Giancoli physics principles with applications 7th c2014 solutions ISM

If the velocity of an object is constant, then the speed and the direction of travel must also be constant.. The magnitude of the change in velocity in each case is the same, 10 km/h, so

INSTRUCTOR SOLUTIONS MANUAL

VOLUME 1 DOUGLAS C GIANCOLI’S

PHYSICS PRINCIPLES WITH APPLICATIONS

7 TH EDITION

BOB DAVIS

TAYLOR UNIVERSITY

J ERIK HENDRICKSON

UNIVERSITY OF WISCONSIN – EAU CLAIRE

San Francisco Boston New York

VOLUMES 1 & 2

President, Science, Business and Technology: Paul Corey

Publisher: Jim Smith

Executive Development Editor: Karen Karlin

Project Manager: Elisa Mandelbaum

Marketing Manager: Will Moore

Senior Managing Editor: Corinne Benson

Managing Development Editor: Cathy Murphy

Production Service: PreMedia Global, Inc

ISBN 10: 0-321-74768-2

ISBN 13: 978-0-321-74768-6

Copyright © 2014, 2009, 1998 Pearson Education, Inc 1301 Sansome St., San Francisco, CA 94111 All rights reserved Manufactured in the United States of America This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise To obtain permission(s) to use material from this work, 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 as trademarks 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

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CONTENTS

PREFACE - iv

Chapter 1 Introduction, Measurement, Estimating - 1-1

Chapter 2 Describing Motion: Kinematics in One Dimension - 2-1

Chapter 3 Kinematics in Two Dimensions; Vectors - 3-1

Chapter 4 Dynamics: Newton’s Laws of Motion - 4-1

Chapter 5 Circular Motion; Gravitation - 5-1

Chapter 6 Work and Energy - 6-1

Chapter 7 Linear Momentum - 7-1

Chapter 8 Rotational Motion - 8-1

Chapter 9 Static Equilibrium; Elasticity and Fracture - 9-1

Chapter 10 Fluids -10-1 Chapter 11 Oscillations and Waves -11-1

Chapter 12 Sound -12-1 Chapter 13 Temperature and Kinetic Theory -13-1

Chapter 14 Heat -14-1 Chapter 15 The Laws of Thermodynamics -15-1

Chapter 16 Electric Charge and Electric Field -16-1

Chapter 17 Electric Potential -17-1

Chapter 18 Electric Currents -18-1

Chapter 19 DC Circuits -19-1

Chapter 20 Magnetism -20-1 Chapter 21 Electromagnetic Induction and Faraday’s Law -21-1

Chapter 22 Electromagnetic Waves -22-1

Chapter 23 Light: Geometric Optics -23-1

Chapter 24 The Wave Nature of Light -24-1

Chapter 25 Optical Instruments -25-1

Chapter 26 The Special Theory of Relativity -26-1

Chapter 27 Early Quantum Theory and Models of the Atom -27-1

Chapter 28 Quantum Mechanics of Atoms -28-1

Chapter 29 Molecules and Solids -29-1

Chapter 30 Nuclear Physics and Radioactivity -30-1

Chapter 31 Nuclear Energy; Effects and Uses of Radiation -31-1

Chapter 32 Elementary Particles -32-1

Chapter 33 Astrophysics and Cosmology -33-1

PREFACE

This Instructor’s Solutions Manual provides answers and worked-out solutions to all end of chapter

questions and problems from chapters 1 – 15 of Physics: Principles with Applications, 7th Edition, by

Douglas C Giancoli At the end of the manual are grids that correlate the 6th edition questions and problems

to the 7th edition questions and problems

We formulated the solutions so that they are, in most cases, useful both for the student and the instructor Accordingly, some solutions may seem to have more algebra than necessary for the instructor Other solutions may seem to take bigger steps than a student would normally take: e.g simply quoting the solutions from a quadratic equation instead of explicitly solving for them There has been an emphasis on algebraic solutions, with the substitution of values given as a very last step in most cases We feel that this helps to keep the physics of the problem foremost in the solution, rather than the numeric evaluation

Much effort has been put into having clear problem statements, reasonable values, pedagogically sound solutions, and accurate answers/solutions for all of the questions and problems Working with us was a team

of five additional solvers – Karim Diff (Santa Fe College), Thomas Hemmick (Stony Brook University), Lauren Novatne (Reedley College), Michael Ottinger (Missouri Western State University), and Trina

VanAusdal (Salt Lake Community College) Between the seven solvers we had four complete solutions for every question and problem From those solutions we uncovered questions about the wording of the problems, style of the possible solutions, reasonableness of the values and framework of the questions and problems, and then consulted with one another and Doug Giancoli until we reached what we feel is both a good statement and a good solution for each question and problem in the text

Many people have been involved in the production of this manual We especially thank Doug Giancoli for his helpful conversations Karen Karlin at Prentice Hall has been helpful, encouraging, and patient as we have turned our thoughts into a manual Michael Ottinger provided solutions for every chapter, and helped in the preparation of the final solutions for some of the questions and problems And the solutions from

Karim Diff, Thomas Hemmick, Lauren Novatne, and Trina VanAusdal were often thought-provoking and always appreciated

Even with all the assistance we have had, the final responsibility for the content of this manual is ours We would appreciate being notified via e-mail of any errors that are discovered We hope that you will find this presentation of answers and solutions useful

Bob Davis (rbdavis@taylor.edu)

Upland, IN

J Erik Hendrickson (hendrije@uwec.edu)

Eau Claire, WI

Responses to Questions

1 (a) A particular person’s foot Merits: reproducible Drawbacks: not accessible to the general public;

not invariable (size changes with age, time of day, etc.); not indestructible

(b) Any person’s foot Merits: accessible Drawbacks: not reproducible (different people have

different size feet); not invariable (size changes with age, time of day, etc.); not indestructible Neither of these options would make a good standard

2 The distance in miles is given to one significant figure, and the distance in kilometers is given to five significant figures! The value in kilometers indicates more precision than really exists or than is meaningful The last digit represents a distance on the same order of magnitude as a car’s length! The sign should perhaps read “7.0 mi (11 km),” where each value has the same number of

significant figures, or “7 mi (11 km),” where each value has about the same % uncertainty

3 The number of digits you present in your answer should represent the precision with which you know a measurement; it says very little about the accuracy of the measurement For example, if you measure the length of a table to great precision, but with a measuring instrument that is not

calibrated correctly, you will not measure accurately Accuracy is a measure of how close a

measurement is to the true value

4 If you measure the length of an object, and you report that it is “4,” you haven’t given enough

information for your answer to be useful There is a large difference between an object that is

4 meters long and one that is 4 feet long Units are necessary to give meaning to a numerical answer

5 You should report a result of 8.32 cm Your measurement had three significant figures When you multiply by 2, you are really multiplying by the integer 2, which is an exact value The number of significant figures is determined by the measurement

6 The correct number of significant figures is three: sin 30.0° =0.500

7 Useful assumptions include the population of the city, the fraction of people who own cars, the average number of visits to a mechanic that each car makes in a year, the average number of weeks a mechanic works in a year, and the average number of cars each mechanic can see in a week

(a) There are about 800,000 people in San Francisco, as estimated in 2009 by the U.S Census

Bureau Assume that half of them have cars If each of these 400,000 cars needs servicing twice

a year, then there are 800,000 visits to mechanics in a year If mechanics typically work

50 weeks a year, then about 16,000 cars would need to be seen each week Assume that on average, a mechanic can work on 4 cars per day, or 20 cars a week The final estimate, then, is

800 car mechanics in San Francisco

Responses to MisConceptual Questions

are inherently very accurate A digital scale is only as accurate as the last digit that it displays

2 (a) The total number of digits present does not determine the accuracy, as the leading zeros in (c)

and (d) are only placeholders Rewriting the measurements in scientific notation shows that (d) has two-digit accuracy, (b) and (c) have three-digit accuracy, and (a) has four-digit accuracy

Note that since the period is shown, the zeros to the right of the numbers are significant

3 (b) The leading zeros are not significant Rewriting this number in scientific notation shows that it

only has two significant digits

4 (b) When you add or subtract numbers, the final answer should contain no more decimal places than

the number with the fewest decimal places Since 25.2 has one decimal place, the answer must

be rounded to one decimal place, or to 26.6

5 (b) The word “accuracy” is commonly misused by beginning students If a student repeats a

measurement multiple times and obtains the same answer each time, it is often assumed to be accurate In fact, students are frequently given an “ideal” number of times to repeat the

experiment for “accuracy.” However, systematic errors may cause each measurement to be inaccurate A poorly working instrument may also limit the accuracy of your measurement

6 (d) This addresses misconceptions about squared units and about which factor should be in the

numerator of the conversion This error can be avoided when students treat the units as algebraic symbols that must be cancelled out

7 (e) When making estimates, students frequently believe that their answers are more significant than

they actually are This question helps the student realize what an order-of-magnitude estimation

is NOT supposed to accomplish

8 (d) This addresses the fact that the generic unit symbol, like [L], does not indicate a specific

7 To add values with significant figures, adjust all values to be added so that their exponents are all the same

When you add, keep the least accurate value, so keep to the “ones” place in the last set of parentheses

8 When you multiply, the result should have as many digits as the number with the least number of significant digits used in the calculation

(3.079 10 m)(0.068 10× × − m) 2.094 m= ≈ 2.1 m

9 The uncertainty is taken to be 0.01 m

2 2

2.483 m

V V

11 To find the approximate uncertainty in the area, calculate the area for the specified radius, the

minimum radius, and the maximum radius Subtract the extreme areas The uncertainty in the area

is then half this variation in area The uncertainty in the radius is assumed to be 0.1 10 cm.× 4

(d) 62.1 ps 62.1 10× −12 s 0.0000000000621 s

Note that in part (f ) in particular, the correct number of significant digits cannot be determined when

you write the number in this format

(d) 18 10 bucks× 2 18 hectobucks =18 hbucks or 1.8 kilobucks

(e) 7 10× −7seconds 700 nanoseconds =700 ns or 0.7 sμ

Note that if more significant figures were used in the original factors, such as 0.6214 miles per

kilometer, more significant figures could have been included in the answers

Since the soccer field is 109.4 yd compared with the 100.0-yd football field, the soccer field

is 9.4% longer than the football field

22 (a) # of seconds in 1.00 yr:

24 The radius of the ball can be found from the circumference (represented by “c” in the equations

below), and then the volume can be found from the radius Finally, the mass is found from the volume

of the baseball multiplied by the density (ρ=mass/volume) of a nucleon

3 3

ball ball ball ball 3 ball 3

nucleon nucleon nucleon

4 1 nucleon 3 nucleon nucleon

3 2 2

4 3

Of course, it would take more time on the clock for a runner to run across the U.S The runner

obviously could not run for 500 hours non-stop If he or she could run for 5 hours a day, then it would take about 100 days to cross the country

28 A commonly accepted measure is that a person should drink eight 8-oz glasses of water each day That is about 2 quarts, or 2 liters of water per day Approximate the lifetime as 70 years

4

(70 yr)(365 d/1 yr)(2 L/1 d)≈ ×5 10 L

29 An NCAA-regulation football field is 360 feet long (including the end zones) and 160 feet wide,

0.5 meters and that a person mowing can walk at about 4.5 km/h, which is about 3 mi/h Thus the distance to be walked is as follows:

30 There are about 3 10× 8 people in the U.S Assume that half of them have cars, that they drive an average of 12,000 miles per year, and that their cars get an average of 20 miles per gallon of gasoline

31 In estimating the number of dentists, the assumptions and estimates needed are:

• the population of the city

• the number of patients that a dentist sees in a day

• the number of days that a dentist works in a year

• the number of times that each person visits the dentist each year

We estimate that a dentist can see 10 patients a day, that a dentist works 225 days a year, and that each person visits the dentist twice per year

(a) For San Francisco, the population as of 2010 was about 800,000 (according to the U.S Census

Bureau) The number of dentists is found by the following calculation:

dentists (of all types, including oral surgeons and orthodontists) listed in the 2012 Yellow Pages

above the surface of the Earth, and the tangent line from the balloon height to the

surface of the Earth indicates the location of the horizon, a distance d away from

the balloon Use the Pythagorean theorem

h r r d

To 1st sunset

To 2nd sunset

AB

Earth center

R R

h d

θθ

34 In the figure in the textbook, the distance d is perpendicular to the radius that is drawn approximately vertically Thus there is a right triangle, with legs of d and R, and a hypotenuse of R h+ Since

35 For you to see the Sun “disappear,” your line of sight

to the top of the Sun must be tangent to the Earth’s

surface Initially, you are lying down at point A, and

you see the first sunset Then you stand up, elevating

your eyes by the height h=130 cm While you stand,

your line of sight is tangent to the Earth’s surface at

point B, so that is the direction to the second sunset

The angle θ is the angle through which the Sun

appears to move relative to the Earth during the time

to be measured The distance d is the distance from

your eyes when standing to point B

Use the Pythagorean theorem for the following relationship:

θ

=

The angle θ can be found from the height change and the radius of the Earth The elapsed time

between the two sightings can then be found from the angle, because we know that a full revolution takes 24 hours

t

θθ

37 (a) For the equation υ=At3−Bt, the units of At3 must be the same as the units of υ So the units

of A must be the same as the units of υ/ ,t3 which would be L T/ 4. Also, the units of Bt must be

the same as the units of υ So the units of B must be the same as the units of υ/ ,t which would

be L T/ 2

(b) For A, the SI units would be m/s ,4 and for B, the SI units would be m/s 2

38 (a) The quantity υt2 has units of (m/s)(s ) m s,2 = i which do not match with the units of meters for x

The quantity 2at has units (m/s )(s) m/s,2 = which also do not match with the units of meters for x

Thus this equation cannot be correct

(b) The quantity υ0t has units of (m/s)(s) m,= and 1 2

2at has units of (m/s )(s ) m.2 2 = Thus, since each term has units of meters, this equation can be correct

(c) The quantity υ0t has units of (m/s)(s) m,= and 2at has units of 2 (m/s )(s ) m.2 2 = Thus, since each term has units of meters, this equation can be correct

39 Using the units on each of the fundamental constants (c, G, and h), we find the dimensions of the

Planck length We use the values given for the fundamental constants to find the value of the Planck length

Thus the order of magnitude is 10−35 m

2 10 m

−

distinguishable from 20,000,002 m, which means that 8 significant figures are needed in the distance measurements

41 Multiply the number of chips per wafer by the number of wafers that can be made from a cylinder

We assume the number of chips per wafer is more accurate than 1 significant figure

42 Assume that the alveoli are spherical and that the volume of a typical human lung is about 2 liters,

3πr

50 km

V d

45 We approximate the jar as a cylinder with a uniform cross-sectional area In counting the jelly beans

in the top layer, we find about 25 jelly beans Thus we estimate that one layer contains about 25 jelly beans In counting vertically, we see that there are about 15 rows Thus we estimate that there

are 25 15 375× = ≈ 400 jelly beans in the jar

π

=⎜ ⎟

volume is calculated from the density, and then the diameter from the volume

The actual value is 3480 km

49 To calculate the mass of water, we need to find the volume of water and then convert the volume to mass The volume of water is the area of the city (48 km )2 times the depth of the water (1.0 cm)

distance around the Earth at the equator is the circumference, 2πREarth We assume that the person can “walk on water,” so ignore the existence of the oceans

Pencil

distance

Moon distance

54 Consider the diagram shown Let represent the distance he walks upstream Then

from the diagram find the distance across the river

0.5

15.0

θθ

0.5

75.0

θθ

sin θ, while the angles around 15° had a 3% error in sin θ

56 Utilize the fact that walking totally around the Earth along the meridian would trace out a circle whose full 360° would equal the circumference of the Earth

d

60o

58 The units for each term must be in liters, since the volume is in liters

This is more than a billion atoms per square meter

60 The density is the mass divided by volume There will be only 1 significant figure in the answer

61 Multiply the volume of a spherical universe times the density of matter, adjusted to ordinary matter

3πr

3 15

Solutions to Search and Learn Problems

1 Both Galileo and Copernicus built on earlier theories (by Aristotle and Ptolemy), but those new theories explained a greater variety of phenomena Aristotle and Ptolemy explained motion in basic terms Aristotle explained the basic motion of objects, and Ptolemy explained the basic motions of astronomical bodies Both Galileo and Copernicus took those explanations of motion to a new level Galileo developed explanations that would apply in the absence of friction Copernicus’s Sun-centered theory explained other phenomena that Ptolemy’s model did not (such as the phases of Venus)

orange krypton-86 light is found from the fact that 1,650,763.73 wavelengths of that light is the definition of the meter

3 The original definition of the meter was that 1 meter was one ten-millionth of the distance from the Earth’s equator to either pole The distance from the equator to the pole would be one-fourth of the circumference of a perfectly spherical Earth Thus the circumference would be 40 million meters:

The value in the front of the textbook is 6.38 10 m.× 6

4 We use values from Table 1–3

2

19 human

17 DNA

4 Moon 3 Moon Moon

(6.38 10 km)

49.3(1.74 10 km)

R

ππ

×

×

Responses to Questions

1 A car speedometer measures only speed It does not give any information about the direction, so it does not measure velocity

2 If the velocity of an object is constant, then the speed and the direction of travel must also be constant

If that is the case, then the average velocity is the same as the instantaneous velocity, because nothing about its velocity is changing The ratio of displacement to elapsed time will not be changing, no matter the actual displacement or time interval used for the measurement

3 There is no general relationship between the magnitude of speed and the magnitude of acceleration For example, one object may have a large but constant speed The acceleration of that object is then zero Another object may have a small speed but be gaining speed and therefore have a positive acceleration So in this case the object with the greater speed has the lesser acceleration

Consider two objects that are dropped from rest at different times If we ignore air resistance, then the object dropped first will always have a greater speed than the object dropped second, but both will have the same acceleration of 9.80 m/s 2

4 The accelerations of the motorcycle and the bicycle are the same, assuming that both objects travel in a straight line Acceleration is the change in velocity divided by the change in time The magnitude of the change in velocity in each case is the same, 10 km/h, so over the same time interval the

accelerations will be equal

5 Yes For example, a car that is traveling northward and slowing down has a northward velocity and a southward acceleration

6 The velocity of an object can be negative when its acceleration is positive If we define the positive direction to be to the right, then an object traveling to the left that is having a reduction in speed will have a negative velocity with a positive acceleration

If again we define the positive direction to be to the right, then an object traveling to the right that is having a reduction in speed will have a positive velocity and a negative acceleration

DESCRIBING MOTION: KINEMATICS IN ONE DIMENSION 2

7 If north is defined as the positive direction, then an object traveling to the south and increasing in speed has both a negative velocity and a negative acceleration Or if up is defined as the positive direction, then an object falling due to gravity has both a negative velocity and a negative acceleration

8 Yes Remember that acceleration is a change in velocity per unit time, or a rate of change in velocity

So velocity can be increasing while the rate of increase goes down For example, suppose a car is traveling at 40 km/h and one second later is going 50 km/h One second after that, the car’s speed is

55 km/h The car’s speed was increasing the entire time, but its acceleration in the second time interval was lower than in the first time interval Thus its acceleration was decreasing even as the speed was increasing

Another example would be an object falling WITH air resistance Let the downward direction be positive As the object falls, it gains speed, and the air resistance increases As the air resistance increases, the acceleration of the falling object decreases, and it gains speed less quickly the longer it falls

9 If the two cars emerge side by side, then the one moving faster is passing the other one Thus car A is passing car B With the acceleration data given for the problem, the ensuing motion would be that car A would pull away from car B for a time, but eventually car B would catch up to and pass car A

10 If there were no air resistance, the ball’s only acceleration during flight would be the acceleration due

to gravity, so the ball would land in the catcher’s mitt with the same speed it had when it left the bat,

120 km/h Since the acceleration is the same through the entire flight, the time for the ball’s speed to change from 120 km/h to 0 on the way up is the same as the time for its speed to change from 0 to

120 km/h on the way down In both cases the ball has the same magnitude of displacement

11 (a) If air resistance is negligible, the acceleration of a freely falling object stays the same as the

object falls toward the ground That acceleration is 9.80 m/s 2 Note that the object’s speed increases, but since that speed increases at a constant rate, the acceleration is constant

(b) In the presence of air resistance, the acceleration decreases Air resistance increases as speed

increases If the object falls far enough, the acceleration will go to zero and the velocity will become constant That velocity is often called the terminal velocity

12 Average speed is the displacement divided by the time Since the distances from A to B and from B to

C are equal, you spend more time traveling at 70 km/h than at 90 km/h, so your average speed should

be less than 80 km/h If the distance from A to B (or B to C) is x km, then the total distance traveled is 2x The total time required to travel this distance is x/70 plus x/90 Then

14 Yes Any time the velocity is constant, the acceleration is zero For example, a car traveling at a constant 90 km/h in a straight line has nonzero velocity and zero acceleration

15 A rock falling from a cliff has a constant acceleration IF we neglect air resistance An elevator moving from the second floor to the fifth floor making stops along the way does NOT have a constant

acceleration Its acceleration will change in magnitude and direction as the elevator starts and stops The dish resting on a table has a constant (zero) acceleration

16 The slope of the position versus time curve is the object’s velocity The object starts at the origin with

a constant velocity (and therefore zero acceleration), which it maintains for about 20 s For the next

10 s, the positive curvature of the graph indicates the object has a positive acceleration; its speed is increasing From 30 s to 45 s, the graph has a negative curvature; the object uniformly slows to a stop, changes direction, and then moves backwards with increasing speed During this time interval, the acceleration is negative, since the object is slowing down while traveling in the positive direction and then speeding up while traveling in the negative direction For the final 5 s shown, the object continues moving in the negative direction but slows down, which gives it a positive acceleration During the

50 s shown, the object travels from the origin to a point 20 m away, and then back 10 m to end up 10 m from the starting position

17 Initially, the object moves in the positive direction with a constant acceleration, until about t=45 s,when it has a velocity of about 37 m/s in the positive direction The acceleration then decreases,

reaching an instantaneous acceleration of 0 at about t = 50 s, when the object has its maximum speed

of about 38 m/s The object then begins to slow down but continues to move in the positive direction

The object stops moving at t = 90 s and stays at rest until about t = 108 s Then the object begins to

move in the positive direction again, at first with a larger acceleration, and then with a lesser acceleration

At the end of the recorded motion, the object is still moving to the right and gaining speed

Responses to MisConceptual Questions

1 (a, b, c, d, e, f, g) All of these actions should be a part of solving physics problems

2 (d) It is a common misconception that a positive acceleration always increases the speed, as in

(b) and (c) However, when the velocity and acceleration are in opposite directions, the speed

will decrease

3 (d) Since the velocity and acceleration are in opposite directions, the object will slow to a stop

However, since the acceleration remains constant, it will stop only momentarily before moving toward the left

4 (c) Students commonly confuse the concepts of velocity and acceleration in free-fall motion At the

highest point in the trajectory, the velocity is changing from positive (upward) to negative (downward) and therefore passes through zero This changing velocity is due to a constant downward acceleration

5 (a) Since the distance between the rocks increases with time, a common misconception is that the

velocities are increasing at different rates However, both rocks fall under the influence of gravity, so their velocities increase at the same rate

6 (c) Since the distances are the same, a common error is to assume that the average speed will be

halfway between the two speeds, or 40 km/h However, it takes the car much longer to travel the

4 km at 30 km/h than at 50 km/h Since more time is spent at 30 km/h, the average speed will be closer to 30 km/h than to 50 km/h

7 (c) A common misconception is that the acceleration of an object in free fall depends upon the

motion of the object If there is no air resistance, the accelerations for the two balls have the same magnitude and direction throughout both of their flights

incorrect because the initial velocity has been inserted for the average velocity Answers (b) and (c)

have the correct signs for each variable and the known values are inserted properly

9 (a) Increasing speed means that the slope must be getting steeper over time In graphs (b) and (e),

the slope remains constant, so these are cars moving at constant speed In graph (c), as time increases x decreases However, the rate at which it decreases is also decreasing This is a car slowing down In graph (d), the car is moving away from the origin, but again it is slowing down The only graph in which the slope is increasing with time is graph (a)

The negative sign indicates the direction

5 The time of travel can be found by rearranging the average velocity equation

approximately 1 km) in 3 seconds So the rule could be approximated as 1 km every 3 seconds

7 The time for the first part of the trip is calculated from the initial speed and the first distance, using d to

The time for the second part of the trip is now calculated

(a) The total distance is then dtotal=d1+d2=180 km 169.3 km 349.3 km+ = ≈ 350 km

(b) The average speed is NOT the average of the two speeds Use the definition of average speed,

Eq 2–1

total total

349.3 km

4 5 h

d t

9 The distance traveled is 3200 m (8 laps 400 m/lap).× That distance probably has either 3 or 4

significant figures, since the track distance is probably known to at least the nearest meter for

competition purposes The displacement is 0, because the ending point is the same as the starting point

14.5 min 60 s

d t

υ = =⎛⎜⎜⎝ × ⎞⎛⎟⎜⎟⎝⎠ ⎞⎛⎟⎜⎠⎝ ⎞⎟⎠= × ≈ ×

11 Both objects will have the same time of travel If the truck travels a distance dtruck, then the distance the car travels will be dcar=dtruck+210 m Using the equation for average speed, υ= Δd t/ , solve for time, and equate the two times

truck car truck truck truck car

Also note that car

To find the average speed, we need the distance traveled (500 km) and the total time elapsed

1

,

x t

υ =Δ

Δ so

1 1 1

250 km

2.632 h

95 km/h

x t

250 km

4.545 h

55 km/h

x t

500 km

61 km/h8.177 h

x t

Δ

To find the average velocity, use the displacement and the elapsed time

υ = Δ Δ =x t/ 0

13 Since the locomotives have the same speed, they each travel half the distance, 4.25 km Find the time

of travel from the average speed

15 The average speed of sound is given by υsound = Δ Δx t/ , so the time for the sound to travel from the

sound

340 m/s

x t

ball to travel from the bowler to the end of the lane is given by Δtball= Δttotal− Δtsound=

x t

Δ

16 For the car to pass the train, the car must travel the length of the train AND the distance the train travels The distance the car travels can thus be written as either dcar=υcart=(95 km/h)t or

car train train 1.30 km (75 km/h)

the distance the car travels

The distance the car travels during this time is d=(95 km/h)(0.065 h) 6.175 km= ≈ 6.2 km

If the train is traveling in the opposite direction from the car, then the car must travel the length of the train MINUS the distance the train travels Thus the distance the car travels can be written as either

The distance the car travels during this time is d =(95 km/h)(7.65 10 h)× −3 = 0.73 km

17 The average acceleration is found from Eq 2–4

Δ

The final velocity is υ =0, and the time to stop is 4.0 s Use Eq 2–11a to find the acceleration

2 0

6.0 m/s4.0 s

2 2

1 m/s(55 km/h)

28 The final velocity of the car is zero The initial velocity is found from Eq 2–11c with υ=0 and solving for υ0 Note that the acceleration is negative

435.2 m/s

44 ’s (9.80 m/s )/

1 2

the equation for motion at constant velocity

(a) Solve Eq 2–11c for the final location

33 Use the information for the first 180 m to find the acceleration and the information for the full motion

to find the final velocity For the first segment, the train has υ =0 0 m/s, υ =1 18 m/s, and a

displacement of x1−x0=180 m Find the acceleration from Eq 2–11c

x t

The runner will accomplish this by accelerating from speed υ0 to speed υ for t seconds, covering a distance d1, and then running at a constant speed of υ for (180−t) seconds, covering a distance d2.

We have these relationships from Eq 2–11a and Eq 2–11b

2 1

Since we must have t<180 s, the solution is t=6.3 s

36 (a) The train’s constant speed is υtrain =5.0 m/s, and the location of the empty box car as a function

of time is given by xtrain =υtraint=(5.0 m/s) t The fugitive has υ0=0 m/s and a=1.4 m/s2

until his final speed is 6.0 m/s The elapsed time during the acceleration is

run The first possibility to consider is, “Can the fugitive catch the empty box car before he reaches his maximum speed?” During the fugitive’s acceleration, his location as a function of

fugitive 0 0 2 0 0 2(1.4 m/s )

x =x +υt+ at = + + t For him to catch

train fugitive (5.0 m/s) 2(1.4 m/s )

0 s, 7.1 s

Now the equation of motion of the fugitive changes After the 4.286 s of acceleration, he runs with a constant speed of 6.0 m/s Thus his location is now given (for times t>5 s) by the following:

1 fugitive 2(1.4 m/s )(4.286 s) (6.0 m/s)( 4.286 s) (6.0 m/s) 12.86 m

2

4.059 s, 20.94 s2

38 Define the origin to be the location where the speeder passes the police car Start a timer at the instant that the speeder passes the police car and find another time that both cars have the same displacement from the origin

For the speeder, traveling with a constant speed, the displacement is given by the following:

For the police car, the displacement is given by two components The first part is the distance traveled

at the initially constant speed during the 1 second of reaction time

The second part of the police car displacement is that during the accelerated motion, which lasts for

(t−1.00) s So this second part of the police car displacement, using Eq 2–11b, is given as follows:

The answer that is approximately 0 s corresponds to the fact that both vehicles had the same

displacement of zero when the time was 0 The reason it is not exactly zero is rounding of previous values The answer of 10.5 s is the time for the police car to overtake the speeder

as a check on the answer, the speeder travels Δ =xs (37.5 m/s)(10.5 s) 394 m,= and the police car travels Δ =xp [26.39 26.39(9.5) 1.30(9.5) ] m 394 m.+ + 2 =

39 Choose downward to be the positive direction, and take y0=0 at the top of the cliff The initial velocity is υ0=0, and the acceleration is a=9.80 m/s 2 The displacement is found from Eq 2–11b,

8.806 s 8.8 s9.80 m/s

41 Choose upward to be the positive direction, and take y0=0 to be the height from which the ball was thrown The acceleration is a= −9.80 m/s 2 The displacement upon catching the ball is 0, assuming

it was caught at the same height from which it was thrown The starting speed can be found from

Eq 2–11b, with x replaced by y

2 1

0 0 2

2 1

υυ

(b) The time of flight can be found from Eq 2–11b, with x replaced by y, using a displacement of 0

for the displacement of the ball returning to the height from which it was hit

trivial amount of air effect on a baseball as it moves through the air—that’s why pitches like the

“curve ball” work, for example So ignoring the effects of air makes this an estimate Another effect is that the problem says “almost” straight up, but the problem was solved as if the initial velocity was perfectly upward Finally, we assume that the ball was caught at the same height as which it was hit That was not stated in the problem either, so that is an estimate

the acceleration is a=9.80 m/s ,2 and the initial velocity is υ0=0 Use Eq 2–11b to calculate the time for the kangaroo to fall back to the ground The total time is then twice the falling time

a

44 Choose upward to be the positive direction, and take y0=0 to be at the floor level, where the jump starts For the upward path, y=1.2 m, υ=0 at the top of the path, and a= −9.80 m/s 2

(a) The initial speed can be found from Eq 2–11c, with x replaced by y

(b) The time of flight can be found from Eq 2–11b, with x replaced by y, using a displacement of 0

for the displacement of the jumper returning to the original height

45 Choose downward to be the positive direction, and take y0=0 to be at the height where the object was released The initial velocity is υ0=0, and the acceleration is a=9.80 m/s 2

(a) The speed of the object will be given by Eq 2–11a with υ0=0, so υ=at=(9.80 m/s ) 2 t This

is the equation of a straight line passing through the origin with a slope of 9.80 m/s 2

(b) The time to reach that height can be found from Eq 2–11b

1

2 2

and once on the way down (t=4.28 s)

47 Choose downward to be the positive direction, and take y0=0 to be the height from which the object

is released The initial velocity is υ0=0, and the acceleration is a=g Then we can calculate the

2

second, the position would be as follows:

The value of (2n+1) is always odd, in the sequence1, 3, 5, 7,…

48 (a) Choose upward to be the positive direction, and y0=0 at the ground The rocket has υ =0 0,

2

3.2 m/s ,

runs out of fuel from Eq 2–11c, with x replaced by y

a

(c) For this part of the problem, the rocket will have an initial velocity υ0=70.43 m/s, an

acceleration of a= −9.80 m/s ,2 and a final velocity of υ=0 at its maximum altitude The altitude reached from the out-of-fuel point can be found from Eq 2–11c

2 2

775 m 2 (a y 775 m)

(d) The time for the “coasting” portion of the flight can be found from Eq 2–11a

0

775 m coast coast 0 70.43 m/s2

7.19 s9.80 m/s

Thus the total time to reach the maximum altitude is t=22.01 s 7.19 s 29.20 s+ = ≈ 29 s

(e) For this part of the problem, the rocket has υ0=0 m/s, a= −9.80 m/s ,2 and a displacement of

1028 m

Earth from Eq 2–11c

Thus the total time for the entire flight is t=29.20 s 14.48 s 43.68 s+ = ≈ 44 s

49 Choose downward to be the positive direction, and take y0=0 to be the height where the object was released The initial velocity is υ0= −5.40 m/s, the acceleration is a=9.80 m/s ,2 and the

Eq 2–11b, with x replaced by y

1

2 2

The correct time is the positive answer, t=5.21 s

0.83 s

2 1

υυ

51 Choose upward to be the positive direction and y0=0 to be the location of the nozzle The initial velocity is υ0, the acceleration is a= −9.80 m/s ,2 the final location is y= −1.8 m, and the time of flight is t=2.5 s. Using Eq 2–11b and substituting y for x gives the following:

2 1

υυ

and the location of the window is y=18 m

(a) Using Eq 2–11c and substituting y for x, we have

Choose the positive value because the initial direction is upward

(b) At the top of its path, the velocity will be 0, so we can use the initial velocity as found above,

along with Eq 2–11c

(velocity υ=14 m/s) Using Eq 2–11a, we have the following:

(d) We want the time elapsed from the window (υ0=14 m/s) to reaching the street

(υ= −23.43 m/s) Using Eq 2–11a, we have:

0

3.819 s 3.8 s9.80 m/s

The total time from throwing to reaching the street again is 0 9622 s 3 819 s 4 8 s + =

dropped Call the location of the top of the window yw, and the time for the stone to fall from release

to the top of the window is tw. Since the stone is dropped from rest, using Eq 2–11b with y

2 w

0.569 s(0.31)

g t

elapsed from dropping the rock to hearing the sound Insert this expression for t1 into the equation for

H from the stone, and solve for H

55 Slightly different answers may be obtained since the data come from reading the graph

(a) The greatest velocity is found at the highest point on the graph, which is at t≈48 s

(b) The indication of a constant velocity on a velocity vs time graph is a slope of 0, which occurs

from t=90 s to t≈108 s

(c) The indication of a constant acceleration on a velocity vs time graph is a constant slope, which

occurs from t=0 s to t≈42 s , again from t≈65 s to t≈83 s , and again from

2

8 s 4 s

a t

υ

57 Slightly different answers may be obtained since the data come from reading the graph

(a) The instantaneous velocity is given by the slope of the tangent line to the curve At t=10.0 s,

58 Slightly different answers may be obtained since the data come from reading the graph

(a) The indication of a constant velocity on a position versus time graph is a constant slope, which

59 Theυvs t graph is found by taking the slope of the x vs t

graph

Both graphs are shown here

60 Choose the upward direction to be positive and y0=0 to be the level from which the object was thrown The initial velocity is υ0 and the velocity at the top of the path is υ=0 The height at the top

of the path can be found from Eq 2–11c with x replaced by y