OpenStax College Physics
Instructor Solutions Manual
Chapter 2
CHAPTER 2: KINEMATICS
2.1 DISPLACEMENT
1.
Solution
Find the following for path A in Figure 2.59: (a) The distance traveled. (b) The
magnitude of the displacement from start to finish. (c) The displacement from start to
finish.
(a) 7 m
(b) 7 m
(c) x  7 m  0 m   7 m
2.
Solution
Find the following for path B in Figure 2.59: (a) The distance traveled. (b) The
magnitude of the displacement from start to finish. (c) The displacement from start to
finish.
(a) 5 m
(b) 5 m
(c) x  7 m  12 m   5 m
3.
Solution
Find the following for path C in Figure 2.59: (a) The distance traveled. (b) The
magnitude of the displacement from start to finish. (c) The displacement from start to
finish.
(a) 8 m  2 m  3 m  13 m
(b) 9 m
(c) x  11 m  2 m   9 m
4.
Find the following for path D in Figure 2.59: (a) The distance traveled. (b) The
magnitude of the displacement from start to finish. (c) The displacement from start to
finish.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 2
(a) 6 m  2 m  8 m
(b) 4 m
(c) x  5 m  9 m   4 m
2.3 TIME, VELOCITY, AND SPEED
5.
Solution
(a) Calculate Earth’s average speed relative to the Sun. (b) What is its average
velocity over a period of one year?
Dist. Traveled 2r

Time
t
11
2 (1.50  10 m)
1 day
1 hour



365.25 days
24 hours 3600 s
(a) Avg. Speed of Earth =
 2.99  10 4 m/s  3.0  10 4 m/s
(b) After one year, Earth has returned to its original position with respect to the Sun.
Thus, v  0 m/s .
6.
A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from
the center of rotation. (a) Calculate the average speed of the blade tip in the
helicopter’s frame of reference. (b) What is its average velocity over one revolution?
Solution
(a) Average speed of blade tip 
distance traveled 2r 2 (5.00 m)


 52.4 m/s
time elapsed
t
60 s/100 rev
(b) v  0 m/s . After one revolution, the blade returns to its original position with total
displacement of 0 m.
7.
Solution
The North American and European continents are moving apart at a rate of about 3
cm/y. At this rate how long will it take them to drift 500 km farther apart than they
are at present?
t 
500 km 5.00  105 m 100 cm


 1.67  107 years  2  107 y
rate
3 cm/y
1m
OpenStax College Physics
Instructor Solutions Manual
Chapter 2
8.
Land west of the San Andreas fault in southern California is moving at an average
velocity of about 6 cm/y northwest relative to land east of the fault. Los Angeles is
west of the fault and may thus someday be at the same latitude as San Francisco,
which is east of the fault. How far in the future will this occur if the displacement to
be made is 590 km northwest, assuming the motion remains constant?
Solution
5.90  105 m 100 cm
t 

 9.83  106 years  1  107 years  10 million years
6 cm/year
1m
9.
On May 26, 1934, a streamlined, stainless steel diesel train called the Zephyr set the
world’s nonstop long-distance speed record for trains. Its run from Denver to Chicago
took 13 hours, 4 minutes, 58 seconds, and was witnessed by more than a million
people along the route. The total distance traveled was 1633.8 km. What was its
average speed in km/h and m/s?
Solution
3600 s  
60 s

t  13 h 
   4 min 
1h  
1 min


  58 s

1h
 4.7098  10 4 s  4.7098  10 4 s 
 13.083 hours
3600 s
average speed in km/h 
average speed in m/s 
10.
Solution
distance traveled 1633.8 km

 124.88 km/h
time elapsed
13.0828 h
1.6338  10 6 m
 34.689 m/s
4.7098  10 4 s
Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is
increasing in radius at a rate of approximately 4 cm/year. Assuming this to be a
constant rate, how many years will pass before the radius of the Moon’s orbit
increases by 3.84 106 m (1%)?
t 
3.84  10 6 m 100 cm

 9.60  10 7 years  1 108 years  100 million years
4 cm/year
1m
OpenStax College Physics
11.
Instructor Solutions Manual
Chapter 2
A student drove to the university from her home and noted that the odometer
reading of her car increased by 12.0 km. The trip took 18.0 min. (a) What was her
average speed? (b) If the straight-line distance from her home to the university is
10.3 km in a direction 25.0 south of east, what was her average velocity? (c) If she
returned home by the same path 7 h 30 min after she left, what were her average
speed and velocity for the entire trip?
Solution
(a) average speed 
(b) v 
12.0 km 60 min

 40.0 km/h
18.0 min
1h
x 10.3 km 60 min


 34.3 km/h, 25 S of E.
t 18.0 min
1h
(c) average speed  2 
12.0 km
 3.20 km/h
7.5 h
v  0 , since the total displacement was 0.
12.
The speed of propagation of the action potential (an electrical signal) in a nerve cell
depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell
connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is
18 m/s, how long does it take for the nerve signal to travel this distance?
Solution
time elapsed 
13.
Conversations with astronauts on the lunar surface were characterized by a kind of
echo in which the earthbound person’s voice was so loud in the astronaut’s space
helmet that it was picked up by the astronaut’s microphone and transmitted back to
Earth. It is reasonable to assume that the echo time equals the time necessary for the
radio wave to travel from the Earth to the Moon and back (that is, neglecting any
time delays in the electronic equipment). Calculate the distance from Earth to the
Moon given that the echo time was 2.56 s and that radio waves travel at the speed of
light (3.00 108 m/s) .
distance traveled
1.1 m

 0.061 seconds.
average speed
18 m/s
OpenStax College Physics
Solution
Chapter 2
dist. traveled = avg. speed  elapsed time = 3.00  108 m/s  2.56 sec  767,000 km
E  M dist. 
14.
Instructor Solutions Manual
dist. traveled 767,000 km

 384,000 km
2
2
A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is
then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and
runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity (a) for
each of the three intervals and (b) for the entire motion.
Solution
(a) For each interval, v 
x
.
t
x 15.0 m

 6.00 m/s
t 2.50 s
x  3.00 m
v2  
  1.71 m/s
t
1.75 s
x 21.0 m
v3  
 4.04 m/s
t
5.20 s
v1 
(b) For the full interval, we need displacement.
displaceme nt 15.0 m  3.00 m  21.0 m 33 m
vavg 


 3.49 m/ s
tot. time
2.50 s  1.75 s  5.20 s
9.45 s
(Note: this is different from the average of the 3 interval velocities, which is 2.77
m/s.)
15.
The planetary model of the atom pictures electrons orbiting the atomic nucleus much
as planets orbit the Sun. In this model you can view hydrogen, the simplest atom, as
having a single electron in a circular orbit 1.06 10 10 m in diameter. (a) If the
average speed of the electron in this orbit is known to be 2.20  106 m/s , calculate the
number of revolutions per second it makes about the nucleus. (b) What is the

electron’s average velocity?
OpenStax College Physics
Solution
(a)
Instructor Solutions Manual
Chapter 2
distance
 2.20  10 6 m/s
time
distance/r evolution  2r/revolutio n  d / revolution
average speed 
  (1.06  10 10 m)/revolut ion  3.33  10 10 m/revoluti on
revolution s
average speed
2.20  10 6 m/s


1s
distance/r evolution
3.33  10 10 m/revoluti on
 6.61  1015 rev/s
(b) v  0 m/s , since there is no net displacement per revolution.
2.4 ACCELERATION
16.
A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its
acceleration?
Solution
a
17.
Professional Application Dr. John Paul Stapp was U.S. Air Force officer who studied
the effects of extreme deceleration on the human body. On December 10, 1954,
Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015
km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his (a)
acceleration and (b) deceleration. Express each in multiples of g (9.80 m/s 2 ) by
vf  v0 30.0 m/s  0 m/s

 4.29 m/s 2
t
7s
taking its ratio to the acceleration of gravity.
Solution
(a) a  v  v0  282 m/s  0 m/s  56.4 m/s 2
t
5.00 s
2
a 56.4 m/s

 5.76  a  5.76 g
g
9.8 m/s 2
(b) a'  v'  v'0  0 m/s  282 m/s   201 m/s 2
t
1.40 s
2
a  201 m/s

 20.55  a   20.6 g
g
9.8 m/s 2
OpenStax College Physics
18.
(a) t 
v  v0 2.00 m/s  0 m/s

 1.43 s
a
1.40 m/s 2
(b) a 
v v 0 0 m/s  2.00 m/s

  2.50 m/s 2
t
0.800 s
Assume that an intercontinental ballistic missile goes from rest to a suborbital speed
of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average
acceleration in m/s 2 and in multiples of g (9.80 m/s 2 ) ?
Solution
Chapter 2
A commuter backs her car out of her garage with an acceleration of 1.40 m/s 2 . (a)
How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a
stop in 0.800 s, what is her deceleration?
Solution
19.
Instructor Solutions Manual
v  v0 6.50 10 3 m/s  0 m/s

 108 m/s 2
t
60.0 s
a 108 m/s 2

 a  11.1 g
g 9.8 m/s 2
a
OpenStax College Physics
Instructor Solutions Manual
Chapter 2
2.5 MOTION EQUATIONS FOR CONSTANT ACCELERATION IN ONE
DIMENSION
20.
Solution
An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s 2 . (a) What is
her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.
(a) v  v0  at  0 m/s  (4.50 m/s 2 )( 2.40 s)  10.8 m/s
(b) Assuming the acceleration is constant, we know x  (½)at 2  2.25t 2 . We can
create a graph by plugging in a few different t-values, say t = 1, 2, 3, 4, 5:
Time (s)
0
1
2
3
4
5
21.
Position
(m)
0
2.25
9
20.25
36
56.25
A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is
2.10  10 4 m/s 2 , and 1.85 ms (1 ms  103 s) elapses from the time the ball first
touches the mitt until it stops, what was the initial velocity of the ball?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 2
v0  v  at  0 m/s  (2.10  10 4 m/s 2 )(1.85  10 3 s)  38.9 m/s (about 87 miles per
hour)
22.
Solution
23.
A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an
average rate of 6.20  105 m/s 2 for 8.10  104 s . What is its muzzle velocity (that is,
its final velocity)?
v  v0  at  0 m/s  (6.20  105 m/s 2 )(8.10  104 s)  502 m/s
(a) A light-rail commuter train accelerates at a rate of 1.35 m/s 2 . How long does it
take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train
ordinarily decelerates at a rate of 1.65 m/s 2 . How long does it take to come to a stop
from its top speed? (c) In emergencies the train can decelerate more rapidly, coming
to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in m/s 2 ?
Solution
24.
(a) t 
v  v0 (80.0 km/h  0 km/h)
1h
1000 m



 16.5 s
2
a
1.35 m/s
3600 s 1 km
(b) t 
v  v0 0 km/h  80.0 km/h
1h
1000 m



 13.5 s
2
a
 1.65 m/s
3600 s 1 km
(c) a 
v  v0 0 km/h  80.0 km/h
1h
1000 m



  2.68 m/s 2
t
8.30 s
3600 s 1 km
While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s 2 for 12.0
s. (a) Draw a sketch of the situation. (b) List the knowns in this problem. (c) How far
does the car travel in those 12.0 s? To solve this part, first identify the unknown, and
then discuss how you chose the appropriate equation to solve for it. After choosing
the equation, show your steps in solving for the unknown, check your units, and
discuss whether the answer is reasonable. (d) What is the car’s final velocity? Solve
for this unknown in the same manner as in part (c), showing all steps explicitly.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 2
(a)
(b) Knowns: a  2.40 m/s 2 ; t  12.0 s; v0  0 m/s; x0  0 m
1 2
at because the
2
only unknown it includes is x , which is what we want to solve for. First we
substitute the knowns into the equation and then we solve for x .
(c) x is the unknown. We can use the equation x  x0  v0 t 
x  x0  v0 t 
1 2
1
at  0 m  (0 m/s)(12.0 s)  (2.40 m/s 2 )(12.0 s) 2  173 m .
2
2
(d) v is the unknown. We need an equation that relates our knowns to the
unknown we want. We can use the equation v  v0  at because in it all of the
variables other than v are known. We substitute the known values into the
equation and then solve for v: v  v0  at  0 m/s  (2.40 m/s)(12.0 s)  28.8 m/s .
25.
Solution
At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of
2.00 m/s 2 . (a) How far does she travel in the next 5.00 s? (b) What is her final
velocity? (c) Evaluate the result. Does it make sense?
1
1
(a) x  x0  v0t  at 2  0 m  (9.00 m/s)(5.00 s)  (2.00 m/s 2 )(5.00 s) 2  20.0 m
2
2
(b) v  v0  at  9.00 m/s  (2.00 m/s 2 )(5.00 s)   1.00 m/s
(c) This result does not really make sense. If the runner starts at 9.00 m/s and
decelerates at 2.00 m/s 2 , then she will have stopped after 4.50 s. If she
continues to decelerate, she will be running backwards.
OpenStax College Physics
26.
Solution
Instructor Solutions Manual
Chapter 2
Professional Application Blood is accelerated from rest to 30.0 cm/s in a distance of
1.80 cm by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List
the knowns in this problem. (c) How long does the acceleration take? To solve this
part, first identify the unknown, and then discuss how you chose the appropriate
equation to solve for it. After choosing the equation, show your steps in solving for
the unknown, checking your units. (d) Is the answer reasonable when compared with
the time for a heartbeat?
(a)
(b) Knowns: v0  0 m/s; v  30.0 cm/s; x  x0  1.80 cm
(c) t 
x  x0
( x  x0 )
2(0.0180 m)


 0.120 s
vavg
(v0  v) / 2 (0 m/s)  (0.300 m/s)
This is the best equation to use because it uses our 3 knowns to determine our
unknown.
(d) Yes, the answer seems reasonable. An entire heartbeat cycle takes about one
second. The time for acceleration of blood out of the ventricle is only a fraction
of the entire cycle.
27.
In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to
40.0 m/s in the same direction. If this shot takes 3.33  102 s , calculate the distance
over which the puck accelerates.
Solution
x  x0  vavg t 
28.
A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90
s. (a) What is its average acceleration? (b) How far does it travel in that time?
1
1
(v0  v)t  (8.00 m/s  40.0 m/s)(3.33  10  2 s)  0.799 m
2
2
OpenStax College Physics
Solution
(a) a 
Instructor Solutions Manual
Chapter 2
v 26.8 m/s  0 m/s

 6.87 m/s 2
t
3.90 s
1
1
(b) x  x0  (v0  v)t  (0 m/s  26.8 m/s)( 3.90 s)  52.3 m
2
2
29.
Freight trains can produce only relatively small accelerations and decelerations. (a)
What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s 2
for 8.00 min, starting with an initial velocity of 4.00 m/s? (b) If the train can slow
down at a rate of 0.550 m/s 2 , how long will it take to come to a stop from this
velocity? (c) How far will it travel in each case?
Solution
(a) v  v0  at  4.00 m/s  (0.0500 m/s 2 )(8.00 min 
(b) t 
60 s
)  28.0 m/s
1 min
v  v0 0 m/s  28.0 m/s

 50.9 s
a
 0.550 m/s 2
(c) For part (a),
x  x0  v0t 
1 2
1
at  (4.0 m/s)(480 s)  (0.0500 m/s 2 )(480 s) 2  7.68  103 m
2
2
For part (b), x  x0 
30.
A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of
0.250 m. (a) How long did the acceleration last? (b) Calculate the acceleration.
Solution
31.
v 2  v02 (0 m/s) 2  (28.0 m/s) 2

 713 m
2a
2(0.550 m/s 2 )
(a) t 
2( x  x0 )
2(0.250 m)

 7.69  10 3 s
v  v0
65.0 m/s  0 m/s
(b) a 
v 2  v02
(65.0 m/s) 2  (0 m/s) 2

 8.45  103 m/s 2
2( x  x0 )
2(0.250 m)
A swan on a lake gets airborne by flapping its wings and running on top of the
water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates
from rest at an average rate of 0.350 m/s 2 , how far will it travel before becoming
airborne? (b) How long does this take?
OpenStax College Physics
Solution
32.
Instructor Solutions Manual
Chapter 2
v 2  v02 (6.00 m/s) 2  (0 m/s) 2
(a) x  x0 

 51.4 m
2a
2(0.350 m/s 2 )
v  v0 6.00 m/s  0 m/s
(b) t 

 17.1 s
a
0.350 m/s 2
Professional Application A woodpecker’s brain is specially protected from large
decelerations by tendon-like attachments inside the skull. While pecking on a tree,
the woodpecker’s head comes to a stop from an initial velocity of 0.600 m/s in a
distance of only 2.00 mm. (a) Find the acceleration in m/s 2 and in multiples of
g g  9.80 m/s 2  . (b) Calculate the stopping time. (c) The tendons cradling the brain
stretch, making its stopping distance 4.50 mm (greater than the head and, hence,
less deceleration of the brain). What is the brain’s deceleration, expressed in
multiples of g ?
Solution
2
2
2
2
(a) a  v  v0  (0 m/s)  (0.600 m/s)   90.0 m/s 2
2( x  x 0 )
2(2.00  10 3 m)
a
g

90.0 m/s 2
 9.18  a  9.18 g .
9.80 m/s 2
(b) x  x0 
(c) a 
a
g
33.

2( x  x0 )
2(2.00  10 3 m)
1

 6.67  103 s
(v0  v)t , so that t 
v0  v
(0.600 m/s)  (0 m/s)
2
v 2  v02
(0 m/s) 2  (0.600 m/s) 2

  40.0 m/s 2
2( x  x 0 )
2(4.50  10 -3 m)
40.0 m/s 2
 4.08  a  4.08 g
9.80 m/s 2
An unwary football player collides with a padded goalpost while running at a
velocity of 7.50 m/s and comes to a full stop after compressing the padding and his
body 0.350 m. (a) What is his deceleration? (b) How long does the collision last?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 2
v 2  v02
(0 m/s) 2  (7.50 m/s) 2

  80.4 m/s 2
2( x  x0 )
2(0.350 m)
2( x  x0 )
2(0.350 m)
(b) t 

 9.33  10 2 s
v  v0
0 m/s  7.50 m/s
(a) a 
34.
Solution
In World War II, there were several reported cases of airmen who jumped from their
flaming airplanes with no parachute to escape certain death. Some fell about 20,000
feet (6000 m), and some of them survived, with few life-threatening injuries. For
these lucky pilots, the tree branches and snow drifts on the ground allowed their
deceleration to be relatively small. If we assume that a pilot’s speed upon impact
was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and
snow stopped him over a distance of 3.0 m.
Knowns: x  3 m; v  0 m/s; v0  54 m/s
We want a , so we can use this equation:
v 2  v02 0 m/s  (54 m/s) 2
2
2
v  v0  2ax  a 

  486 m/s 2 . Negative acceleration
2x
2(3 m)
means that the pilot was decelerating at a rate of 486 m/s every second.
35.
Solution
Consider a grey squirrel falling out of a tree to the ground. (a) If we ignore air
resistance in this case (only for the sake of this problem), determine a squirrel’s
velocity just before hitting the ground, assuming it fell from a height of 3.0 m. (b) If
the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its
deceleration with that of the airman in the previous problem.
(a) v 2  v02  2ax  v  2ax  2(9.8 m/s 2 )( 3.0 m)  7.7 m/s
v 2  v02 (0 m/s) 2  (7.7 m/s) 2
(b) a 

  1.5 103 m/s 2
2x
2(0.02 m)
This is ~ 3 times the deceleration of the pilots, who were falling from thousands
of meters high!
OpenStax College Physics
36.
Solution
Instructor Solutions Manual
Chapter 2
An express train passes through a station. It enters with an initial velocity of 22.0
m/s and decelerates at a rate of 0.150 m/s 2 as it goes through. The station is 210 m
long. (a) How long is the nose of the train in the station? (b) How fast is it going
when the nose leaves the station? (c) If the train is 130 m long, when does the end of
the train leave the station? (d) What is the velocity of the end of the train as it
leaves?
1
(a) x  x0  v0t  at 2 , and rearranging
2
t

 v0  v02  2a ( x  x0 )
a
 22.0 m/s  (22.0 m/s) 2  2(0.150 m/s 2 )(210 m)
 0.150 m/s 2
 9.88 s
(b) v 2  v02  2a( x  x0 ) , and rearranging
v  v02  2a( x  x0 )  (22.0 m/s) 2  2(0.150 m/s 2 )(210 m)  20.6 m/s
(c) Here, we use the fact that the train will leave the station when the nose is
210 m + 130 m = 340 m away from the beginning of the station.
1
x  x0  v0t  at 2
2
t

 v 0  v02  2a ( x  x0 )
a
 22.0 m/s  (22.0 m/s) 2  2(0.150 m/s 2 )(340 m)
 0.150 m/s 2
2
2
(d) v  v0  2a( x  x0 )
v  (22.0 m/s) 2  2(0.150 m/s 2 )(340 m)  19.5 m/s
 16.4 s
OpenStax College Physics
37.
Instructor Solutions Manual
Chapter 2
Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less
time than given in Example 2.10 and Example 2.11. (a) Calculate the average
acceleration for such a dragster. (b) Find the final velocity of this dragster starting
from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without
using any information on time. (c) Why is the final velocity greater than that used to
find the average acceleration? Hint: Consider whether the assumption of constant
acceleration is valid for a dragster. If not, discuss whether the acceleration would be
greater at the beginning or end of the run and what effect that would have on the
final velocity.
Solution
(a) a 
v  v0 145 m/s  0 m/s

 32.6 m/s 2
t
4.45 s
(b) v 2  v02  2a( x  x0 ) , and rearranging
v  (0 m/s) 2  2(32.6 m/s 2 )(402 m)  162 m/s
(c) v  vmax because the assumption of constant acceleration is not valid for a
dragster. A dragster changes gears, and would have a greater acceleration in
first gear than second gear than third gear, etc. The acceleration would be
greatest at the beginning, so it would not be accelerating at 32.6 m/s 2 during the
last few meters, but substantially less, and the final velocity would be less than
162 m/s .
38.
Solution
A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial
velocity of 11.5 m/s and accelerates at the rate of 0.500 m/s 2 for 7.00 s. (a) What is
his final velocity? (b) The racer continues at this velocity to the finish line. If he was
300 m from the finish line when he started to accelerate, how much time did he
save? (c) One other racer was 5.00 m ahead when the winner started to accelerate,
but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How
far ahead of him (in meters and in seconds) did the winner finish?
(a) v  v0  at  11.5 m/s  (0.500 m/s 2 )(7.00 s)  15.0 m/s
(b) Let t const be the time it takes the rider to reach the finish line without
x
300 m
accelerating: t const 

 26.09 s  26.1s
v0 11.5 m/s
Now let d be the distance traveled during the 7 seconds of acceleration.
We know t  7.00 s so
OpenStax College Physics
d  v0 t 
Instructor Solutions Manual
Chapter 2
1 2
at  (11.5 m/s)(7.00 s)  0.5(0.500 m/s 2 )(7.00 s) 2  92.75 m  92.8 m
2
Let t be the time it will take the rider at the constant final velocity to complete
x  d 300 m  92.75 m
the race: t  

 13.82 s  13.8 s .
v
15.0 m/s
So the total time T it will take the accelerating rider to reach the finish line is
T  t  t  7 s  13.82 s  20.82 s  20.8 s.
Finally, let T* be the time saved. So T*  26.09 s  20.82 s  5.27 s .
(c) Let t 2 be the time it takes for rider 2 to reach the finish line.
t2 
x
295 m

 25.0 s; time difference  t 2  T  25.0 s  20.817 s  4.2 s
v0 11.8 m/s
Therefore he finishes 4.2 s after the winner.
When the other racer reaches the finish line, the winner has been traveling at
15 m/s for 4.2 seconds, so the other racer finishes x  (4.2 m/s)(15 m/s) = 63 m
behind the other racer.
39.
Solution
In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle,
on the Bonneville Salt Flats in Utah, of 183.58 mi/h. The one-way course was 5.00 mi
long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h
from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached
his maximum speed, how long did it take Burt to complete the course?
There are two parts to the race: an acceleration part and a constant speed part.
First, we need to determine how long (both in distance and time) it takes the
60 mph
motorcycle to finish accelerating. During acceleration, a 
.
4s
 4s 
  12.2 s
v max  183 mph  at1  t1  183 mph 
 60 mph 
 183 mph  0 mph 
 1h 
x1  vavg t1  
(12.2 s)  (91.5 mph) 
(12.2 s)  0.31 mi
2


 3600 s 
At constant velocity, x2  5 mi  0.31 mi  4.7 mi .
Now, we complete the calculation by determining how much time is spent on the
course at max speed.
OpenStax College Physics
t2 
Instructor Solutions Manual
Chapter 2
x2
4.7 mi
3600 s

 0.026 h
 92 s, so
v max 183 mph
1h
t total  t1  t 2  12.2 s  92 s  104 s
40.
(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in
Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of
9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed,
and maintained that speed for the rest of the race, calculate his maximum speed
and his acceleration. (b) During the same Olympics, Bolt also set the world record in
the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m
dash, what was his maximum speed for this race?
Solution
(a) There are two parts to the race and must be treated separately since
acceleration is not uniform over the race. We will divide the race into x1 (while
accelerating) and x2 (with constant speed), where x1  x2  100 m .
When the speed is constant, t  6.69 s , so x2  vt  (6.69 s)v  100 m  x1 .
1
1
When accelerating, t  3.00 s , so x1  at 2  vt  (1.50 s)v . Plugging x1 into
2
2
the previous equation, we get his maximum speed:
100 m  1.50 s v  6.69 s v  v 
100 m
 12.2 m/s
6.69 s  1.50 s
Therefore, his acceleration was a 
v 12.2 m/s

 4.07 m/s 2
t
3.00 s
(b) Similar to part (a), we can plug in the different values for time and total distance:
200 m  1.50 s v  16.30 s v  v 
2.7 FALLING OBJECTS
200 m
 11.2 m/s
16.30 s  1.50 s
OpenStax College Physics
41.
Instructor Solutions Manual
Chapter 2
Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d)
2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of
release to be y0  0 .
Solution
Knowns: a  accelerati on due to gravity  g   9.8 m/s 2 ; y0  0 m; v0   15.0 m/s
1
To find displacement we use y  y0  v0t  at 2 , and to find velocity we use
2
v  v0  at .
(a) y1  y0  v0t1 
1 2
at1
2
1
(9.8 m/s 2 )(0.500 s) 2  6.28 m
2
v1  v0  at1  (15.0 m/s)  (9.8 m/s 2 )(0.500 s)  10.1 m/s
 0 m  (15.0 m/s)(0.500 s) 
(b) y 2  y0  v0 t 2 
1 2
at 2
2
1
(9.8 m/s 2 )(1.00 s) 2  10.1 m
2
v2  v0  at 2  (15.0 m/s)  (9.8 m/s 2 )(1.00 s)  5.20 m/s
 0 m  (15.0 m/s)(1.00 s) 
(c) y3  y0  v0t3 
1 2
at3
2
1
(9.8 m/s 2 )(1.50 s) 2  11.5 m
2
v3  v0  at3  (15.0 m/s)  (9.8 m/s 2 )(1.50 s)  0.300 m/s
 0 m  (15.0 m/s)(1.50 s) 
The ball is almost at the top.
(d) y 4  y0  v0 t 4 
1 2
at 4
2
1
(9.8 m/s 2 )(2.00 s) 2  10.4 m
2
v4  v0  at 4  (15.0 m/s)  (9.8 m/s 2 )(2.00 s)   4.60 m/s
 0 m  (15.0 m/s)(2.00 s) 
The ball has begun to drop.
OpenStax College Physics
42.
Instructor Solutions Manual
Chapter 2
Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d)
2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0
m/s from the Verrazano Narrows Bridge in New York City. The roadway of this
bridge is 70.0 m above the water.
Solution
Knowns: a   9.8 m/s 2 ; v0   14.0 m/s; y 0  0 m
1
To find displacement we use y  y0  v0t  at 2 , and to find velocity we use
2
v  v0  at .
(a) y1  y0  v0t1 
1 2
at1
2
1
(9.8 m/s 2 )(0.500 s) 2   8.23 m
2
v1  v0  at1  (14.0 m/s)  (9.8 m/s 2 )(0.500 s)   18.9 m/s
 0 m  (14.0 m/s)(0.500 s) 
(b) y 2  y0  v0t 2 
1 2
at 2
2
1
(9.8 m/s 2 )(1.00 s) 2   18.9 m
2
v2  v0  at 2  (14.0 m/s)  (9.8 m/s 2 )(1.00 s)   23.8 m/s
 0 m  (14.0 m/s)(1.00 s) 
(c) y3  y0  v0t3 
1 2
at3
2
1
(9.8 m/s 2 )(1.50 s) 2   32.0 m
2
v3  v0  at3  (14.0 m/s)  (9.8 m/s 2 )(1.50 s)   28.7 m/s
 0 m  (14.0 m/s)(1.50 s) 
(d) y 4  y0  v0 t 4 
1 2
at 4
2
1
(9.8 m/s 2 )(2.00 s) 2   47.6 m
2
v4  v0  at 4  (14.0 m/s)  (9.8 m/s 2 )(2.00 s)   33.6 m/s
 0 m  (14.0 m/s)(2.00 s) 
(e) y5  y0  v0t5 
1 2
at5
2
1
(9.8 m/s 2 )(2.50 s) 2   65.6 m
2
v5  v0  at5  (14.0 m/s)  (9.8 m/s 2 )(2.50 s)   38.5 m/s
 0 m  (14.0 m/s)(2.50 s) 
OpenStax College Physics
Instructor Solutions Manual
Chapter 2
43.
A basketball referee tosses the ball straight up for the starting tip-off. At what
velocity must a basketball player leave the ground to rise 1.25 m above the floor in
an attempt to get the ball?
Solution
v 2  v02  2a( y  y0 )  v02  v 2  2a( y  y0 )
v0  v 2  2a( y  y0 )  (0 m/s) 2  2(9.80 m/s 2 )(1.25 m)  4.95 m/s
44.
A rescue helicopter is hovering over a person whose boat has sunk. One of the
rescuers throws a life preserver straight down to the victim with an initial velocity of
1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in
this problem. (b) How high above the water was the preserver released? Note that
the downdraft of the helicopter reduces the effects of air resistance on the falling life
preserver, so that an acceleration equal to that of gravity is reasonable.
Solution
(a) Knowns: a   9.80 m/s 2 ; v0   1.40 m/s; t  1.8 s; y  0 m
1
(b) y0   y  v0t  at 2
2
 0 m  (1.40 m/s)(1.8 s)  0.5(9.80 m/s 2 )(1.8 s) 2  18 m
45.
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0
m/s. (a) List the knowns in this problem. (b) How high does his body rise above the
water? To solve this part, first note that the final velocity is now a known and
identify its value. Then identify the unknown, and discuss how you chose the
appropriate equation to solve for it. After choosing the equation, show your steps in
solving for the unknown, checking units, and discuss whether the answer is
reasonable. (c) How long is the dolphin in the air? Neglect any effects due to his size
or orientation.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 2
(a) Knowns: a   9.80 m/s 2 ; v0  13.0 m/s; y0  0 m
(b) At the highest point in the jump, v  0 m/s . We can use the equation
v 2  v02
because the only unknown it includes is y , which is what we
y  y0 
2a
want to solve for. First we substitute the knowns into the equation and then we
solve for y .
v 2  v02
v 2  v02
(0 m/s) 2  (13.0 m/s) 2
y  y0 
y
 y0 
 0 m  8.62m .
2a
2a
2(9.80 m/s 2 )
Dolphins measure about 2 meters long and can jump several times their length
out of the water, so this is a reasonable result.
(c) If t is the time for the dolphin to reach its peak height, then 2t is the time the
dolphin is out of the water.
v  v0  at  t 
46.
Solution
v  v0 0 m/s  13.0 m/s

 1.3265 s, and 2t  2.65 s
a
 9.8 m/s 2
A swimmer bounces straight up from a diving board and falls feet first into a pool.
She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool.
(a) How long are her feet in the air? (b) What is her highest point above the board?
(c) What is her velocity when her feet hit the water?
(a) Knowns: y0  1.80 m, y  0 m, a  9.80 m/s 2 , v0  4.00 m/s , so we use the
1
equation y  y 0  v0 t  at 2 . Rearranging,
2
t
 v0  v02  2a ( y0  y )
a
 4.00 m/s  (4.00 m/s) 2  2(9.80 m/s 2 )(1.80 m)

 1.14 s
 9.80 m/s 2
v 2  v02 (0 m/s) 2  (4.00 m/s) 2

 0.816 m
(b) y  y0 
2a
2(9.80 m/s 2 )
(c) v 2  v02  2a( y  y0 ) and rearranging
OpenStax College Physics
Instructor Solutions Manual
Chapter 2
v   v02  2a( y  y0 )   (4.00 m/s) 2  2(9.80 m/s 2 )( 1.80 m)
  51.28 m 2 /s 2   7.16 m/s
Since the diver must be moving in the negative direction, v   7.16 m/s.
47.
(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is
thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long would
it take to reach the ground if it is thrown straight down with the same speed?
Solution
(a) Knowns: t  2.35 s; y  0 m; v0   8.00 m/s; a   9.8 m/s 2
Since we know t , y , v0 , and a and want to find y 0 , we can use the equation
1
y  y 0  v 0 t  at 2 .
2
y  (0 m)  (8.00 m/s)(2.35 s) 

1
(9.80 m/s 2 )(2.35 s) 2  8.26 m , so the cliff is
2
8.26 m high.
(b) Knowns: y  0 m; y0  8.26 m; v0   8.00 m/s; a   9.80 m/s 2
Now we know y , y 0 , v0 , and a and want to find t , so we use the equation
1
y  y 0  v 0 t  at 2 again. Rearranging,
2

t
 v0  v02  4(0.5a )( y 0  y )
2(0.5a )
 (8.00 m/s)  (8.00 m/s) 2  2(9.80 m/s 2 )(8.26 m  0 m)
(9.80 m/s 2 )
8.00 m/s  15.03 m/s

 9.80 m/s 2
t  0.717 s or  2.35 s  t  0.717 s
t
OpenStax College Physics
48.
Solution
Instructor Solutions Manual
Chapter 2
A very strong, but inept, shot putter puts the shot straight up vertically with an
initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot
was released at a height of 2.20 m, and he is 1.80 m tall?
Knowns: y0  2.20 m; y  1.80 m; v0  11.0 m/s; a  9.80 m/s 2
t

 v0  v02  2a ( x  x0 )
a
 11.0 m/s  (11.0 m/s) 2  2(9.80 m/s 2 )(0.40 m)
 2.28 s
 9.80 m/s 2
49.
You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree
branch on the way up at a height of 7.00 m. How much additional time will pass
before the ball passes the tree branch on the way back down?
Solution
Knowns: a   9.80 m/s 2 ; v0  15.0 m/s; y  7.00 m
y  v0 t 
t

1 2
at 
2
 v0  v02  2ay
a
 15.0 m/s  (15.0 m/s) 2  2(9.80 m/s 2 )(7.00 m)
 9.80 m/s
2

 15.0 m/s  9.37 m/s
 9.80 m/s 2
So t1  0.58 s and t2  2.49 s , so the total time between passing the branch is 1.91 s .
50.
A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed
when it leaves the ground. (b) How long is it in the air?
Solution
(a) v 2  v02  2ay  v0  v 2  2ay  0  2(9.80 m/s 2 )(2.50 m)  7.00 m/s
2y
2(2.50 m)
 v  v0 
(b) y  vavgt  

 0.714 s
t  t 
v  v0 7.00 m/s
 2 
OpenStax College Physics
Instructor Solutions Manual
Chapter 2
51.
Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker
hears a rock break loose from a height of 105 m. He can’t see the rock right away
but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see
it? (b) How much time does he have to move before the rock hits his head?
Solution
(a) y  v0t 
1 2 1
at  (9.80 m/s 2 )(1.50 s) 2  11.0 m
2
2
So the rock falls 11.0 m in the 1.50 s before the hiker sees the rock. When he
finally sees the rock, it is 94.0 m above his head.
1 2
2y
2(105 m)
at  t 

 4.63 s , so the rock will take 4.63 s to fall
2
a
9.80 m/s 2
the full distance. Thus, the hiker will have 4.63 s  1.50 s  3.13 s to move out of
(b) y 
the way before the rock strikes the hiker's location, ignoring the height of
person.
52.
Solution
An object is dropped from a height of 75.0 m above ground level. (a) Determine the
distance traveled during the first second. (b) Determine the final velocity at which
the object hits the ground. (c) Determine the distance traveled during the last
second of motion before hitting the ground.
(a) x  v0t 
1 2 1
at  (9.80 m/s 2 )(1.00 s) 2  4.90 m
2
2
(b) v 2  v02  2ax  v  2(9.80 m/s 2 )(75.0 m)  38.3 m/s
(c) First find the total time to fall: y 
1 2
at  t 
2
2y
2(75.0 m)

 3.91 s
a
9.80 m/s 2
Next, we find the distance traveled up to the last 1 second of flight:
1
1
y  at 2  (9.80 m/s 2 )(2.91 s) 2  41.5 m , so the distance traveled in the last
2
2
second will be the difference: x  75.0 m  41.5 m  33.5 m .
OpenStax College Physics
Instructor Solutions Manual
Chapter 2
53.
There is a 250-m-high cliff at Half Dome in Yosemite National Park in California.
Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going
when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long will a
tourist at the bottom have to get out of the way after hearing the sound of the rock
breaking loose (neglecting the height of the tourist, which would become negligible
anyway if hit)? The speed of sound is 335 m/s on this day.
Solution
(a) Knowns: v0  0 m/s, y0  0 m, y   250 m, a   9.80 m/s 2 .
v   v02  2a( y  y0 )   (0 m/s) 2  2(9.80 m/s 2 )(250 m)   70.0 m/s ,
downward.
(b) Let t s = the time for the sound to travel to the tourist.
ts 
height of cliff
250 m

 0.7463 s
vs
335 m/s
Let t = the total time before the tourist can react.
t  t s  reaction t ime  0.7463 s  0.300 s  1.046 s
Let t b = the time it takes the rock to reach the bottom.
tb 
v  v0  7.00 m/s  0 m/s

 7.143s.
g
 9.8 m/s 2
Now subtract tb  t  7.143 s  1.046 s  6.10 s
54.
A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground
on its path up and takes 1.30 s to go past the window. What was the ball’s initial
velocity?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 2
The time it takes the ball to reach the top of the window, at a height of 9.5 m, will
be the amount of time it takes to reach the bottom of the window, plus the time to
traverse the window, 1.30 s. So let's denote the time to the top of the window as
tTOP , and the time to the bottom of the window t B . We then have t TOP = t B + 1.30 s .
We will call the height to the bottom of the window y1 , and the height to the top of
the window y2 .
1
1
Then, we have y 2  y 2, 0  v0 t  at 2  9.50 m = 0 + v0 (t TOP )  (9.80 m/s 2 )(t TOP ) 2
2
2
We also have an equation for y1 ,
y1 = v0 t B  ( 4.90 m/s 2 )(t B ) 2  7.50 m = v0 t B  (4.90 m/s 2 )(t B ) 2
So now we have 3 equations in 3 unknowns ( t B , t BOT , and v0 ). Solving, we get a
value of v 0  14.5 m/s .
55.
Solution
Suppose you drop a rock into a dark well and, using precision equipment, you
measure the time for the sound of a splash to return. (a) Neglecting the time
required for sound to travel up the well, calculate the distance to the water if the
sound returns in 2.0000 s. (b) Now calculate the distance taking into account the
time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.
1
(a) y  y0  v0 t  at 2  0.5(9.80 m/s 2 )(2.0000 s) 2   19.6 m.
2
(b) Let h be the depth of the well, y0  0 m, y  h at the bottom of the well. Let
v2 = the speed of sound and t 2 = the time for the sound to travel from the
bottom to the top of the well. Let t1 = the time for the rock to reach the bottom.
t1 + t 2 = t = 2.0000 s
h 
1 2
at1
2
h = v2t 2 = v2 (2.0000 s  t1 )
Adding equations (ii) and (iii):
(i)
(ii)
(iii)
OpenStax College Physics
Instructor Solutions Manual
Chapter 2
1
0  at12  v2 (2.0000 s  t1 )
2
 0.5(9.80 m/s 2 )t12  (2.0000 s)(332.00 m/s)  (232.00 m/s) t1
0  (4.90 m/s 2 )t12  (332.00 m/s) t1  664m
Using the quadratic formula:
332 m/s  (332 m/s) 2  4(4.90 m/s 2 )(664 m)
 1.944 s.
 9.80 m/s 2
1
h   at12  0.5(9.80 m/s 2 )(1.944 s) 2  18.5 m
2
t1 
56.
A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a
height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate
its velocity just after it leaves the floor on its way back up. (c) Calculate its
acceleration during contact with the floor if that contact lasts 0.0800 ms
(8.00 10 5 s) . (d) How much did the ball compress during its collision with the floor,
assuming the floor is absolutely rigid?
Solution
2
2
(a) v  v0  2a( y  y0 ) 
v   v02  2a( y  y0 )   (0 m/s) 2  2(9.80 m/s 2 )( 1.50 m)   5.42 m/s
2
2
(b) v  v0  2a( y  y0 ) 
v0   v 2  2a( y  y0 )   (0 m/s) 2  2(9.80 m/s 2 )(1.45 m)   5.33 m/s
(c) v  v0  at 
v  v0 5.331 m/s  (5.422 m/s)
a

 1.34 105 m/s 2
5
t
8.00 10 s
(d) The period of compression occurs when the ball goes from v0 =  5.42 m/s to
v = 0 m/s . From part (c), a = 1.344 × 105 m/s 2 . So,
v 2  v 2 (5.422 m/s) 2  (0 m/s) 2
y0  y  0

 1.09 10 4 m
5
2
2a
2(1.344 10 m/s )
57.
A coin is dropped from a hot-air balloon that is 300 m above the ground and rising
at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its
position and velocity 4.00 s after being released, and (c) the time before it hits the
ground.
OpenStax College Physics
Instructor Solutions Manual
Chapter 2
Solution
(a) Knowns: v0 = 10.0 m/s; a =  9.80 m/s 2 ; v = 0 m/s (at top of ascent)
v 2  v02 0 m/s  (10.0 m/s) 2

 5.10 m .
2a
2(9.80 m/s 2 )
Maximum height = 300 m + 5.1 m = 305 m .
1
1
(b) y  v0t  at 2  (10.0 m/s)(4.00 s)  (9.80 m/s 2 )(4.00 s) 2  38.4 m , which is
2
2
a height of 262 m.
y
v  v0  at  10.0 m/s  (9.80 m/s 2 )(4.00 s)   29.2 m/s
1 2
1 2
(c) y  y0 + v0 t  at  ( y0  y )  v0 t  at 
2
2
t
 v0  v02  2a ( y0  y )
 10.0 m/s  (10.0 m/s) 2  2(9.80 m/s 2 )(300 m)

 9.80 m/s 2
a
 10 m/s  77 m/s
t
 8.91s.
 9.8 m/s 2
58.
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds
to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b)
Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its
acceleration during contact with the floor if that contact lasts 3.50 ms
(3.50  10 3 s). (d) How much did the ball compress during its collision with the floor,
assuming the floor is absolutely rigid?
Solution
2
2
(a) v  v0  2a( y  y0 ) 
v   v02  2a( y  y0 )   (0 m/s) 2  2(9.80 m/s 2 )( 1.50 m)   5.42 m/s
2
2
(b) v  v0  2a( y  y0 ) 
v0   v 2  2a( y  y0 )   (0 m/s) 2  2(9.80 m/s 2 )(1.10 m)   4.64 m/s
(c) v  v0  at 
v  v0 4.643 m/s  (5.422 m/s)
a

 2.88  10 3 m/s 2  2880 m/s 2
3
t
3.50  10 s
(d) The period of compression occurs when the ball goes from v0 =  5.422 m/s to
v = 0 m/s . From part (c), a = 2.88 × 10 3 m/s 2 .
v02  v 2 (5.422 m/s) 2  (0 m/s) 2

 5.11  10 3 m  0.00511 m
So, y 0  y 
3
2
2a
2(2.88  10 m/s )
OpenStax College Physics
Instructor Solutions Manual
Chapter 2
2.8 GRAPHICAL ANALYSIS OF ONE-DIMENSIONAL MOTION
59.
(a) By taking the slope of the curve in Figure 2.60, verify that the velocity of the jet
car is 115 m/s at t  20 s . (b) By taking the slope of the curve at any point in Figure
2.61, verify that the jet car’s acceleration is 5.0 m/s 2 .
Solution
position (meters)
position vs. time
4000
3000
2000
1000
0
0
10
20
30
time (seconds)
t (s)
0
10
20
30
rise (2900  600) m

 115 m/s
run
(30  10) s
velocity (meters per
second)
(a) v 
x (m)
200
600
1500
2900
velocity vs. time
200
150
100
50
0
0
10
20
30
time (seconds)
t (s)
0
5
10
v (m/s)
15
40
65
40
40
OpenStax College Physics
Instructor Solutions Manual
15
20
25
30
(b) a 
60.
Chapter 2
90
115
140
165
rise (140-65) m

 5.0 m/s 2
run
(25-10) s
Using approximate values, calculate the slope of the curve in Figure 2.62 to verify
that the velocity at t  10 s is 0.208 m/s. Assume all values are known to 3
significant figures.
Solution
position (meters)
position vs. time
30
20
10
0
0
20
40
60
80
time (seconds)
t (s)
0
10
20
30
40
50
60
70
v
61.
x (km)
3
4.8
7
9.3
11.7
14.2
16.7
19.2
(7  3) km
 .200 m/s  0.208 m/s
(20 - 0) s
Using approximate values, calculate the slope of the curve in Figure 2.62 to verify
that the velocity at t  30.0 s is 0.238 m/s. Assume all values are known to 3
significant figures.
OpenStax College Physics
Solution
62.
v
Instructor Solutions Manual
Chapter 2
(11.7  7) m
 .235  0.238 m/s
(40 - 20) s
By taking the slope of the curve in Figure 2.63, verify that the acceleration is
approximately 3.2 m/s 2 at t  10 s .
Solution
velocity (meters per
second)
velocity vs. time
300
200
100
0
0
63.
40
60
80
time (seconds)
t (s)
0
10
20
30
40
50
60
70
a
20
v (m/s)
165
207
228
239
246
249
250
250
(228  165) m/s
 3.15 m/s 2  3.2 m/s 2
(20  0) s
Construct the displacement graph for the subway shuttle train as shown in Figure
2.48(a). You will need to use the information on acceleration and velocity given in
the examples for this figure.
OpenStax College Physics
Instructor Solutions Manual
Chapter 2
Solution
position (km)
time vs. position
4.85
4.8
4.75
4.7
4.65
0
10
20
30
time (seconds)
64.
(a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at
t  2.5 s . (b) Repeat at 7.5 s. These values must be consistent with the graph in
Figure 2.65.
Solution
position (meters)
position vs. time
30
20
10
0
0
5
10
15
time (seconds)
t (s)
0
2.5
5
7.5
10
12.5
15
20
x (m)
0
10
17.5
10
2.5
2.5
5
25
20
25
OpenStax College Physics
Instructor Solutions Manual
Chapter 2
velocity (meters
per seocnd)
velocity vs. time
10
0
-10
time (seconds)
velocity (meters per
second)
velocity vs. time
5
0
0
5
10
-5
time (seconds)
time (s)
0
2.5
5
7.5
10
12.5
15
20
v (m/s)
4
4
0
-3.6
-0.6
.7
3
4.4
(a) v 
(17.5  0) m
 3.5 m/s
(5 - 0) s
(b) v 
(2.5  17.5) m
  3 m/s
(10  5) s
15
20
25
OpenStax College Physics
65.
Instructor Solutions Manual
Chapter 2
A graph of vt  is shown for a world-class track sprinter in a 100-m race. (See Figure
2.68). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous
velocity at t  5 s ? (c) What is his average acceleration between 0 and 4 s? (d)
What is his time for the race?
Solution
(a) For the first 4 s, vavg = 6 m/s .
(b) At 5 s, v = 12 m/s .
(c) a 
12 m/s
 3 m/s 2 .
4s
(d) For the first 4 s, we know a and v, so x = 24 m .
For the second interval, distance remaining = 100 m  24 m = 76 m and
76 m
 6.3 s. So t total = t1 + t 2 = 4 s + 6.3 s = 10 s .
v = 12 m/s , so t2 
12 m/s
66.
Figure 2.68 shows the displacement graph for a particle for 5 s. Draw the
corresponding velocity and acceleration graphs.
OpenStax College Physics
Instructor Solutions Manual
Solution
position (meters)
time vs. position
5
0
0
2
4
6
8
-5
time (seconds)
time (s)
0
2
3
5
6
x (m)
0
2
-3
-3
-2
(a)
velocity (meters per
second)
time vs. velocity
2
0
-2
0
2
4
-4
-6
time (s)
0
2
2
3
3
5
5
6
6
time (seconds)
v (m/s)
1
1
-5
-5
0
0
1
1
0
6
8
Chapter 2
OpenStax College Physics
Instructor Solutions Manual
(b)
Time (s)
0
1
2
3
4
5
6
7
a (m/s2)
0
0
-6
6
0
1
-1
0
This file is copyright 2016, Rice University. All Rights Reserved.
Chapter 2