Solution manual for essential college physics 1st edition by andrew rex richard wolfson pages deleted

R EFLECT Even if the object’s instantaneous velocity has not been zero throughout the time interval, a velocity of zero implies that it has returned to its starting point, so   0.x If

R EFLECT Since acceleration is the change in velocity with time,   

 0 0.

a

t Velocity can be positive,

negative, or zero Acceleration can be zero in any of these cases An object with zero velocity and zero acceleration

is at rest

4 S OLVE Yes.Think of throwing an apple straight up in the air Establish a coordinate system in which “up” is positive On its way up, the apple’s velocity is positive While it is falling back down, the velocity is negative During the entire time the apple is in the air, the acceleration due to gravity is negative

R EFLECT While the apple is moving upward, it is slowing down because acceleration points downward After it has reached its highest point, acceleration still points downward, and the apple “speeds up” in the negative direction

5 S OLVE Yes If you throw the apple of Question 4 straight upward, it slows down to a velocity of zero at its highest point The acceleration due to gravity is still negative at this highest point

R EFLECT Think about a drag race with the positive x-direction toward the finish line At the instant the race starts, the vehicles have zero velocity At the same instant, the acceleration has a large positive value As a result, the velocities of the vehicles increase rapidly as they move down the track

6 S OLVE The speed at the end of each time interval is

distance traveled distance traveled

distance traveled distance traveled

t a

Number of time interval Total distance traveled

distance traveled a t

where a 2m s2and t increases by one second for each time interval

Motion in One Dimension 2.3

7 S OLVE

R EFLECT Look at a point on a velocity-versus-time graph This point gives the slope of the line on the

corresponding position-versus-time graph at that same time The slope of the velocity-versus-time graph gives the location of a point on the acceleration-versus-time graph at that same time

8 S OLVE Suppose your car has a maximum speed of 100mi h, or about 45m s At an acceleration of 3.0m s ,2 it would take   

2

45 m s

15 s3.0 m s

t to reach that maximum speed

R EFLECT This answer assumes that this maximum acceleration remains constant In practice, acceleration depends on the gear you are in and the speed of the engine (we’ll study more about what is known as “torque” in Chapter 8 After the transmission shifts into higher gears, the acceleration decreases As you speed up, air

resistance also increases This also contributes to decreasing the acceleration

9 S OLVE The velocities of each of the two cars are given as constant, with no mention of change in velocity The acceleration of each car must be zero

R EFLECT If the problem had stated that the two velocities were given only for a specific point in time, then the answer might have been ambiguous An alternative is that the faster car has zero acceleration while the slower car has a positive acceleration The faster car passes the slower car, but in a short time, the car that was slower will accelerate to a speed faster than25 m sand will pass the first car This is what happens when you pass a stationary police car while you are driving at constant speed over the speed limit!

10 S OLVE At a given position, velocity is different, speed is the same, and acceleration is the same

R EFLECT At any particular position or height, velocity has the same magnitude, but opposite sign Going up, the sign on velocity is positive Coming back down, the sign is negative Speed does not change because v v y, so the sign of v does not matter Acceleration is constant during free fall, with y a y   g 9.80 m s 2

11 S OLVE If an object’s average velocity is zero, then its displacement must be zero

x

x v t

   Forv x 0,     x 0 t 0 m

If the object’s acceleration is zero, then the displacement might be zero, positive, or negative, depending on the value and sign of the velocity

R EFLECT Even if the object’s instantaneous velocity has not been zero throughout the time interval, a velocity of zero implies that it has returned to its starting point, so   0.x If acceleration is zero and velocity is zero, then the

object is at rest and has not moved If velocity is positive or negative, then displacement constantly increases in the same direction as velocity

12 S OLVE With a sufficient number of steps, you will eventually reach your destination If we graph displacement

versus time, we see that even though we divide the distance traveled by two in each successive step, we also divide the time it takes by two The speed therefore remains constant Therefore, if you cover half the distance to your goal in1

2 you will also cover the remaining half in an additional time of , 1

2 no matter the number of parts ,into which it is divided

R EFLECT In fact, the distance is divided into an infinite number of steps Mathematicians have found this to be a

“summable” infinite series, written as

M ULTIPLE -C HOICE P ROBLEMS

13 O RGANIZE AND P LAN We are to find average speed We use the definition of average speed,v   The x t

responses are all in m s, so we must convert distance to meters and time to seconds

Known:   385,000 km;x   2.5 day.t

S OLVE Using the definition of speed,

v t

x

v

The answer is response (C)

R EFLECT The escape velocity from Earth is about11,000 m s A spacecraft traveling toward the moon initially achieves escape velocity, and then slows down due to Earth’s gravity If the intention is to make a “soft” landing

on the moon withv x 0 m s, it makes sense that the average speed must be less than the escape velocity

14 O RGANIZE AND P LAN We are given time and speed and are asked to find distance We use the definition of

speed, v   and rearrange it to solve for distance x t

Known:v 32 m s;   35 s.t

S OLVE The definition of speed is

x v t





Rearranging, we get

x v t

   Substituting known values,

 x 32 m s 35 s 1100 m The answer is response (D)

R EFLECT Even though the problem states that the cheetah’s top speed is 32 m s we consider it to be the average speed for the duration of this problem A distance of 1100 m is about 1200 yards, the length of 12 football fields This is a reasonable distance for a cheetah to chase its prey

15 O RGANIZE AND P LAN In this problem, the runner has already run part of the race with known quantities We

have to figure out what’s going to happen during the rest of the race We’ll use subscript “1” for the first part of the race, subscript “2” for the remainder of the race, and no subscript if the variable applies to the entire race We need

to find the speedv the runner must maintain until the finish line We know that it’s a 1500 m race, with 1200 m 2

Motion in One Dimension 2.5

already run Since the runner must finish the race in under 4 minutes, our strategy is to find the time already elapsed and then the allowed time remaining From that, we can calculate the necessary speed

Known:  1500 m;x  x1 1200 m;v16.14 m s;  t 4 min

S OLVE For the part of the race already completed, we use

1 1 1

x v t





Solving for , we get t1



1 1

1200 m

195.44 s6.14 m s

x t v

300 m

6.73 m s44.56 s

x v t

The answer is response (a)

R EFLECT This is nearly a 1-mile race It is normal for runners to sprint, or run faster, at the end of a long race The required average speed to finish under 4 minutes is only a little faster than the average speed in the first part of the race, which is reasonable

16 O RGANIZE AND P LAN In this problem, the runner travels forward and backward along the x-axis We are asked to

calculate the runner’s average velocity We can’t just average the two velocities given; this would violate math rules for the definition of velocity Our strategy is to calculate the total displacement and the total time From these

we can calculate the average velocity for the entire run We’ll use subscript “1” for running forward, subscript “2” when jogging backward, and no subscript for a variable that applies to the entire run Since we are calculating average velocity rather than speed, we carefully note the signs of the values

Known: v x19.2 m s;v x2 3.6 m s;  x1 100 m;  x2 50 m

S OLVE First, the total displacement

     x x1 x2 100 m 50 m 50 mThe runner ends up only 50 m from the starting point

Next, the total time



1 1

100 m

10.87 s9.2 m s

x

x t v



2 2 2

50 m

13.89 s3.6 m s

x

x t v

     t t1 t2 10.87 s 13.89 s 24.76 s Finally, the average velocity

x

x v t

The answer is response (a)

R EFLECT Velocity is based upon displacement, not distance The farther backward (toward the origin) the runner jogs, the smaller the displacement and the closer the average velocity will approach zero

17 O RGANIZE AND P LAN This problem is about the properties of displacement in one-dimensional motion and

comparison of displacement with distance

S OLVE The definition of displacement in one-dimensional motion is the difference between final and initial

positions in a coordinate system:

0

x x x

  

A one-dimensional coordinate system extends without bound in both the positive and negative directions from the

origin Therefore both x and x can have values that are positive, negative, or zero, independently of each other 0

Since this is the case, the difference between the two values can also be positive, negative, or zero

The answer is response (b)

R EFLECT Distance is the sum of the lengths of all the straight-line segments of a trip, without regard to sign Distance is always positive The displacement value may be equal to the distance value, but does not have to be Displacement is never greater than distance

18 O RGANIZE AND P LAN In this problem we are to find average speed from two different speeds and distances run

at those speeds We don’t just average the speeds We must divide the total distance by the total time, so we have

to find each of these first We’ll use the formula for average speedv   x t

Known: v14.0 m s;v26.0 m s;    x1 x2 60 m

S OLVE The total distance is just the sum of the two distances

     x x1 x2 60 m 60 m 120 m Now we find the times for each distance run, using the formula for speed



1 1

60 m

15 s4.0 m s

x t v



2 2

60 m

10 s6.0 m s

x t v

The answer is response (a)

R EFLECT The answer is an average The value we found lies between the higher and lower values, as it should However, note that the answer is not just the arithmetic mean of the two given velocity values

19 O RGANIZE AND P LAN We must find average acceleration from initial velocity, final velocity, and elapsed time

This means we use the definition of average acceleration, a x v x  t

The answer is response (c)

R EFLECT The acceleration of the spacecraft may seem modest, about a 50% increase in velocity But the velocity values themselves are not important Only their difference matters, v In this case the acceleration is about 180% x

of that due to gravity If this spacecraft has human occupants, acceleration of this magnitude would certainly be noticeable By comparison, a sports car accelerating from zero to 60 miles per hour in 4 seconds only accelerates at about 68% of that due to gravity

20 O RGANIZE AND P LAN This problem requires us to find displacement from initial velocity, final velocity, and

The answer is response (c)

Motion in One Dimension 2.7

R EFLECT Note that squaring the velocities results in a positive numerator The sign of the acceleration is

negative, producing a negative displacement, as we would expect from an object starting from rest

21 O RGANIZE AND P LAN This problem requires us to find acceleration from initial velocity, final velocity, and



The answer is response (c)

R EFLECT The arrow has a relatively high speed and stops in a very short distance This is like an automobile traveling 50 mi/h stopping in about 1.5 inches This requires a very large negative acceleration The sign on the acceleration is negative because the initial velocity of the arrow is positive, and it slows down as it travels into the target

22 O RGANIZE AND P LAN This is a free-fall problem, with the object in constant acceleration We want to know how

long it takes to strike the ground, in seconds We will have to use one of the kinematic equations that contain time

as a variable Since we know that initial velocity is zero, but we don’t know final velocity, we will use

9.5 s9.80 m s

y

y t a

The answer is response (c)

R EFLECT The Sears Tower is one of the world’s tallest buildings It takes an object less than 10 seconds to fall to the ground, ignoring air resistance This emphasizes the important aspect of constant acceleration The longer the object falls, the faster it goes, and the greater the distance it covers each second We see that the height of the building appears under the radical Doubling the height of the building will not double the length of time the object takes to fall from the top Rather, the time increases by a factor of 2

23 O RGANIZE AND P LAN This is a free-fall problem Notice that only an algebraic solution is needed We need to

find the effect on final velocity if we double the distance an object falls We won’t use a subscript for the original height, but we’ll use the subscript “2” for the doubled height Since time is not included in the problem, we’ll use the kinematic equation v2yv2y 2a y y

Known:   y h; v y 0 m s; v y v; a y  g

S OLVE Substituting the variables given in the problem, we get

v v  a h

Since v  for both the original height and the doubled height, y0 0

The answer is response (c)

R EFLECT Displacement in the y-direction is a function of the square of velocity It makes sense that velocity is a function of the square root of height during free-fall

24 O RGANIZE AND P LAN This is a constant-acceleration problem Since time is not part of the problem, we can use

the kinematic formula v2xv2x 2a x x

The answer is response (b)

R EFLECT We can think of this problem as the opposite of a free-fall problem (and in the horizontal direction) Since the car initially has positive velocity and is slowing down, the acceleration must be negative

25 O RGANIZE AND P LAN We are asked to find the shape of the graph of velocity versus time for a moving object

under constant acceleration Since the object is moving, we know that velocity can be positive or negative, but it cannot be zero No calculation is needed

We see that a is the slope and x v is the vertical intercept (the intercept on the velocity axis) We are told 0

thata x 0 The slope of the line must be either positive, sloping upward, or negative, sloping downward Either of these possibilities gives a diagonal line on the graph

The answer is response (b)

R EFLECT An acceleration of zero would give us a horizontal line (slope = zero) In order for the graph to have the shape of a parabola, velocity would have to be a function oft Rather, it is a function of the first power of2 t, a linear function

26 O RGANIZE AND P LAN This problem contains two objects! We must find some common value between them In

this case, both the time at which they pass and their height y as they pass one another are common We solve for

the height at which each ball passes the other and set the two equations equal to each other From this, we find a numeric value for  Substituting this value back into the equation for height for either ball 1 or ball 2 gives us t.the value for height

Known: v  y1 10m s; a y   g 9.80 m s ;2 v y2 10 m s;   y2 10 m

S OLVE The two balls not only pass each other at the same height, the must pass each other at the same time Our

key is to find that time For the first ball,

y y v  t a t

Motion in One Dimension 2.9

For the second ball,

y y v  t a t Since the balls meet at the same height y, we can set these two equations equal:

y v  t a t y v  t a t Canceling like terms and substituting known values, we get

The answer is response (c)

R EFLECT The two balls have to meet somewhere between ground level and10 m above the ground, which they

do Checking our work with the equation for the position of the second ball,

27 O RGANIZE AND P LAN In this problem we must show the difference between the distance and the displacement

for a round trip between two points We’ll use “1” as the subscript for the first part of the trip and “2” as the subscript for the second part of the trip

Known: d1200 m

S OLVE The distance between two points is always positive regardless of the direction traveled For a round trip to the video store, the distance from your friend’s house isd1200 m The distance from the video store back to your friend’s house is alsod 2 200 m. So the total distance for the round trip is

 1 2200 m 200 m 400 m 

But for displacement, we must take into account the sign of the direction of travel for each part of the trip

Traveling in the positive direction, from your friend’s house to the video store,

 x1 200 mReturning from the video store in the negative direction,

  x2 200 mThe total displacement for the round trip is

     x x1 x2 200 m 200 m 0 m

R EFLECT A round trip, with the ending position the same as the starting position, always has a positive distance and a zero displacement

28 O RGANIZE AND P LAN This problem uses the definition of displacement We choose a coordinate system with the

positive direction to the east Since Lincoln is our starting point, we’ll choose the position of Lincoln to bex Our 0

final position will be Grand Island, 160 km to the west Its position isx

Known: x00 km;  160 km.x

S OLVE We travel in the negative direction from Lincoln to Grand Island Using the definition of displacement,

  x x x0 160 km 0 km  160 km

R EFLECT When you give someone directions to a destination, you must tell them which direction to travel if you want them to arrive where they intend to go When traveling in the negative direction, the value of displacement will have a negative sign

29 O RGANIZE AND P LAN In this problem, we have to consider what effect a fractional round trip has on distance and

displacement We set up a coordinate system with the positive direction to the east We start at Grand Island, so we can declare this position to bex The position of Lincoln will be 0 x

S OLVE (a) Here we have three round trips from Grand Island to Lincoln The displacement  traveling from x1

Grand Island to Lincoln is

  x1 x x0160 km 0 km 160 km Traveling from Lincoln back to Grand Island, the displacement is

  x2 160 kmThe total displacement for one round trip is

 x 160 km 160 km 0 kmTherefore, the displacement for three round trips is

3 x 3 0 km 0 kmThe distance traveled from Grand Island to Lincoln is 160 km Since distance is always positive, the distance traveled from Lincoln back to Grand Island is also 160 km The distance traveled in three round trips is

  x x x0160 km 0 km 160 km However, the distance for 1

3 round trips From (b), the displacement for 3 round trips

is zero During the remaining 3

4 round trip, we travel from Grand Island to Lincoln in the positive direction, and then halfway back to Grand Island in the negative direction The displacement for this trip is

30 O RGANIZE AND P LAN We must find the average speed in m s of a runner under three different conditions of time

and distance For each of these, we’ll use the definition of average speed, v distance traveled

t



 However, in each scenario, we’ll have to convert time or distance, or both, to the proper units

Known: (a) distance traveled 41 km;   2 h 25 min;t (b) distance traveled 1500 m;   3 min 50 s;t

Motion in One Dimension 2.11

Then we convert 41 kmto m,

distance traveled 41 km 1000 m km 41,000 km Now we use the definition of average speed

31 O RGANIZE AND P LAN To calculate elapsed time, we need to know average distance between the Earth and the

sun, and the speed of light We use 3.00 10 m s 8 for the speed of light and 1.50 10 m for the orbital radius of  11

Earth

Known:distance traveled 1.50 10 m;  11 v3.00 10 m s. 8

S OLVE Using the definition of average speed,

11 8

distance traveled 1.50 10 m

500 s3.00 10 m s

60 s or about 8 minutes and 20 seconds

32 O RGANIZE AND P LAN This problem requires us to rearrange the definition of average speed, solving for time

Then we are given two instances in which we must find time from speed and distance

Known:(A) distance traveled 18.4 m; v 44 m s;(B)distance traveled 18.4 m; v 32 m s

S OLVE Rearranging the definition of average speed we obtain

33 O RGANIZE AND P LAN In this problem there are two parts to the motion, at different speeds We do not simply

average the speeds, even though the distances traveled are both 100.m We must go back to the definition of

average speed, v distance traveled.

t



 We’ll use subscript (1) for the 4.0 m s run and subscript (2) for the 5.0 m s run

Known:v14.0 m s; distance traveled1100 m; v25.0 m s; distance traveled2100 m

S OLVE First we find the distance traveled

34 O RGANIZE AND P LAN In this problem we are asked to compare average speeds calculated under different

conditions We use the definition of average speed, v distance traveled

t



 In part (a) time is constant In part (b) distance is constant We will use subscript (1) for the first run and subscript (2) for the second run in each set of conditions

Known:(A) v 1 10.m s;  t1 100.s; v 2 20.m s;  t2 100.s;(B)v 1 10.m s; distance traveled 1000.m;

v

(b) We must find the time it takes to run each leg of the total distance

1 1

35 O RGANIZE AND P LAN We are given the distances traveled and the speeds for two legs of a flight, and the layover

time between legs We calculate the times for each leg using the definition of velocity Then we add the layover

Motion in One Dimension 2.13

time to find t From the total time and total distance, we find average speed We’ll use subscript (1) for the first leg of the flight and subscript (2) for the second leg

Known: distance traveled11100 km; distance traveled2550 km; v 800.km/h; tlayover80.min

distance traveled distance traveled 1100 km 550 km

490 km/h1.375 h 0.688 h 1.333 h

36 O RGANIZE AND P LAN In a race, the winner is the contestant that starts at positionx and first reaches the finish 0

line x Finishing first means reaching x with a lower time t than the other contestant Here we will compare the

elapsed times of the two boats The boat with the lower t is the winner We will use subscripts (1) and (2) for boats 1 and 2, respectively, and a second subscript stating the first or second variable of the boat during the race

Known:distance traveled 2.000 km; Boat1: v114.0 m s; v123.1 m s; distance traveled111500 m; Boat2:v213.6 m s; v223.9 m s; distance traveled211200 m

S OLVE Since speed is given in m s, we will need to convert the length of the race to meters

2.000 km 1000 m km 2000.m Now, for boat 1, the first leg requires time

t

v

The second leg of the race requires

12 12

12

distance traveled 2000 m 1500 m

161 s3.1 m s

t

v

The total time for boat 1 is

    t2 t21 t22333 s 205 s 538 s Boat 1 reaches the finish line first and wins the race

R EFLECT Another way of determining the winner is to calculate the higher average speed of each boat and to compare these values The boat with the higher average speed wins However, average speed is a function of total time, so this alternate method requires two additional math steps

37 O RGANIZE AND P LAN We are to find both speed and velocity First we must find both distance and displacement

To find total distance, we use the definition of speed to find the distance traveled on each leg of the trip To find displacement, we take into account the signs of the distances For this purpose, our coordinate system will establish east as positive and west as negative We’ll use subscript (1) for the first leg and subscript (2) for the second leg

Known:v1210 km h east;  t1 3.0 h; v2170 km h west, or 170 km h;  t2 2.0 h

S OLVE First we find the displacement for each leg

 x1 630 kmThe second leg of the trip is in the negative direction:

  x2 170 km h 2.0 h  340 km

     x x1 x2 630 km 340 km 290 kmDividing displacement by time to obtain velocity,

v t

Then we find the distance for each leg:

distance traveled v t 210 km h 3.0 h 630 kmRemember that distance is always positive, so we have to use the absolute value of velocity

distance traveled v t 170 km h 2.0 h 340 kmThe total distance is

v

t

R EFLECT Since the plane reverses its path and flies back toward its starting point, the distance is greater than the displacement and the speed is greater than the velocity

38 O RGANIZE AND P LAN In this problem, we help a runner plan strategy for a 10K (10-km) race We know the total

length of the race, but not the lengths of the individual legs We have enough information to find average speed There is only one combination of distances at each of the two given speeds that will result in this calculated average speed We can express the distance of the first leg asdistance traveled The second leg will then be 1

1

10,000 m distance traveled We will use subscript (1) for the first leg and subscript (2) for the second leg

Known:distance traveled 10.0 km; v14.10 m s; v27.80 m s;   40.0 min.t

S OLVE First we convert the time to seconds:

40.0 min 60 s min 2400 s Then we convert the distance of the race to meters:

10.0 km 1000 m km 10,000 m Finding expressions for the elapsed times of the two legs,

1 1

Motion in One Dimension 2.15

Now we clear the denominators by multiplying byv v , group similar terms, and solve for1 2 distance traveled : 1

 v2 distance traveled1   v1 10,000 distance traveled 1v v t1 2

 v2 distance traveled1   v1 distance traveled1   v1 10,000 mv v t 1 2

7.80 m s 4.10 m s



1

distance traveled 9663 m This is the distance the runner must cover before starting the sprint Now we subtract this value from the length of the race:

2

distance traveled 10,000 m 9663 m 337 m The runner must start her sprint 337 m before the end of the race

R EFLECT Starting the sprint at 337 m before the finish line will cause the runner to finish the race in exactly 40.0 min If she starts the sprint sooner, she will finish the race in less time

39 O RGANIZE AND P LAN This problem emphasizes that velocity takes into account all the elapsed time, not just the

time an object is in motion Here we will use the definition of velocity The dogsled goes “straight” so we are free

to establish our own coordinate system, with the “straight” direction of travel being in the positive x-direction We’ll use the subscript (1) for the time the dogsled is in motion, and no subscript for the variables pertaining to the entire 24-hour period

Known:  24 h;t  t1 10 h; v x19.5 m s

S OLVE First, we convert the two known times to seconds

 t 24 h 3600 s h 86,400 s 

 t1 10 h 3600 s h 36,000 s Then we find the displacement during the 10-hour period when the dogsled is moving x1

   x1 v t1 1 9.5 m s 36,000 s 342,000 mThis gives us velocity

x

x v t

R EFLECT A velocity of 9.5 m s is about the highest velocity a human can achieve for short periods of time No wonder that humans in snowy regions of the Earth use dogsleds for transportation

40 O RGANIZE AND P LAN This problem contrasts speed and velocity When we calculate speed, we use the absolute

distance traveled on each leg of a trip When we want velocity, all we care about is how far we end up away from the starting point This means we have to use the sign of each direction traveled We’ll use the subscripts (1), (2), and (3) for the three legs of the trip

Known:v  x1 100.km h;  t1 30.0 min; v x260.0 km h;  t2 10.0 min; v x3 80.0 km h;

Adding the three separate displacements,

 x 50.0 km 10.0 km  26.6 km 33.4 kmand the average velocity is

v t

Then we find the total distance traveled:

v

t

R EFLECT Displacement is never greater than distance, and is always less than distance if a change in direction takes place during the motion Likewise, average velocity is never greater than average speed

41 O RGANIZE AND P LAN In this problem, we have to find the error in an observation Error

is observed value true value The observed value is the speedometer reading, 60.0mi h We must calculate the true speed from the true distance between highway mileposts and the true elapsed time (measured by your clock)

Known:vobserved60.0 mi h; distance traveledtrue5.00 mi; (a):ttrue4 min 45 s; (b): vtrue65.0 mi h

S OLVE (a) First we convert time to hours:

v

t

error v v 60.0 mi h 63.2 mi h 3.2 mi h(b) If our true speed is 65.0mi h, then

true true

distance traveled 1.00 mi

0.154 h 55.4 s65.0 mi h

42 O RGANIZE AND P LAN This problem emphasizes that we must calculate values for each leg of a trip In the first

case (a) there is no wind, so the bird’s speed in the same whether flying east or west In the second case (b) with a tailwind outbound and a headwind inbound, the two speeds are very different We must calculate the time of each leg We must not simply average the two speeds We’ll use subscript (1) for the eastbound leg and subscript (2) for the westbound (returning) leg

Known: x1 10.0 km;   x2 10.0 km; (a): v110.0 m s; v2 10.0 m s; (b) v115.0 m s; v2 5.0 m s

S OLVE First we convert the displacements to meters:

 x1 10.0 km 1000 m 1km 10,000 m 

  x2 10.0 km 1000 m 1km  10,000 m

Motion in One Dimension 2.17

(a) Flying east,



1 1

10,000 m

1000 s10.0 m s

x t v

Flying west toward home,

 



2 2 2

10,000 m

1000 s10.0 m s

x t v

 t 1000 s 1000 s 2000 s (b) Flying east,



1 1

10,000 m

667 s15.0 m s

x t v

 



2 2 2

10,000 m

2000 s5.0 m s

x t v

 t 667 s 2000 s 2667 s (c) The two times are different We can’t just average two velocities, because time is inversely proportional to velocity We would have to average the reciprocals of the velocities to find the reciprocal of the average velocity in this case This is the same strategy we would use in a rate problem

R EFLECT The elapsed time to fly to a location and return to your origin depends on the wind If you have a constant wind during a trip, the elapsed time will be greater than for no wind The higher the wind speed, the greater the elapsed time Aircraft pilots must be very careful of this, because the aircraft consumes the same amount of fuel per hour, regardless of groundspeed

43 O RGANIZE AND P LAN In this problem, two moving objects start at different positions Here we have to use the full form of

a kinematic equation for position, not just displacement We’ll establish a coordinate system with both animals traveling in the positive x-direction The cheetah starts at the origin and the zebra starts at a position of 35 m Since both animals end up

at the same spot and at the same time, we can set the equations for each animal’s position equal to each other and solve for

R EFLECT The answer is just the zebra’s initial lead divided by the difference in velocities Think of the animals

on a treadmill that is moving at the zebra’s velocity of 14 m s To a stationary observer standing on the ground next to the treadmill, the zebra appears to be standing still while the cheetah approaches at

30.m s 14 m s 16 m s The time it takes the cheetah to reach the zebra under these conditions is given by solving the definition of velocity   This gives the exact answer we obtained above We’ll discuss this x v t.notion of relative motion in a later chapter

44 O RGANIZE AND P LAN The parachutist’s average velocity v while her chute is open is her y

displacement y divided by the time her chute is open We are given the displacement while the chute is open We have to find the time by subtracting 10.0 seconds from the elapsed time for the entire trip We’ll calculate this from the average velocity for the entire trip We’ll use the subscripts “ff” for free fall and “chute” for the time her parachute is open

Now, the time for which the chute is open:

chute

1350 m

2.65 m s508.8 s

y

y v

t





R EFLECT The velocity while the chute is open is between the average free-fall velocity and the overall velocity,

as we would expect To check how reasonable our answer is, we can use the kinematic equation v2yv2y   2g y.Solving for  we find that a final velocity of 2.65 m s is how fast we would strike the ground after jumping y,from a height of about 0.36 m, or about 14 inches This seems to be a reasonable velocity

45 O RGANIZE AND P LAN We are to find an experimental speed of light and compare it with today’s accepted value

of 3.00 10 m s. 8 We must convert the experimental distance from kilometers to meters and the experimental time from minutes to seconds

Known: distance traveled 299,000,000 km;=   22 min.t

1761 Römer’s value was quite a scientific feat for his time

46 O RGANIZE AND P LAN Average velocity is displacement divided by the time interval Each time interval is 2.0 s

We must read initial and final positions from the graph shown in Figure P2.46 in the text to find displacement

S OLVE We summarize the data read from the graph in a table Then we use the definition of average

velocityv x   to complete the last column of the table x t

Motion in One Dimension 2.19

Time interval, s x, m x0, m  x v x

0.0–2.0 4.17 10.0 –5.83 –2.92 2.0–4.0 5.00 4.17 0.83 0.42 4.0–6.0 12.5 5.00 7.50 3.75 6.0–8.0 25.7 12.5 14.2 7.08

R EFLECT The average velocity is the slope of a tangent to the curve of the graph at the midpoint of each time interval, for small intervals

47 O RGANIZE AND P LAN Here we are to construct a graph of velocity versus time for the 8-second time interval

From Problem 46, we already have the average velocity for each time interval Velocity is the slope of a line

tangent to curve of the position-versus-time graph when t is small

S OLVE The table below summarizes the data from Problem 46 that we need to construct our graph

Time interval, s Midpoint values, s v x

Using the values from the table below, we plot the following:

R EFLECT Velocity increases steadily with time, from an initial negative (downward) value to positive values The original path of the object (from Problem 46) is a parabola, concave upward It shows an initial downward

velocity, decreasing to zero, and then increasing as t increases

48 O RGANIZE AND P LAN Average acceleration is found using 0

Known:

S OLVE A systematic way to set up this problem is to create a table Each line of the table represents one time period We calculate average acceleration from the values in each line Reading time and velocity values from the

49 O RGANIZE AND P LAN Here we have to draw a graph of instantaneous acceleration We’ll need to look at the

graph in Problem 48 This graph has three regions of constant slope We’ll find the slope for each region using

x x

S OLVE We’ll indicate the three line segments using the subscripts 1, 2, and 3 The acceleration values for the

three segments are

1

17.5 m s 0 m s

2.33 m s7.5 s

x x v a t

2 3

2.50 m s 17.5 m s

2.67 m s

20 s 12.5 s

x x v a t

50 O RGANIZE AND P LAN Here we are to draw a motion diagram for the car in Problem 48 We’ll refer to the figure

in that problem for our information A motion diagram shows the position of an object after equal intervals of time Known: Velocity increases steadily from  0 st tot 6 s Then velocity becomes constant from  8 st tot 12 s.From  16 st to  20 st velocity steadily decreases to a negative value

S OLVE

Motion in One Dimension 2.21

R EFLECT In the first part of the graph from Problem 48, the car starts at zero velocity and speeds up at a constant rate We know this because the slope is constant and positive In the middle of the graph, acceleration is zero because the slope is zero During the last part of the graph, the car slows down at a constant rate, because the slope

is constant and negative At the very end of the graph, we see that the car reaches zero velocity and then reverses direction We know this because the velocity falls below zero

51 O RGANIZE AND P LAN Acceleration is the value of the slope of a graph of velocity versus time We are to

consider the graph in Problem 48 The greatest acceleration occurs when the slope of the graph is most positive The least acceleration occurs when the slope is most negative Acceleration is zero where the graph is a horizontal line We use the equation x x0

(a) Acceleration is greatest between 0 s and 7.5 s

(b) Acceleration is least between 12.5 s and 20 s

(c) Acceleration is zero between 7.5 s and12.5 s

(d) For the greatest acceleration,

2

0 12.5 m s 0 m s

2.33 m s 7.5 s

2

0 2.5 m s 17.5 m s

2.67 m s 20.0 s 12.5 s

52 O RGANIZE AND P LAN The race lasts10.0 s, so the first half of the race ends at  5.0 s.t We use data read from

the graph in Figure 2.13(b) to find the average acceleration, x x0

S OLVE During the time intervalt t , velocity increases from 0 v x00 m stov x11.0 m s

53 O RGANIZE AND P LAN The stock car’s velocity is related to time by the equationv x1.4t21.1 t We are asked

to find v after 4.0 s This means we have to evaluate the given equation at 4.0 s For the units to cancel properly x

the value 1.4 must have units of m s3and the value 1.1 must have units of m s 2

54 O RGANIZE AND P LAN We must graph the velocity of the stock car versus time We create a table containing

ordered pairs of time and velocity values We only know that the functionv x1.4t21.1tis valid

Plotting time on the horizontal axis and velocity on the vertical axis, we obtain a graph like this

R EFLECT This graph is a segment of a parabola, as we expect from seeing thet term in the equation Velocity 2

increases more and more rapidly as elapsed time increases

55 O RGANIZE AND P LAN The instantaneous velocity is the y-value on a graph of velocity versus time This is the

graph we constructed in Problem 54

Known: t 2.0 s

S OLVE In the present problem, we raise a perpendicular from t 2.0 s on the horizontal axis At the point where this perpendicular intersects the curve, we draw a horizontal line segment left to the vertical axis The point on the vertical axis where the horizontal line intersects it represents the value of the instantaneous velocity

We see that the horizontal line intersects the vertical axis at the point 0,7.8 m s , which is the instantaneous velocity

R EFLECT This value is the velocity at the point in time t 2.0 s We can see from the graph that for any other time on the graph, the velocity will have a different value

Motion in One Dimension 2.23

56 O RGANIZE AND P LAN We are to draw a motion diagram based on Figure 2.15(a) in the text, which is also shown

below In this figure, the car speeds up during the first equal time interval Then it moves with constant speed during the second equal time interval Finally the car slows down during the third equal time interval

Known:

S OLVE

R EFLECT We see that the positions of the car are closer together at lower velocities than at higher velocities Some safety professionals suggest that a driver should maintain a 2-second time interval between his or her car and the car ahead This corresponds to a smaller spacing on the roadway at lower speeds, while allowing for a constant reaction time to apply the brakes

57 O RGANIZE AND P LAN We are to draw a motion diagram based on Figure 2.15(b) in the text, which is also shown

below The car speeds up in the negative direction during the first equal time interval Then it moves with constant velocity during the second equal time interval Finally the car slows down and stops during the final equal time interval

Known:

S OLVE

R EFLECT We cannot tell from the information given whether the vehicle is pointed in the positive or negative direction It doesn’t matter, because we model the car as a point, as shown in Section 2.1 We only know that the car is moving in the negative direction, regardless of whether it is in a forward gear or reverse gear

58 O RGANIZE AND P LAN We are to find the average acceleration between points on the figure

This figure is a graph of velocity versus time The slope of a secant line between any two points on this graph is the average acceleration between these points We are asked to find average acceleration between the positive diastolic peak and the negative systolic peak, and then again to the next diastolic peak We’ll use the kinematic

 We’ll need the velocity and time values from the graph

Known: From the graph, the diastolic peak occurs at the ordered pair0.45 s, 0.60 m s The systolic peak occurs 

at 0.97 s, 0.55 m s   The next diastolic peak occurs at 1.27 s, 0.60 m s 

S OLVE In the first case, we use the diastolic values as the initial values and the systolic values as the final values

2 0

0

0.55 m s 0.60 m s

2.2 m s0.97 s 0.45 s

0.60 m s 0.55 m s

3.8 m s1.27 s 0.97 s

59 O RGANIZE AND P LAN We know our initial velocity v  x0 50.km hand the distance to the stoplight,  x 40.m

We see that we need to convert the initial velocity to m s to agree with the units of We need to find both x.acceleration and how long it takes to stop We can use v2xv2x 2a x x to find the acceleration Then,

x

t a

The acceleration is about the same as that of a car accelerating from 0 to 60mi h in 11 seconds This seems to be a reasonable acceleration

R EFLECT According to the sign of our initially velocity, we are traveling in the positive x-direction We are slowing down, so the sign of acceleration must be negative, as we calculated

Motion in One Dimension 2.25

60 O RGANIZE AND P LAN To calculate acceleration, we need the change in velocity and the elapsed time Then we

can use the definition of acceleration, x x0

S OLVE (a) First we convert the cheetah’s speed to m/s

61 O RGANIZE AND P LAN This is a comparison problem We are given the distance to the hole Now we have to

calculate how far the ball goes, and compare that value to the distance to the hole We assume for the moment that the ball comes to rest and usev2xv2x 2a x x with a final velocity of zero If we find that the ball makes it to the hole, then we can recalculate using the same formula, but instead using x 4.8 m and solving for final

velocityv at the position of the hole x

Known: v x02.52 m s; a x 0.65 m s ;2 distance to the hole 4.8 m.

S OLVE (a) To find whether the ball makes it to the hole,

v v  a x Rearranging,

So the ball does reach the hole!

(b) Now, at the hole,

v  as the ball passes the hole

R EFLECT Since the problem tells us that the golfer putts straight toward the hole, we have confidence that the ball goes into the hole at this modest speed If the ball’s speed were too great, it could strike the far edge of the cup and bounce out When we study Chapters 3 and 6, we’ll learn about projectile motion and collisions, which will help us understand why the ball might not end up in the cup!

62 O RGANIZE AND P LAN We’re given acceleration and time We have to find the sled’s final speed We can use

(b) Then we can find the distance traveled

v v  a x Rearranging,

R EFLECT A velocity of188 m s is about 420 mi h The sled travels 822 m just to reach its top speed After the end of this problem, the sled will require even more distance to slow down and stop We would want a very long and straight path for this sled run When we study rotational motion in Chapter 8, we’ll learn more about what would happen to the sled if the path were not straight

63 O RGANIZE AND P LAN We have a car that speeds up from some initial speed, then slows down and stops in

two distinct time intervals We need to find the car’s maximum speed, and the total time and distance it

travels We’ll use the subscripts 1 and 2 to represent what happens during the first and second time intervals Between the time intervals, acceleration changes from a to1 a We can use the formulas 2 v v 0  a t x

S OLVE (a) The car’s maximum speed occurs at the end of the first time interval, that is, after the maximum time

with positive acceleration:

0 m s 25.3 m s

21.1 s1.2 m s

x

t a

ttotal    t1 t2 6.2 s 21.1 s 27.3 s (c) Since we now know both time intervals, we can calculate the distance traveled for both intervals:

distance traveled distance traveled distance traveled 120.1 m 266.7 m 387 m

R EFLECT To solve graphically, we could plot distance traveled versus time, and find distance as the area under the line We’d divide the graph into a trapezoid and a triangle and add the areas of each The numeric equivalent of this is to multiply the average speed for each time interval by the length of each interval, and add the two values

64 O RGANIZE AND P LAN To calculate stopping time from an initial speed and acceleration, we can use the

x

t a

Motion in One Dimension 2.27

For the second car,

x

t a

(b) For the first car, the stopping distance is

t t

The ratio of stopping distances is

2 1

distance traveled 110.4 m

4distance traveled 27.6 m

R EFLECT The answers make sense since stopping time has only one factor of t in the equation, but stopping distance uses two factors of Therefore, for a doubling of ,t t  we would expect stopping distance to increase by

a factor of 4, which we see here

65 O RGANIZE AND P LAN We’ll model the bullet as a point at the front of the bullet Since we know displacement

but not time, we can use v2v22a x x for both parts of this problem

66 O RGANIZE AND P LAN Since we don’t know time, we can usev2xv2x 2a x x to find the length of the runway,