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

2.2

Chapter 2

vx  vx 0
 0. Velocity can be positive,
t
negative, or zero. Acceleration can be zero in any of these cases. An object with zero velocity and zero acceleration
is at rest.
REFLECT

Since acceleration is the change in velocity with time, a x 

4.

SOLVE 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.
REFLECT 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.

SOLVE 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.
REFLECT 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.

SOLVE The speed at the end of each time interval is

v

distance traveled
.
t

The change in speed, v  v0 , is
v 

distance traveled 2 distance traveled1 distance traveled2  distance traveled1


t
t
t

The change in speed with time is acceleration, so
distance traveled 2  distance traveled1
distance traveled 2  distance traveled1
t
a

.
t
t 2

 

Since the time intervals are all equal, their squares are all equal. We are told that the differences in the distances
traveled are also equal. Therefore acceleration is constant.
REFLECT Let us go one step further and add the distances in this series of observations. After each succeeding
time interval, we have traveled the distance obtained by adding all the distances traveled up through the current
time interval, as shown in the table below.

Number of time interval
0
1
2
3
4
5

Total distance
traveled
00
0 1 1
0 1 3  4
0 1 3  5  9
0  1  3  5  7  16
0  1  3  5  7  9  25

With an initial speed v0 = 0, these values match the results of the kinematic equation for distance with constant
acceleration:
distance traveled  21 a  t 

2

where a  2m s2 and t increases by one second for each time interval.


Motion in One Dimension

7.

2.3

SOLVE

REFLECT 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.

SOLVE Suppose your car has a maximum speed of 100mi h, or about 45m s. At an acceleration of 3.0m s 2 , it
45 m s
would take t 
 15 s to reach that maximum speed.
3.0 m s2
REFLECT 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.

SOLVE 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.

REFLECT 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 than 25 m s and 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.

SOLVE At a given position, velocity is different, speed is the same, and acceleration is the same.
REFLECT 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  vy , so
the sign of v y does not matter. Acceleration is constant during free fall, with ay   g  9.80 m s 2 .

11.

SOLVE If an object’s average velocity is zero, then its displacement must be zero.

x  vx t
For vx  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.
REFLECT 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 x  0. If acceleration is zero and velocity is zero, then the


2.4

Chapter 2

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.

SOLVE 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 in 21 t, you will also cover the remaining half in an additional time of

1
t,
2

no matter the number of parts

into which it is divided.
REFLECT In fact, the distance is divided into an infinite number of steps. Mathematicians have found this to be a
“summable” infinite series, written as
1 1 1 1
1
  


2 4 8 16 32



1
1
2n

So the sum of this infinite number of steps toward the goal is simply the distance to the goal. Since the paradox
gives no limit on the number of steps and velocity remains constant, you can reach your goal in exactly twice the
time it takes you to go half the distance toward the goal.

MULTIPLE-CHOICE PROBLEMS
13.

ORGANIZE AND PLAN We are to find average speed. We use the definition of average speed, v  x t . The
responses are all in m s, so we must convert distance to meters and time to seconds.
Known: x  385,000 km; t  2.5 day.
SOLVE Using the definition of speed,

vx 

x
t

vx 

385,000 km 1000 m km
 1800 m s
2.5 day  24 h day  3600 s h

The answer is response (C).
REFLECT The escape velocity from Earth is about 11,000 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 with vx  0 m s, it makes sense that the average speed must be less than the escape velocity.
14.

ORGANIZE AND PLAN We are given time and speed and are asked to find distance. We use the definition of

speed, v  x t and rearrange it to solve for distance.
Known: v  32 m s; t  35 s.
SOLVE The definition of speed is

v

x
t

Rearranging, we get
x  v t

Substituting known values,

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

ORGANIZE AND PLAN 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 speed v2 the runner must maintain until the finish line. We know that it’s a 1500 m race, with 1200 m


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: x  1500 m; x1  1200 m; v1  6.14 m s; t  4 min.
SOLVE For the part of the race already completed, we use
x
v1  1
t1
Solving for t1 , we get

t1 

x1 1200 m

 195.44 s
v1
6.14 m s

t2  4 min 60 s min  195.44 s  44.56 s
x2  1500 m  1200 m  300 m
v2 

x2 300 m

 6.73 m s
t2 44.56 s

The answer is response (a).
REFLECT 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.

ORGANIZE AND PLAN 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: vx1  9.2 m s; vx 2  3.6 m s;  x1  100 m; x2  50 m.
SOLVE First, the total displacement
x  x1  x2  100 m   50 m   50 m

The runner ends up only 50 m from the starting point.
Next, the total time
x
100 m
t1  1 
 10.87 s
v x1 9.2 m s
t 2 

x2
50 m

 13.89 s
v x 2 3.6 m s

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

vx 

x
50 m

 2.0 m s
t 24.76 s

The answer is response (a).
REFLECT 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.

ORGANIZE AND PLAN This problem is about the properties of displacement in one-dimensional motion and
comparison of displacement with distance.
SOLVE The definition of displacement in one-dimensional motion is the difference between final and initial
positions in a coordinate system:

x  x  x0


2.6

Chapter 2

A one-dimensional coordinate system extends without bound in both the positive and negative directions from the
origin. Therefore both x and x0 can have values that are positive, negative, or zero, independently of each other.
Since this is the case, the difference between the two values can also be positive, negative, or zero.
The answer is response (b).

REFLECT 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.

ORGANIZE AND PLAN 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 speed v  x t .
Known: v1  4.0 m s; v2  6.0 m s; x1  x2  60 m.
SOLVE 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.

t1 

x1
60 m

 15 s
v1
4.0 m s

t2 

x2
60 m

 10 s
v2

6.0 m s

t  t1  t2  15 s  10 s  25 s
v

x 120 m

 4.8 m s
t
25 s

The answer is response (a).
REFLECT 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.

ORGANIZE AND PLAN We must find average acceleration from initial velocity, final velocity, and elapsed time.
This means we use the definition of average acceleration, ax  vx t .
Known: vx 0  1250 m s; vx 1870 m s; t  35 s.
SOLVE We use the definition of average acceleration

ax 

vx vx  vx 0 1870 m s  1250 m s


 17.7 m s 2
t
t
35 s

The answer is response (c).
REFLECT 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, vx . In this case the acceleration is about 180%
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.

ORGANIZE AND PLAN This problem requires us to find displacement from initial velocity, final velocity, and
acceleration.
Known: a x  1.4 m s 2 ; vx  10 m s; vx0  0 m s.
SOLVE Since we don’t know time, we will make use of the kinematic equation
v2x  v x20  2a x x

Rearranging to find x ,
v x2  v x20
 10 m s    0 m s   35.7 m
 x 
2a x
2  1.4 m s 2
2



The answer is response (c).

2



Motion in One Dimension

2.7

REFLECT 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.

ORGANIZE AND PLAN This problem requires us to find acceleration from initial velocity, final velocity, and
displacement.
Known: vx0  21.4 m s; vx  0 m s x  3.75 cm.
SOLVE Since we don’t know time, we will make use of the kinematic equation
v2x  v x20  2a x x

Since x is given in centimeters, we need to convert it to meters.
Rearranging to find ax ,

 0 m s    21.4 m s   6100 m s2
v2x  v2x 0
 ax 
2x
2  3.75 cm  1 m 100 cm 
2

2

The answer is response (c).
REFLECT 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.

ORGANIZE AND PLAN 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
y  vy0 t  21 a y t 2 and solve for t.
Known: y  442 m; ay   g  9.81 m s2 ; vy 0  0 m s.
SOLVE We start with the formula we chose.

y  vy0 t  21 a y t 2
Since v y 0  0,

y  21 ay t 2
2y
 t 2
ay
t 

2y

ay

2   441 m 
9.80 m s 2

 9.5 s

The answer is response (c).
REFLECT 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.

ORGANIZE AND PLAN 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 vy2  vy20  2ay y.
Known: y  h; v y  0 m s; vy  v; a y   g.
SOLVE Substituting the variables given in the problem, we get

v2  v02  2a yh


2.8

Chapter 2

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

v2  2ah
v  2 ah

If we now double y to the value 2h, then the new velocity is
v2  2a y  2h   2 2ah  2 v

The answer is response (c).
REFLECT 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.

ORGANIZE AND PLAN This is a constant-acceleration problem. Since time is not part of the problem, we can use
the kinematic formula v 2x  v x20  2a x x.
Known: vx  0 m s; vx 0  12.8 m s; x  16.0 m.
SOLVE Starting with our chosen formula and solving for acceleration,
v2x  v x20  2a x x

ax 

v2x  vx20
2x

Substituting known values,

ax 

0 m s  12.8 m s
 5.1 m s 2
2  16.0 m

The answer is response (b).
REFLECT 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.

ORGANIZE AND PLAN 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.
Known: v  0 m s; a x  0 m s 2 .
SOLVE We know that velocity is a function of time and can be expressed as

vx  vx 0  ax t
We rearrange this formula to put it in the slope-intercept form of the equation of a straight line.

vx  ax t  vx 0
We see that ax is the slope and vx 0 is the vertical intercept (the intercept on the velocity axis). We are told
that ax  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).
REFLECT 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 of t 2 . Rather, it is a function of the first power of t , a
linear function.
26.

ORGANIZE AND PLAN 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 t . Substituting this value back into the equation for height for either ball 1 or ball 2 gives us
the value for height.
Known: vy1  10m s; ay  g  9.80 m s2 ; v y 2  10 m s; y2  10 m.
SOLVE 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  y1  vy1t  21 a y t 2

Motion in One Dimension

2.9

For the second ball,

y  y2  vy2 t  21 ay t 2
Since the balls meet at the same height y, we can set these two equations equal:

y1  vy1t  21 ay t 2  y2  vy2t  21 a y t 2
Canceling like terms and substituting known values, we get

10 m s  t  10 m+  10
t 

m s  t

10 m
 0.5 s
20 m s

We know that the height is the same for both balls, so we choose the equation for the position of the first ball.
Substituting known values including the time,





y  10 m s  0.5 s+ 12  9.80 m s 2   0.5 s   3.8 m
2

The answer is response (c).
REFLECT The two balls have to meet somewhere between ground level and 10 m above the ground, which they
do. Checking our work with the equation for the position of the second ball,

y  10 m   10 m / s   0.5 s 

1
2
 9.80 m / s 2   0.5 s 
2





y  3.8m

PROBLEMS
27.

ORGANIZE AND PLAN 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.
SOLVE 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 is d1  200 m. The distance from the video store back to your
friend’s house is also d2  200 m. So the total distance for the round trip is
d  d1  d2  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 m
Returning from the video store in the negative direction,

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

x  x1  x2  200 m   200 m   0 m
REFLECT A round trip, with the ending position the same as the starting position, always has a positive distance
and a zero displacement.
28.

ORGANIZE AND PLAN 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 be x0 . Our
final position will be Grand Island, 160 km to the west. Its position is x.
Known: x0  0 km; x  160 km.
SOLVE 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


2.10

Chapter 2

REFLECT 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.

ORGANIZE AND PLAN 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 be x0 . The position of Lincoln will be x.
SOLVE (a) Here we have three round trips from Grand Island to Lincoln. The displacement x1 traveling from
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 km
The total displacement for one round trip is

x  160 km   160 km   0 km
Therefore, the displacement for three round trips is
3x  3  0 km  0 km
The 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

d  3  160 km  160 km   960 km
(b) Here we have 3 21 round trips. This means three round trips plus one last trip from Grand Island to Lincoln.
Since the displacement from (a) for exactly three round trips was 0 km, the displacement here is the same as onehalf of a round trip, or the displacement from Grand Island to Lincoln, so

x  x  x0  160 km  0 km  160 km
However, the distance for 3 21 round trips from Grand Island to Lincoln is

d  3  160 km  160 km   160 km  7 160 km  1120 km

(c) This part asks about displacement and distance for 3 34 round trips. From (b), the displacement for 3 round trips
is zero. During the remaining 34 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

x  160 km  12   160 km   80 km
The distance for each leg of the trip is always positive, so

d  7  160 km  12  160 km  1200 km
REFLECT The displacement for a round trip is always zero. The distance for each leg of a trip is positive,
regardless of direction. Think about how often a car’s owner must fill the fuel tank just from driving round trips to
school!
30.

ORGANIZE AND PLAN 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
. However, in each
t

scenario, we’ll have to convert time or distance, or both, to the proper units.
Known: (a) distance traveled  41 km; t  2 h 25 min; (b) distance traveled  1500 m; t  3 min 50 s;
(c) distance traveled  100. m; t  10.4 s
SOLVE
(a) Start by converting 2 h 25 min to seconds. We think of this as 2 h  25 min and convert each term separately:

t  2 h  3600 s h   25 min  60 s min   8700 s


Motion in One Dimension

2.11

Then we convert 41 km to m,
distance traveled  41 km 1000 m km   41,000 km

Now we use the definition of average speed.
distance traveled 41,000 m
v

 4.7 m s
t
8700 s
(b) The distance is already given in meters. We convert the time to seconds

t  3 min  50 s  3 min  60 s min   50 s  230 s
Then

v

distance traveled 1500 m

 6.5 m s
t
230 s

(c) Here the distance is already in meters and the time is already in seconds.
distance traveled 100.m
v

 9.62 m s
t

10.4 s
REFLECT The lengths of these races systematically decrease from very long (a marathon!) to moderate (just
under a mile) and finally to a short dash. We can expect the average speeds to increase from a bit more than a fast
walk ( 4.7 m s ) to a fast run ( 9.62 m s ).
31.

ORGANIZE AND PLAN 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  108 m s for the speed of light and 1.50 1011 m for the orbital radius of
Earth.
Known: distance traveled  1.50 1011 m; v  3.00  108 m s.
SOLVE Using the definition of average speed,
distance traveled
v
t
and rearranging for t ,
t 

distance traveled 1.50  1011 m

 500 s
v
3.00  108 m s

REFLECT
One occasionally hears that if the sun suddenly stopped shining, it would take the inhabitants of Earth about 8
minutes to realize this. Checking this value,
1 min 
 500 s  
  8.33 min or about 8 minutes and 20 seconds.
 60 s 

32.

ORGANIZE AND PLAN 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.
SOLVE Rearranging the definition of average speed we obtain

t 

distance traveled
v

(a) t 

18.4 m
 0.42 s
44 m s

(b) t 

18.8 m
 0.58 s
32 m s

REFLECT Major league batters have about one-half second to see, track, and hit a pitch. The faster the pitch, the
less time the batter has.
33.

ORGANIZE AND PLAN 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

2.12

Chapter 2

average speed, v 

distance traveled
. We’ll use subscript (1) for the 4.0 m s run and subscript (2) for the
t

5.0 m s run.
Known: v1  4.0 m s; distance traveled1  100. m; v2  5.0 m s; distance traveled2  100. m.
SOLVE First we find the distance traveled
distance traveled  distance traveled1  distance traveled2
Then we find total time
t  t1  t2 

distance traveled1 distance traveled2

v1
v2

Finally we substitute known values into the definition of average speed
distance traveled 100.m  100.m
v

 4.4 m s
t

25 s  20 s
REFLECT To understand why we can’t just average the speeds, we see that this is a rate problem. We have to add
the reciprocals of the individual speeds to get the reciprocal of the average speed:
t
20.s
25 s
45 s
1




200.m 200 m 200 m 200 m v

v
34.

ORGANIZE AND PLAN

200 m
 4.4 m s
45 s

In this problem we are asked to compare average speeds calculated under different

conditions. We use the definition of average speed, v 

distance traveled
. In part (a) time is constant. In part
t

(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) v1  10.m s; t1  100.s; v2  20.m s; t2  100.s; (B) v1  10.m s; distance traveled  1000.m;
v2  20.m s; distance traveled  1000.m.
SOLVE (a) We must find the distances run at each speed.
distance traveled1  v1t1  10.m s 100.s   1000 m

distance traveled 2  v2t2   20.m s 100.s   2000 m
Then we use the definition of average speed:

v

distance traveled1  distance traveled 2 1000.m  1000.m

 15 m s
t1  t2
100.s  100.s

(b) We must find the time it takes to run each leg of the total distance.
distance traveled1 1000.m
t1 

 100.s
v1
10.m s
t2 

distance traveled 2 1000.m


 50.s
v2
20.m s

Using the definition of average speed,
distance traveled distance traveled1  distance traveled 2 2000.m
v


 13 m s
t
t1  t2
150.s
(c) The speeds in (a) and (b) are not identical. In (a) you ran longer at 20.m s than you did in (b). You ran for the
same time at 10.m s in both cases.
REFLECT The only circumstance in which you can find average speed by taking the arithmetic means of two
different speeds is if the times run at each speed are equal. You cannot do so if the distances run are equal.
35.

ORGANIZE AND PLAN 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.

SOLVE
(a) First we find the time for the first leg of the flight. Notice that distance is in kilometers and time is in hours.

t1 

distance traveled1 1100 km

 1.375 h
v1
800.km h

t2 

distance traveled 2
550 km

 0.688 h
v2
800.km h

Then we convert the layover time to hours:

tlayover  80.min 1 h 60 min   1.333 h
t  t1  t 2  tlayover  1.375 h  0.688 h  1.333 h  3.4 h

(b) Average speed is total distance divided by total time, so
1100 km + 550 km
v  distance traveled1 + distance traveled 2 
 490 km/h
t1 + t2 + tlayover

1.375 h + 0.688 h + 1.333 h
REFLECT This is one kind of problem in physics where it is not necessary to convert all units to SI base or
derived units. Since the problem does not ask for specific units, we are free to express speed in km/h and time in
hours. These units allow us to use values of reasonable magnitude for travel by air.
36.

ORGANIZE AND PLAN In a race, the winner is the contestant that starts at position x0 and first reaches the finish
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; Boat 1: v11  4.0 m s; v12  3.1 m s; distance traveled11  1500 m;
Boat 2: v21  3.6 m s; v22  3.9 m s; distance traveled21  1200 m.
SOLVE 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

t11 

distance traveled11 1500 m

 375 s
v11
4.0 m s

The second leg of the race requires
t12 

distance traveled12 2000 m  1500 m


 161 s
v12
3.1 m s

The total time for boat 1 is then

t1  t11  t12  375 s  161 s  536 s
Doing the same calculations for boat 2,

t21 
t21 

distance traveled 21 1200 m

 333 s
v21
3.6 m s

distance traveled 21 2000 m  1200 m

 205 s
v21
3.9 m s

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.
REFLECT 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.


2.14

37.

Chapter 2

ORGANIZE AND PLAN 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.
SOLVE First we find the displacement for each leg
x1  630 km
The 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 km
Dividing displacement by time to obtain velocity,
x 290 km
vx 

 58 km h
t
5.0 h
Then we find the distance for each leg:

distance traveled1  v1t1   210 km h    3.0 h   630 km

Remember that distance is always positive, so we have to use the absolute value of velocity

distance traveled 2  v2 t2  170 km h    2.0 h   340 km
The total distance is
distance traveled  630 km  340 km  970 km

So the average speed is

v

distance traveled 970 km

 194 km h
t
5.0 h

REFLECT 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.

ORGANIZE AND PLAN 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 as distance traveled1 . The second leg will then be
10,000 m  distance traveled1 .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; t  40.0 min.
SOLVE 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,
distance traveled1
t1 
v1
and

t2 

10,000 m  distance traveled1
v2

For the entire race,

t  t1  t2 

distance traveled1 10,000 m  distance traveled1

v1
v2


Motion in One Dimension

2.15

Now we clear the denominators by multiplying by v1v2 , group similar terms, and solve for distance traveled1 :

 v2  distance traveled1    v1 10,000  distance traveled1   v1v2t

 v2  distance traveled1    v1  distance traveled1    v1 10,000 m   v1v2t
 distance traveled1  v2  v1   v1v2t  v1 10,000 m 
distance traveled1 
distance traveled1 

 4.10 m s  7.80

v1v2t  v1 10,000 m 
v2  v1
m s  2400 s    4.10 m s 10,000 m 
7.80 m s  4.10 m s

distance traveled1  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:

distance traveled2  10,000 m  9663 m  337 m
The runner must start her sprint 337 m before the end of the race.
REFLECT 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.

ORGANIZE AND PLAN 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: t  24 h; t1  10 h; vx1  9.5 m s.
SOLVE 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 x1 during the 10-hour period when the dogsled is moving.
x1  v1t1   9.5 m s  36,000 s   342,000 m
This gives us velocity.
x 342,000 m
vx  1 
 4.0 m s
t
86,400 s
REFLECT 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.

ORGANIZE AND PLAN 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: vx1  100.km h; t1  30.0 min; vx 2  60.0 km h; t2  10.0 min; vx3  80.0 km h;
t3  20.0 min.
SOLVE We must first convert the elapsed times to hours to be consistent with the velocity units.

t1  30.0 min 1 h 60 min   0.500 h
t2  10.min 1 h 60 min   0.167 h
t3  20.0 min 1 h 60 min   0.333 h

t  t1  t2  t3  0.500 h  0.167 h  0.333 h  1.000 h
First we find the average velocity, taking into account the negative sign on v x 3

x3  vx 3t3   80 km h  0.333 h   26.6 km

x1  distance traveled1 and x2  distance traveled2


2.16

Chapter 2

Adding the three separate displacements,
x  50.0 km  10.0 km   26.6 km   33.4 km

and the average velocity is

vx 

x 33.4 km

 33 km h
t 1.000 h

Then we find the total distance traveled:

distance traveled1  vx1t1  100.km h  0.500 h   50.0 km
distance traveled 2  vx 2t2   60.0 km h  0.167 h   10.0 km

Since we are calculating distance, not velocity, we use the absolute value of v x 3

distance traveled3  vx3  t3    80.0 km h   0.333 h   26.6 km

distance traveled  distance traveled1  distance traveled2  distance traveled3
distance traveled  50.0 km  10.0 km  26.6 km  86.6 km

So the average speed is

v

distance traveled 86.6 km

 87 km h
t
1.000 h

REFLECT 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.

ORGANIZE AND PLAN 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.
SOLVE (a) First we convert time to hours:
t true  4.75 min  1 h 60 min   0.792 h

Then we calculate our true speed
v

distance traveled true
5.00 mi

 63.2 mi h
t true

0.0792 h

error  vobserved  vactual  60.0 mi h  63.2 mi h  3.2 mi h
(b) If our true speed is 65.0mi h, then
t 

distance traveled true
1.00 mi

 0.154 h  55.4 s
vtrue
65.0 mi h

REFLECT Error is one of many statistical functions that we use as tools to compare data from a limited number of
observations to data from a large population.
42.

ORGANIZE AND PLAN 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.
SOLVE 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,
x1 10,000 m
 1000 s
v1 10.0 m s

t1 

Flying west toward home,

t2 

x2 10,000 m
 1000 s
v2 10.0 m s

t  1000 s  1000 s  2000 s

(b) Flying east,

t1 
t2 

x1 10,000 m
 667 s
v1 15.0 m s

x2 10,000 m

 2000 s
v2 5.0 m s

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.
REFLECT 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.

ORGANIZE AND PLAN 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
t . We’ll use subscript (1) for the cheetah and subscript (2) for the zebra.
Known: x10  0 m; x20  35 m; vx1  30.m s; vx 2  14 m s.
SOLVE This is a constant velocity problem. For the cheetah,

x  x10  vx1t
For the zebra,

x  x20  vx 2t
Setting these equal,

x10  vx1t  x20  vx2t
Solving for t ,

x10  x20   vx 2  vx1  t

t 

 x10  x20   0 m  35 m  2.2 s
 vx 2  vx1  14 m s  30 m s

REFLECT 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 x  vt. This gives the exact answer we obtained above. We’ll discuss this
notion of relative motion in a later chapter.


2.18

44.

Chapter 2

ORGANIZE AND PLAN The parachutist’s average velocity v y while her chute is open is her
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.
Known: yFF  440.m; tFF 10.0 s; vy  3.45 m s; ychute  1350 m.
SOLVE First, the time for the entire trip: We see that the total displacement is the sum of the two displacements
given.

t 

y 440.m  1350 m

 518.8 s
vy
3.45 m s

Now, the time for which the chute is open:

tchute  t  tff  518.8 s  10.0 s  508.8 s
and finally the velocity while the chute is open:
v ychute 

ychute 1350 m

 2.65 m s
tchute 508.8 s

REFLECT 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  v2y0  2gy.
Solving for y, we find that a final velocity of 2.65 m s is how fast we would strike the ground after jumping
from a height of about 0.36 m, or about 14 inches. This seems to be a reasonable velocity.
45.

ORGANIZE AND PLAN We are to find an experimental speed of light and compare it with today’s accepted value
of 3.00  108 m s. 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; t  22 min.
SOLVE

v







2.99  108 km 103 m km
distance traveled

 2.3  108 m s
t
 22 min  60 s min 

REFLECT Today we find Römer’s value to be significantly in error, low by 70,000,000 m s , or 23% . We must
remember that only shortly before his work, scientists were still considering the speed of light to be infinite. Only
about 65 years earlier, Galileo Galilei had developed the modern telescope. Before this, Jupiter’s moons had not
been seen at all. In 1675, timekeeping devices were still not accurate. Sixty years later, in 1735, John Harrison
started his work on the first accurate marine chronometer, which was not completed until a voyage to Jamaica in
1761. Römer’s value was quite a scientific feat for his time.
46.

ORGANIZE AND PLAN 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.

SOLVE We summarize the data read from the graph in a table. Then we use the definition of average
velocity vx  x t to complete the last column of the table.

Motion in One Dimension

Time interval, s

x, m

0.0–2.0
2.0–4.0
4.0–6.0
6.0–8.0

4.17
5.00
12.5
25.7

x0 , m
10.0
4.17
5.00
12.5

x

2.19

vx
–2.92
0.42
3.75

7.08

–5.83
0.83
7.50
14.2

REFLECT 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.

ORGANIZE AND PLAN 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.
SOLVE The table below summarizes the data from Problem 46 that we need to construct our graph.
Time interval, s

Midpoint values, s

vx

0.0-2.0

1.0
3.0
5.0
7.0

2.8
0.2

3.8
6.2

2.0-4.0
4.0-6.0
6.0-8.0

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

REFLECT 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.

ORGANIZE AND PLAN Average acceleration is found using a x 

vx  vx 0 .
We’ll read the velocity values from
t  t0

the vertical axis of the graph shown in Figure P2.48 in the text. Final and initial values for time come from the
horizontal axis.
Known:


2.20

Chapter 2

SOLVE 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
graph, we obtain:

t0 , s
0
5
10
15

t, s

5
10
15
20

vx 0 , m s
0
12
18
13

vx , m s
12
18
13
–3

a x , m s2

2.4
1.2
–1.0
–3.2

REFLECT This graph consists of three straight line segments connected by smooth curves. The slope of each line
segment is the acceleration. However, the second and third time intervals include parts of the curves, so the
average acceleration values for those intervals do not correspond to any straight line segment.
49.

ORGANIZE AND PLAN 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
v
ax  x .
t
Known: From t  0 s to t  7.5 s velocity changes from 0 m s to 17.5 m s. From t  8 s to t  12 s velocity is
constant at 17.5 m s. From t  12.5 s to t  20 s velocity decreases from 17.5 m s to 2.5 m s.
SOLVE We’ll indicate the three line segments using the subscripts 1, 2, and 3. The acceleration values for the
three segments are
v
17.5 m s  0 m s
a x1  x 
 2.33 m s2
t
7.5 s
ax 2 

ax 3 

vx 18 m s  18 m s


 0 m s2
t
12 s  8 s

vx 2.50 m s  17.5 m s

 2.67 m s2
t
20 s  12.5 s

REFLECT Acceleration is the slope of the graph of velocity versus time. The graph of instantaneous acceleration
consists of three horizontal line segments, each indicating a constant acceleration during that time interval.
50.

ORGANIZE AND PLAN 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 t  0 s to t  6 s. Then velocity becomes constant from t  8 s to t  12 s.
From t  16 s to t  20 s velocity steadily decreases to a negative value.
SOLVE


Motion in One Dimension

2.21

REFLECT 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.

ORGANIZE AND PLAN 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
v  vx 0
line. We use the equation a x  x
to calculate acceleration.
t
Known: From t  0 s to t  7.5 s velocity changes from 0 m s to 12.5 m s. From t  7.5 s to t  12.5 s velocity is
constant at 17.5 m s. From t  12.5 s to t  20 s velocity decreases from 17.5 m s to 2.5 m s.
SOLVE
(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 and 12.5 s.
(d) For the greatest acceleration,

ax 

vx  vx 0 12.5 m s  0 m s
2

 2.33 m s .
t
7.5 s

For the least acceleration,

ax 

vx  vx 0 2.5 m s  17.5 m s
2

 2.67 m s .
t
20.0 s  12.5 s

REFLECT This motion is like starting from rest in your car, accelerating to some constant speed, realizing that
you should have turned at the corner, then slowing down and backing up toward the corner. Actually turning the
corner is not included in this problem.
52.

The race lasts 10.0 s, so the first half of the race ends at t  5.0 s. We use data read from
v  vx 0 .
the graph in Figure 2.13(b) to find the average acceleration, a x  x
t
Known: t0  0 s; t  5.0 s; vx 11.0 m s.
SOLVE During the time interval t  t0 , velocity increases from vx 0  0 m s to vx 11.0 m s .
ORGANIZE AND PLAN

ax 

v x  v x 0 v x  v x 0 11.0 m s  0 m s


 2.2 m s2
t
t  t0
5.00 s  0 s

REFLECT Average acceleration is just that — an average. It does not tell us how acceleration has varied during a
time interval.
53.

ORGANIZE AND PLAN The stock car’s velocity is related to time by the equation v x  1.4t 2  1.1t. We are asked
to find vx after 4.0 s. This means we have to evaluate the given equation at 4.0 s. For the units to cancel properly
the value 1.4 must have units of m s3 and the value 1.1 must have units of m s2 .
Known: t0  0 s; t  4.0 s.
SOLVE Substituting 4.0 s for t , we get









v x  1.4t 2  1.1t  1.4 m s3  4.0 s   1.1 m s 2  4.0 s   26.8 m s

REFLECT
gives

2

We know the car starts from rest at time t  0 s since substituting this value into the given equation









v x  1.4t 2  1.1t  1.4 m s3  0 s   1.1 m s 2  0 s   0 m s
2

During the time interval of this problem, velocity increases as a quadratic function of time. This situation can’t last.
The engine’s ability to accelerate the car will decrease as engine speed increases past a certain point. Air resistance
also reduces acceleration as the car moves faster.


2.22

54.

Chapter 2

ORGANIZE AND PLAN 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 function vx  1.4t 2  1.1t is valid
between t  0 s and t  4.0 s.
Known: t0  0 s; t  4.0 s.
SOLVE Using the function v x  1.4t 2  1.1t , we find these values for t and vx :
t ,s

vx ,m s

0

1
2
3
4

0
2.5
7.8
15.9
26.8

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

REFLECT This graph is a segment of a parabola, as we expect from seeing the t 2 term in the equation. Velocity
increases more and more rapidly as elapsed time increases.
55.

ORGANIZE AND PLAN 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
SOLVE 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.
REFLECT 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

56.

2.23

ORGANIZE AND PLAN 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:

SOLVE

REFLECT 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.

ORGANIZE AND PLAN 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:

SOLVE

REFLECT 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.

2.24

58.

Chapter 2

ORGANIZE AND PLAN 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
v  vx 0
. We’ll need the velocity and time values from the graph.
equation a x  x
t  t0
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  .
SOLVE In the first case, we use the diastolic values as the initial values and the systolic values as the final values.
ax 

vx  vx 0 0.55 m s  0.60 m s
2

 2.2 m s
t  t0
0.97 s  0.45 s

In the second case, the systolic value is the initial value:

ax 

vx  vx 0 0.60 m s   0.55  m s
2

 3.8 m s
t  t0
1.27 s  0.97 s

REFLECT We can use the same formula for average acceleration even though the acceleration varies
considerably.
59.

ORGANIZE AND PLAN We know our initial velocity vx0  50.km h and 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 x. We need to find both
acceleration and how long it takes to stop. We can use v2x  vx20  2a x x to find the acceleration. Then,

vx  vx 0
to find the stopping time, t .
t
Known: x  40.m; vx 0  50 km h.
SOLVE First, convert the initial velocity to m s
knowing a x , we use a x 

 1000 m  1 h 
50.km/h 

  13.9 m/s
 1 km  3600 s 

(a) Then rearranging
v2x  vx20  2a x x

We get
ax 

vx2  vx20 0 m s  13.9 m s

 2.4 m s 2
2x
2  40.m 

(b) To find the time it takes to stop, we rearrange the definition of acceleration

ax 
t 

vx  vx 0
t

v x  v x 0 0  13.9 m s

 5.8 s
ax
2.4 m s 2

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.
REFLECT 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

60.

ORGANIZE AND PLAN

2.25

To calculate acceleration, we need the change in velocity and the elapsed time. Then we

can use the definition of acceleration, a x 

vx  vx 0
.
t

Known: (a) vx 0  0 m s; vx  90 km h; t  3.0 s. (b) vx 0  0 m s; vx 10 m s; t  2.0 s.
SOLVE (a) First we convert the cheetah’s speed to m/s.

 90

 1000 m  3600 s 
km h  

  25 m s
 1 km  1 h 

ax 

vx  vx 0 25 m s  0 m s

 8.3 m s 2
t
3.0 s

Then,

(b) Since the human’s speed is already in m s ,

ax 

vx  vx 0 10 m s  0 m s

 5.0 m s 2
t
2.0 s

REFLECT It may surprise us that a human’s acceleration can be as much as 60% of a cheetah’s acceleration.
However, the cheetah can maintain its acceleration for 50% more time than the human, giving the cheetah a final
velocity of 2.5 times that of the human.
61.

ORGANIZE AND PLAN 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 use v2x  vx20  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
velocity vx at the position of the hole.

Known: vx0  2.52 m s; a x  0.65 m s 2 ; distance to the hole  4.8 m.
SOLVE (a) To find whether the ball makes it to the hole,
v2x  vx20  2a x x

Rearranging,
v x2  v x20  0 m s    2.52 m s 

 4.9 m
2a x
2 0.65 m s 2
2

x 

2





So the ball does reach the hole!
(b) Now, at the hole,

vx2  vx20  2ax x



vx  2 a x x  vx20  2 0.65 m s

2

  4.8 m    2.52 m s 

2

vx  0.33 m s as the ball passes the hole.
REFLECT 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.

ORGANIZE AND PLAN We’re given acceleration and time. We have to find the sled’s final speed. We can use
vx  vx0  ax t. Then we can use v2x  vx20  2a x x to find the distance the sled has traveled.
Known: vx 0  0 m s; ax  21.5 m s 2 ; t  8.75 s.
SOLVE (a) First calculate the final velocity:





vx  vx 0  ax t  0 m s  21.5 m s 2 8.75 s 

vx 188 m s


2.26

Chapter 2

(b) Then we can find the distance traveled.
v2x  vx20  2a x x

Rearranging,
v 2  v x20 188 m s    0 m s 
x  x

 822 m
2a x
2 21.5 m s 2
2



2



REFLECT A velocity of 188 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.

ORGANIZE AND PLAN 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 x1 to ax 2 . We can use the formulas v  v0  ax t
and distance traveled  v0 t  21 a x  t  . We have to be careful because the car slows down to less than its
2

original speed.
Known: t1  6.2s; a x1  1.9 m s 2 ; ax 2  1.2 m s 2 ; v0  13.5 m s; v2  0 m s.
SOLVE (a) The car’s maximum speed occurs at the end of the first time interval, that is, after the maximum time
with positive acceleration:





v1  v0  ax t  13.5 m s  1.9 m s 2  6.2 s   25.3 m s

(b) We use the answer to (a) to find the length of the second time interval:
t 

v2  v1 0 m s  25.3 m s

 21.1 s
ax 2
1.2 m s 2

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:



 1.2 m s  21.1 s

distance traveled1  v0t1 12 a x1  t1   13.5 m s  6.2 s   12 1.9 m s 2  6.2 s   120.1 m
2

2

distance traveled 2  v1t2 12 ax 2  t2    25.3 m s  21.1 s   12
2

2

2

 266.7 m

distance traveled  distance traveled1  distance traveled2  120.1 m  266.7 m  387 m
REFLECT 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.

ORGANIZE AND PLAN

To calculate stopping time from an initial speed and acceleration, we can use the
v  v0
. To find stopping distance, we use distance traveled  21  v0  v  t.
definition of acceleration a x 
t
Known: ax1  ax 2  3.50 m s 2 ; v10  50.km h; v20  1.0  102 km h; v1  v2  0 km h.
SOLVE First, convert the initial velocities to m s.

 1ms 
v10   50.km h  

  13.9 m s
 3.6 km h 
 1ms 
v20  1.0  102 km h 
  27.8 m s
 3.6 km h 





(a) The stopping time for the first car is
t1 

v1  v10 0 m s  13.9 m s

 3.97 s  4.0 s
a x1
3.50 m s 2