................................................................25-1Chapter 26The Electric Field...............................................................................................26-1Chapter 27Gauss’s Law.......................................................................................................27-1Chapter 28The Electric Potential.........................................................................................28-1Chapter29Potential and Field.............................................................................................29-1Chapter 30Current and Resistance......................................................................................30-1Chapter 31Fundamentals of Circuits...................................................................................31-1Chapter 32The Magnetic Field............................................................................................32-1Chapter 33Electromagnetic Induction.................................................................................33-1Chapter 34Electromagnetic Fields and Waves....................................................................34-1Chapter 35AC Circuits........................................................................................................35-1PART VIIRelativity and Quantum PhysicsChapter 36Relativity............................................................................................................36-1Chapter 37The Foundations of Modern Physics.................................................................37-1Chapter 38Quantization.......................................................................................................38-1Chapter 39Wave Functions and Uncertainty.......................................................................39-1Chapter 40One-Dimensional Quantum Mechanics.............................................................40-1Chapter 41Atomic Physics..................................................................................................41-1Chapter 42Nuclear Physics..................................................................................................42-1PART VWaves and Optics

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PrefaceThisInstructor Solutions Manualhas a twofold purpose. First, and most obvious, is to pro-vide worked solutions for the use of instructors. Second, but equally important, is to provideexamples of good problem-solving techniques and strategies that will benefit your students ifyou post these solutions.Far too many solutions manuals simply plug numbers into equations, thereby reinforcing oneof the worst student habits. The solutions provided here, by contrast, attempt to:•Follow, in detail, the problem-solving strategies presented in the text.•Articulate the reasoning that must be done before computation.•Illustrate how to use drawings effectively.•Demonstrate how to utilize graphs, ratios, units, and the many other“tactics”that must besuccessfully mastered and marshaled if a problem-solving strategy is to be effective.•Show examples of assessing the reasonableness of a solution.•Comment on the significance of a solution or on its relationship to other problems.Most education researchers believe that it is more beneficial for students to study a smallernumber of carefully chosen problems in detail, including variations, than to race through alarger number of poorly understood calculations. The solutions presented here are intendedto provide a basis for this practice.So that you may readily edit and/or post these solutions,they are available for downloadaseditable Worddocumentsand as pdffiles via the “Resources” tab in the textbook’s Instruc-tor Resource Center (www.pearsonhighered.com/educator/catalog/index.page) or from thetextbook’s Instructor Resource Area in MasteringPhysics®(www.masteringphysics.com).We have made every effort to be accurate and correct in these solutions. However, if you dofind errors or ambiguities, we would be very grateful to hear from you.Please contact yourPearson Education sales representative.

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viP R E F A C EAcknowledgments for the First EditionWe are grateful for many helpful comments from Susan Cable, Randall Knight, and SteveStonebraker. We express appreciation to Susan Emerson, who typed the word-processingmanuscript, for her diligence in interpreting our handwritten copy. Finally, we would like toacknowledge the support from the Addison Wesley staff in getting the work into a publisha-ble state. Our special thanks to Liana Allday, Alice Houston, and Sue Kimber for their will-ingness and preparedness in providing needed help at all times.Pawan KaholMissouri State UniversityDonald FosterWichita State UniversityAcknowledgments fortheSecond EditionI would like to acknowledge the patient support of my wife, Holly, who knows what is im-portant.Larry SmithSnow CollegeI would like to acknowledge the assistance and support of my wife, Alice Nutter, who helpedtype many problems and was patient while I worked weekends.Scott NutterNorthern Kentucky UniversityAcknowledgments for the Third EditionTo Holly, Ryan, Timothy, Nathan, Tessa, and Tyler, who make it all worthwhile.Larry SmithSnow CollegeI gratefully acknowledge the assistance of the staff at Physical Sciences Communication.Brett KraabelPhD-University of Santa Barbara

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ConceptualQuestions1.1.(a)3 significant figures.(b)2significant figures.This is more clearly revealed byusingscientific notation:2 sig. figs.10.535.310−=(c)4significant figures. Thetrailingzerois significantbecause it indicates increased precision.(d)3significant figures.The leading zeros are not significantbut just locate the decimal point.1.2.(a)2significant figures. Trailing zerosin front of the decimal point merely locate the decimal point and arenot significant.(b)3significant figures.Trailing zerosafter the decimal pointare significantbecause they indicate increasedprecision.(c)4 significant figures.(d)3 significant figures.Trailing zeros after the decimal point are significant because they indicate increasedprecision.1.3.Without numbers on the dots we cannot tell if the particle in the figure is moving left or right, so we can’t tellif it is speeding up or slowing down. If the particle is moving to the right it is slowing down. If it is moving to the leftit is speeding up.1.4.Because the velocity vectors get longer foreachtime step, the objectmust bespeeding up as ittravelsto theleft. The accelerationvectormust thereforepointin the same direction as the velocity, so the acceleration vector alsopoints to the left. Thus,xais negative as per our convention (see Tactics Box 1.4).1.5.Because the velocity vectors get shorter for each time step, the objectmust beslowing down as it travels inthey2direction(down). The acceleration vector must therefore point in the direction opposite to the velocity;namely, in the+ydirection (up). Thus,yais positive as per our convention (see Tactics Box 1.4).1.6.The particle position is to the left of zero on thex-axis, so its position is negative. The particle is moving to theright, so its velocity is positive. The particle’s speed is increasing as it moves to the right, so its acceleration vectorpoints in the same direction as its velocity vector (i.e., to the right). Thus, the acceleration is also positive.1.7.The particle position is below zero on they-axis, so its position is negative. The particle is moving down, soits velocity is negative. The particle’s speed is increasing as it moves in the negative direction, so its accelerationvector points in the same direction as its velocity vector (i.e., down). Thus, the acceleration is also negative.CONCEPTS OFMOTION1

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1-2Chapter 11.8.The particle position is above zero on they-axis, so its position ispositive. The particle is moving down, so itsvelocity is negative.The particle’s speed is increasing as it moves in the negative direction, so its acceleration vectorpoints in the same direction as its velocity vector (i.e., down). Thus, the acceleration is also negative.Exercisesand ProblemsSection 1.1Motion Diagrams1.1.Model:Imagine a car moving in the positive direction (i.e., to the right). As it skids, it covers less distancebetween each movie frame (or between each snapshot).Solve:Assess:As we go from left to right, the distance between successive images of the cardecreases. Because the timeinterval between eachsuccessiveimage is the same, the car must be slowing down.1.2.Model:We have no information about the acceleration of the rocket, so we will assume that it acceleratesupwardwith a constant acceleration.Solve:Assess:Notice that the length of the velocity vectors increases each step byapproximatelythe length of theacceleration vector.

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Concepts of Motion1-31.3.Model:We will assume that the term “quickly” used in the problem statement means a time that is shortcompared to 30 s.Solve:Assess:Notice that the acceleration vector points in the direction oppositeto the velocity vector because the car isdecelerating.Section 1.2The Particle Model1.4.Solve:(a)The basic idea of the particle model is that we will treat an objectas ifall its mass is concentratedinto a single point. The size and shape of the object will not be considered. This is a reasonable approximation ofreality if (i) the distance traveled by the object is large in comparison to the size of the object and (ii) rotations andinternal motions are not significant features of the object’s motion. The particle model is important in that it allowsus tosimplifya problem. Complete reality—which would have to include the motion of every single atom in theobject—is too complicated to analyze. By treating an object as a particle, we can focus on the most important aspectsof its motion while neglecting minor and unobservable details.(b)The particle model is valid for understanding the motion of a satellite or a car traveling a large distance.(c)The particle model is not valid for understanding how a car engine operates, how a person walks, how a bird flies,or how water flows through a pipe.Section 1.3Position and TimeSection 1.4Velocity1.5.Model:We model the ball’s motion from the instant after it is released, when it has zero velocity, to theinstant before it hits the ground, when it will have its maximum velocity.Solve:

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1-4Chapter 1Assess:Notice that the “particle”we have drawn has a finite dimensions,soitappears as if the bottom half of this“particle”haspenetrated intothe ground in the bottom frame. This is not really the case;ourmentalparticlehas nosize andislocatedatthetip of the velocity vector arrow.1.6.Solve:The player starts from rest and movesfaster and faster.1.7.Solve:Theplayerstarts with an initial velocity but asheslideshemoves slower and sloweruntilcoming torest.Section 1.5Linear Acceleration1.8.Solve:(a)Let0vbe the velocity vector between points 0 and 1 and1vbe the velocity vector between points1 and 2. Speed1vis greater than speed0vbecause more distance is covered in the same interval of time.(b)To find the acceleration, use the method of Tactics Box 1.3:Assess:The acceleration vector points in the same direction as the velocity vectors, which makes sense because thespeed is increasing.1.9.Solve:(a)Let0vbe the velocity vector between points 0 and 1 and1vbe the velocity vector between points 1and 2. Speed1vis greater than speed0vbecause more distance is covered in the same interval of time.(b)Acceleration is found by the method of Tactics Box 1.3.Assess:The acceleration vector points in the same direction as the velocity vectors, which makes sense because thespeed is increasing.

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Concepts of Motion1-51.10.Solve:(a)(b)1.11.Solve:(a)(b)1.12.Model:Represent the car as a particle.Visualize:The dots are equally spaced until brakes are applied to the car. Equidistant dotson a single lineindicateconstant averagevelocity. Upon braking, the dots get closeras the averagevelocitydecreases,andthe distancebetweendotschanges by a constant amount because theaccelerationis constant.1.13.Model:Represent the (child+ sled) system as a particle.Visualize:The dots in the figure are equally spaced until the sled encounters a rocky patch. Equidistant dotson asingle lineindicate constant averagevelocity. On encountering a rocky patch, the averagevelocitydecreases and thesled comes to a stop. This part of the motion is indicated by adecreasingseparation between the dots.1.14.Model:Represent the wad of paper as a particle. Ignore air resistance.

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1-6Chapter 1Visualize:The dots become more closely spaced because the particle experiences a downwardacceleration. Thedistance between dots changes by a constant amount because theaccelerationis constant.1.15.Model:Represent the tile as a particle.Visualize:Starting from rest, the tile’s velocity increases until it hits the water surface. This part ofthe motion isrepresented by dots with increasing separation, indicating increasing average velocity. After the tile enters the water,it settles to the bottom at roughly constant speed, so this part of the motion is represented by equally spaced dots.1.16.Model:Represent the tennis ball as a particle.Visualize:The ball falls freely for three stories. Upon impact, it quickly decelerates to zero velocity while comp-ressing, then accelerates rapidly while re-expanding. As vectors, both the deceleration and acceleration are an upwardvector.The downward and upward motions of the ball are shownseparatelyin the figure. The increasing length betweenthe dots during downward motion indicates an increasing average velocity or downward acceleration. On the other hand,the decreasing length between the dots during upward motion indicates acceleration in a direction opposite to themotion,so the average velocity decreases.Assess:Forfree-fall motion, acceleration due to gravity is always vertically downward.Notice that the accelerationdue to the ground is quite large (although not to scale—that would take too much space) because in a time intervalmuch shorter than the timeintervalbetweenthepoints, thevelocity of the ball isessentiallycompletelyreversed.

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Concepts of Motion1-71.17.Model:Represent the toy car as a particle.Visualize:As the toy car rolls down the ramp, its speed increases. This is indicated by the increasing length of thevelocity arrows. That is, motion down the ramp is under a constant accelerationaAt the bottom of the ramp, the toycar continues with aconstant velocityand no acceleration.Section 1.6Motion in One Dimension1.18.Solve:(a)DotTime (s)x(m)(b)1002230349546215584006105107126008146709167201.19.Solve:A forgetful physics professoriswalking from one class to the next. Walking at a constant speed, hecovers a distance of 100 m in 200 s. He then stops and chats with a student for 200 s.Suddenly, herealizes he isgoing to be late for his next class, so the hurries on and covers the remaining 200 m in 200 s to get to class on time.1.20.Solve:Forty miles into a car trip north from his home in El Dorado, an absent-minded English professorstopped at a rest area one Saturday. After staying there for one hour, he headed back home thinking that he wassupposed to go on this trip on Sunday. Absent-mindedly he missed his exit and stopped after one hour of driving atanother rest area 20 miles south of El Dorado. After waiting there for one hour, he drove back very slowly, confusedand tired as he was, and reached El Dorado two hourslater.Section 1.7Solving Problems in Physics1.21.Visualize:The bicycle move forward with an acceleration of21.5 m/s .Thus, the velocity will increase by1.5 m/s each second of motion.

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1-8Chapter 11.22.Visualize:The rocket moves upward with a constant acceleration.aThe final velocity is 200 m/s and isreached at a height of 1.0 km.Section 1.8Units and Significant Figures1.23.Solve:(a)331s6.15 ms(6 15 ms)6.1510s10ms−==(b)3310m27.2 km(27 2 km)27.210m1 km==(c)3km10m1 hour112 km/hour11231.1 m/shour1 km3600 s==(d)326m1 m10ms72m/ms727.210m/sms1 s10m−==1.24.Solve:(a)22.54 cm10m8.0 inch(8 0 inch)0.20 m1 inch1 cm−==(b)feet12 inch1 m66 feet/s6620 m/ss1 foot39.37 inch==(c)3miles1.609 km10m1 hour60 mph6027 m/shour1 mile1 km3600 s==(d)223231 m14 square inches(14 inch )9.010m9.010square meters39.37 inches−−===

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Concepts of Motion1-91.25.Solve:(a)43600 s3 hour(3 hour)10,800 s1 10s1 hour== (b)5524 hours3600 s2 day(2 day)1.7310s210s1 day1 hour==(c)477365.25 days8.6410s1 year(1 year)3 1610s310s1 year1 day==(d)ft12 inch1 m215 ft/s21565.5 m/ss1 ft39.37 inch==Assess:The results are given to appropriate number of significant figures.1.26.Solve:(a)1 m20 ft(20 ft)7 m3 ft(b)1 km1000 m60 miles(60 miles)100,000 m0.6 miles1 km(c)1 m/s60 mph(60 mph)30 m/s2 mph(d)21 cm10m8 in(8 in)0.2 m1/2 in1 cm−1.27.Solve:(a)4 in30 cm(30 cm)12 in10 cm(b)2 mph25 m/s(25 m/s)50 mph1 m/s(c)0.6 mi5 km(5 km)3 mi1 km(d)1/2 in0.5 cm(0 5 cm)0.3 in1 cm1.28.Solve:(a)33.325.4846=(b)33.325.47.9−=(c)33.35.77=(d)333.325.413.1=1.29.Solve:(a)33(12.5)1.9510 .=(b)12.55.2165.1=(c)12.51.23.541.22.3−=−=(d)1212.51.00/12.50.08008.0010−−===1.30.Solve:The length of a typical car is 15 ft or12 inch1 m(15 ft)4.6 m1 ft39.37 inch =This length of 15 ft is approximately two-and-a-half times my height.

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1-10Chapter 11.31.Solve:The height of a telephone pole isestimated to be around 50 ft or(using 1 m ~ 3 ft) about15 m. Thisheight is approximately8 times my height.1.32.Solve:I typically take 15 minutes in my car to cover a distance of approximately 6 miles from home tocampus. My average speed is6 miles60 min0.447 m/s24 mph(24 mph)11 m/s15 min1 hour1 mph ===1.33.Solve:My barber trims about an inch of hair when I visit him every month for a haircut. The rate of hairgrowth is29691inch2.54 cm10m1 month1 day1 h9.810m/s1 month1 inch1 cm30 days24 h3600 sm10m3600 s9.81040m/hs1 m1 h−−− ==1.34.Model:Represent the Porsche as a particle for the motion diagram. Assume the car moves at a constantspeed when it coasts.Visualize:1.35.Model:Represent thejetas a particle for the motion diagram.Visualize:

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Concepts of Motion1-111.36.Model:Represent (Sam+car) as a particle for the motion diagram.Visualize:1.37.Model:Represent the wad as a particle for the motion diagram.Visualize:

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1-12Chapter 11.38.Model:Represent the speed skater as a particle for the motion diagram.Visualize:1.39.Model:RepresentSanta Clausas a particle for the motion diagram.Visualize:

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Concepts of Motion1-131.40.Model:Represent the motorist as a particle for the motion diagram.Visualize:1.41.Model:Represent the car as a particle for the motion diagram.Visualize:1.42.Model:Represent Bruce and the puck as particles for the motion diagram.Visualize:

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1-14Chapter 11.43.Model:Representthe cars of David and Tinaand as particles for the motion diagram.Visualize:1.44.Solve:Isabelis the first car in line at a stop light. When it turns green, she accelerates, hoping to make thenext stop light 100 m away before it turns red. When she’s about30m away, the light turns yellow, so she starts tobrake, knowing that she cannot make the light.1.45.Solve:A carcoasts alongat 30m/sandarrives at ahill.The car decelerates as it coasts up the hill.At thetop, the road levels and the car continues coasting along the road at a reduced speed.1.46.Solve:A skier starts from rest down a 25°slope with very little friction. At the bottom of the 100-m slopethe terrain becomes flat and the skier continues at constant velocity.1.47.Solve:A ball is dropped from a height to check its rebound properties. It rebounds to 80% of its originalheight.1.48.Solve:Two boards lean against each other at equal angles to the vertical direction. A ball rolls up theincline, over the peak, and down the other side.1.49.Solve:(a)(b)A train moving at 100 km/hour slows down in 10 s to a speed of 60 km/hour as it enters a tunnel. The drivermaintains this constant speed for the entire length of the tunnel that takes the train a time of 20 s to traverse. Find thelength of the tunnel.

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Concepts of Motion1-15(c)1.50.Solve:(a)(b)Sue passes 3rdStreetdoing 30 km/h, slows steadily to the stop sign at 4thStreet, stops for 1.0s, then speeds upand reaches her original speed as she passes 5thStreet. If the blocks are 50 m long, how long does it take Sue to drivefrom 3rdStreet to 5thStreet?(c)1.51.Solve:(a)(b)Jeremy has perfected the art of steady acceleration and deceleration. From a speed of 60 mph he brakes his car torest in 10swith a constant deceleration. Then he turns into an adjoining street. Starting from rest, Jeremy accelerateswith exactly the same magnitude as his earlier deceleration and reaches the same speed of 60 mph over the samedistance in exactly the same time. Find the car’s acceleration or deceleration.

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1-16Chapter 1(c)1.52.Solve:(a)(b)A coyote (A) sees a rabbit and begins to run toward it with an acceleration of 3.02m/s .At the same instant, therabbit (B) begins to run away from the coyote with an acceleration of 2.02m/s .The coyote catches the rabbit afterrunning 40 m. How far away was the rabbit when the coyote first saw it?(c)1.53.Solve:Since area equals length3width, the smallest area will correspond to the smaller length and thesmaller width. Similarly, the largest area will correspond to the larger length and the larger width. Therefore, thesmallest area is (64 m)(100 m)=6.43210m3and the largest area is (75 m)(110 m)=8.333210m .1.54.Solve:(a)We need3kg/m .There are 100 cm in 1 m. If we multiply by33100 cm(1)1 m=we do not change the size of the quantity, but only the number in terms of the new unit. Thus, the mass density ofaluminum is33333kg100 cmkg2.7102.7101 mcmm−=

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Concepts of Motion1-17(b)Likewise, the mass density of alcohol is333g100 cm1 kgkg0.818101 m1000 gcmm= 1.55.Model:In the particle model, the car is represented as a dot.Solve:(a)Timet(s)Positionx(m)(b)0120010975208253075040700506506060070500803009001.56.Solve:Susan enters a classroom, sees a seat 40 m directly ahead, and begins walking toward it at a constantleisurely pace, covering the first 10 m in 10 seconds. But then Susan notices that Ella is heading toward the sameseat, so Susan walks more quickly to cover the remaining 30 m in another 10 seconds, beating Ella to the seat. Susanstands next to the seat for 10 seconds to remove her backpack.1.57.Solve:A crane operatorstarts lifting a ton of bricks off the ground. In8s, he has lifted them to a height of15 m, thenhetakes4sto makeasafety check. He then continuesraisingthe bricks the remaining 15 m,which takes4 s.

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ConceptualQuestions2.1.It was a typical summer day on the interstate. I started 10 mi east of town and drove for 20 min at 30 mph westto town. My stop for gas took 10 min. Then I headed back east at 60 mph before I encountered a construction zone.Traffic was at a standstill for 10 min and then I was able to move forward (east) at 30 mph until I got to mydestination 30 mi east of town.2.2.With a slow start out of the blocks, a super sprinter reached top speed in about 5 s, having gone only 30 m. Hewas still able to finish his 100 m in only just over 9 s by running a world record pace for the rest of the race.2.3.The baseball team is warming up. The pitcher (who is 50 feet from home plate) lobs the ball at 100 ft/s to thesecond baseman who is 100 ft from home plate. The second baseman then fires the ball at 200 ft/s to the catcher athome plate.2.4.(a)At1 s,t=the slope of the line for object A is greater than that for object B. Therefore, object A’s speed isgreater. (Both are positive slopes.)(b)No, the speeds areneverthe same. Each has a constant speed (constant slope) and A’s speed is always greater.2.5.(a)A’s speed is greater at1 s.t=The slope of the tangent to B’s curve at1 st=is smaller than the slope ofA’s line.(b)A and B have the same speed at just about3 s.t=At that time, the slope of the tangent to the curve representingB’s motion is equal to the slope of the line representing A.2.6.(a)B. The object is still moving, but the magnitude of the slope of the position-versus-time curve is smaller than at D.(b)D. The slope is greatest at D.(c)At points A, C, and E the slope of the curve is zero, so the object is not moving.(d)At point D the slope is negative, so the object is moving to the left.2.7.(a)The slope of the position-versus-time graph is greatest at D, so the object is moving fastest at this point.(b)The slope is negative at points C, D, and E, meaning the object is moving to the left at these points.(c)At point C the slope is increasing in magnitude (getting more negative), meaning that the object is speeding up tothe left.(d)At point B the object is not moving since the slope is zero. Before point B, the slope is positive, while after B it isnegative, so the object is turning around at B.2.8.(a)The positions of the third dots of both motion diagrams are the same, as are the sixth dots of both, so cars Aand B are at the same locations at the time corresponding to dot 3 and again at that of dot 6.(b)The spacing of dots 4 and 5 in both diagrams is the same, so the cars are traveling at the same speeds betweentimes corresponding to dots 4 and 5.KINEMATICS INONEDIMENSION2

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2-2Chapter22.9.No, though you have the same position along the road, his velocity is greater because he is passing you. If hisvelocity were not greater, then he would remain even with the front of your car.2.10.Yes. The acceleration vector will point west when the bicycle is slowing down while traveling east.2.11.(a)As a ball tossed upward moves upward, its vertical velocity is positive, while its vertical acceleration isnegative, opposite the velocity, causing the ball to slow down.(b)The same ball on its way down has downward (negative) velocity. The downward negative acceleration ispointing in the same direction as the velocity, causing the speed to increase.2.12.For all three of these situations the acceleration is equal togin the downward direction. The magnitude anddirection of the velocity of the ball do not matter. Gravity pulls down at constant acceleration. (Air friction isignored.)2.13.(a)The magnitude of the acceleration while in free fall is equal togat all times, independent of the initialvelocity. The acceleration only tells how the velocity is changing.(b)The magnitude of the acceleration is stillgbecause the rock is still in free fall. The speed is increasing at the samerate each instant, that is, by the sameveach second.2.14.The ball remains in contact with the floor for a small but noticeable amount of time. It is in free fall when notin contact with the floor. When it hits the floor, it is accelerated very rapidly in the upward direction as it bounces.

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Kinematics in One Dimension2-3Exercisesand ProblemsSection 2.1Uniform Motion2.1.Model:Cars will be treated by the particle model.Visualize:Solve:Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows:10101010xxxxxvtttttv−−===+−Using the known values identified in the pictorial representation, we find:Alan 1Alan 0Alan 1Alan 0Beth 1Beth 0Beth 1Beth 0400 mile8:00 AM8:00 AM8 hr4:00 PM50 miles/hour400 mile9:00 AM9:00 AM6.67 hr3:40 PM60 miles/hourxxttvxxttv−=+=+=+=−=+=+=+=(a)Beth arrives first.(b)Beth has to waitAlan 1Beth 120 minutestt−=for Alan.Assess:Times of the order of 7 or 8 hours are reasonable in the present problem.2.2.Model:We will consider Larry to be a particle.Visualize:

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2-4Chapter2Solve:Since Larry’s speed is constant, we can use the following equation to calculate the velocities:fisfissvtt−=−(a)For the interval from the house to the lamppost:1200 m600 m200 m/min9 : 079 : 05v−== −−Forthe interval from the lamppost to the tree:21200 m200 m333 m/min9 :109 : 07v−== +−(b)For the average velocity for the entire run:avg1200 m600 m120 m/min9 :109 : 05v−== +−2.3.Solve:(a)The time for each segment is150 mi/40 mph5/4 hrt==and250 mi/60 mph5/6hr.t==Theaverage speed to the house is100 mi48 mph5/6 h5/4 h=+(b)Julie drives the distance1xin time1tat 40 mph. She then drives the distance2xin time2tat 60 mph. Shespends the same amount of time at each speed, thus121212/40 mph/60 mph(2/3)ttxxxx=  =  =But12100 miles,xx+ =so22(2/3)100 miles.xx+ =This means260 milesx=and140 miles.x=Thus,the times spent at each speed are140 mi/40 mph1.00 ht==and260 mi/60 mph1.00 h.t==The total time forher return trip is122.00 h.tt+ =So, her average speed is100 mi/2 h50 mph.=2.4.Model:The jogger is a particle.Solve:The slope of the position-versus-time graph at every point gives the velocity at that point. The slope at10 st=is50 m25 m1.25 m/s20 ssvt−===The slope at25 st=is50 m50 m0.0 m/s10 sv−==The slope at35 st=is0 m50 m5.0 m/s10 sv−== −Section 2.2Instantaneous VelocitySection 2.3Finding Position from Velocity2.5.Solve:(a)We can obtain the values for the velocity-versus-time graph from the equation/.vst=  

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Kinematics in One Dimension2-5(b)There is only one turning point. At1 st=the velocity changes from20 m/s+to5 m/s,−thus reversing the directionof motion. At3 s,t=there is an abrupt change in motion from5 m/s−to rest, but there is no reversal in motion.2.6.Visualize:Please refer to Figure EX2.6. The particle starts at010 mx=at00.t=Its velocity is initially inthe–xdirection. The speed decreases as time increases during the first second, is zero at1 s,t=and then increasesafter the particle reverses direction.Solve:(a)The particle reverses direction at1 s,t=whenxvchanges sign.(b)Using the equationf0xx=+area of the velocity graph between1tandf,t2s3s10 m(area of triangle between 0 s and 1 s)(area of triangle between 1 s and 2 s)1110 m(4 m/s)(1 s)(4 m/s)(1 s)10 m2210 marea of trapazoid between 2 s and 3 s110 m(4 m/s8 m/s)(3 s22xx=−+=−+==+=++−4s3ss)16 marea between 3 s and 4 s116 m(8 m/s12 m/s)(1 s)26 m2xx==+=++=2.7.Model:The graph shows the assumption that the blood isn’t moving for the first 0.1 s nor at the end of thebeat.Visualize:The graph is a graph of velocity vs. time, so the displacement is the area under the graph—that is, thearea of the triangle. The velocity of the blood increases quickly and decreases a bit more slowly.Solve:Call the distance traveled.yThe area of a triangle is12.BH11 (0.20 s)(0.80 m/s)8.0 cm22yBH===Assess:This distance seems reasonable for one beat.2.8.Solve:(a)We can calculate the position of the particle at every instant with the equationfixx=+area under the velocity-versus-time graph betweenitandftThe particle starts from the origin at0 s,t=so0 m.ix=Notice that the each square of the grid in Figure EX2.8 has“area”(5 m/s)(2 s)10 m.=We can find the area under the curve, and thus determinex, by counting squares. Youcan see thatx=35 m at4 st=because there are 3.5 squares under the curve. In addition,35 mx=att=8 sbecause the 5 m represented by the half square between 4 and 6 s is cancelled by the–5 m represented by the halfsquare between 6 and 8s. Areas beneath the axis are negative areas. The particle passes through35 mx=at4 st=and again at8 s.t=(b)The particle moves to the right for0 s6 s,twhere the velocity is positive. It reaches a turning point at40 mx=at6 st=The motion is to the left for6 stThis is shown in the motion diagram below.

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2-6Chapter2Section 2.4Motion with Constant Acceleration2.9.Visualize:The object has a constant velocity for 2 s and then speeds up between2t=and4.t=Solve:A constant velocity from0 st=to2 st=means zero acceleration. On the other hand, a linear increase invelocity between2 st=and4 st=implies a constant positive acceleration which is the slope of the velocity line.2.10.Visualize:The graph is a graph of velocity vs. time, so the acceleration is the slope of the graph.Solve:When the blood is speeding up the acceleration is2m/sm/s0.05syyat=== When the blood is slowing down the acceleration is20.80 m/s5.3 m/s0.15 syyat−=== −Assess:216 m/sis an impressive but reasonable acceleration.2.11.Solve:(a)At2 0 s,t=the position of the particle is2 s2.0 marea under velocity graph from0 s to2.0 s12.0 m(4.0 m/s)(2.0 s)6.0 m2xtt=+===+=(b)From the graph itself at2 0 s,t=4 m/sv=(c)The acceleration is2fi6 m/s0 m/s2 m/s3 sxxxxvvvatt−−====2.12.Solve:(a) Using the equationfixx=+area under the velocity-versus-time graph betweenitandftwe have(at1 s)(at0 s)area between0 s and1 s2.0 m(4 m/s)(1 s)6 mxtxttt===+===+=Reading from the velocity-versus-time graph,(at1 s)4 m/s.xvt==Also,2slope/0 m/s .xavt==  =(b)(at3.0 s)(at0 s)xtxt===+area between0 st=and3 st=2.0 m4 m/s2 s2 m/s1 s(1/2)2 m/s1 s13.0 m=+++=Reading from the graph,(3 s)2 m/s.xvt==The acceleration is2(at4 s)(at2 s)(3 s)slope2 m/s2 sxxxvtvtat=−===== −

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