Solutions to selected problems in ch.2
PROBLEMS
Section 2.1
1. (I) What is the displacement of a shopper in a mall who first walks 100 m
straight east and then walks 125 m straight west?
Solution: Let East be positive and West be negative.
100m+-125m= - 25m Therefore, the answer is 25 m west.
2.2. (1) On a long distance car trip a person travels 800 km north (straight line
distance on a map) in 12 hr. (a) What was the average speed? (b) What was the
average velocity?
total distance 800km
km
=
= 66.7
Solution: Average speed =
time
12 hr
hr
Solution: 2.2b. velocity=
displacement 800 km
km
=
North = 66.7
North
time
12 hr
hr
2.4. (I) A cheetah can reach a final speed of 30 m/sec from rest in 6.0 sec. Calculate
the average acceleration of the cheetah.
Solution: a =
ΔV V2 − V1
=
=
ΔT
ΔT
30
m
m
−0
sec
sec = 5 m forward
6 sec
sec 2
2.5. (I) An elephant can run at a steady speed of 10 m/sec and can stop in 2.5 sec.
What is the average acceleration of the elephant while stopping?
ΔV V2 − V1
=
=
ΔT
ΔT
Solution: a =
Final answer is 4
0
m
m
− 10
sec
sec = −4 m
2.5 sec
sec 2
m
backward
s2
2.6 (II) A Grand Prix racer completes a 500 km race in exactly 2 hr and 15 min.
What is the average speed of the car in kilometers per hour and meters per second?
distance 500km
km
=
= 222
Solution: 2.6a. speed =
time
2.25 hr
hr
Solution: 2.6b.
500x10 3m
m
= 61.7
sec ⎞
⎛
sec
2.25hr ⎜ 3600
⎟
⎝
hr ⎠
Solutions to selected problems in ch.2
2.15. (I) What is the final speed of a bicyclist initially moving at 10 m/sec who
accelerates at a rate of 1.5 m/sec2 for 3.0 sec?
10m
m
m
m
m
+1.5 2 ( 3sec ) = 10
+ 4.5
= 14.5
Solution: V = Vo + at =
sec
sec
sec
sec
sec
2.16. (I) A dolphin accelerates backward for 2.0 sec from an initial speed of 7.5 m/
sec in order to take a look at a dolphin of the opposite sex. Calculate the dolphin's
final speed if the backward acceleration had a magnitude of 1.5 m/sec2.
7.5m
m
m
m
m
−1.5 2 ( 2sec ) = 7.5
−3
= 4.5
Solution: V = Vo + at =
sec
sec
sec
sec
sec
2.17. (I) How far will a car travel from a standing start if it accelerates at 4.0 m/
sec2 for 9.0 sec?
1
2
2
Solution: Displacement = Vot + at = 0 +
1 ⎛ 4m ⎞
2
9
sec
=162 meters
(
)
⎜⎝
⎟
2 sec 2 ⎠
Section 2.3
2.29. (I) If you drop an aspirin and it takes 0.18 sec to hit the table, how high
above the table was the aspirin when it was released?
Solution: Choosing the down direction as negative:
1 2
1 ⎛ - 9.8m ⎞
2
.18 sec ) = −.159 meters
Displacement = Vot + at = 0 + ⎜
2 ⎟(
2
2 ⎝ sec ⎠
The negative sign means that the aspirin is below the point of release,
therefore we know that the aspirin was .159 meters above the table when it was
dropped.
Solutions to selected problems in ch.2
2.30. (I) A child throws a ball straight down off the balcony of an apartment
building at a speed of 7.5 m/sec. (a) Calculate the position and velocity of the ball
1.0 and 2.0 sec later. (b) Make the same calculations if the ball is thrown straight
up.
Solution: Let’s let negative indicate the down direction and positive indicate
the up direction. The initial velocity V0 = - 7.5 m/s and the acceleration,
a = - 9.8m/s2.
2.30a. After 1 sec. with an initial velocity of 7.5 m/s down:
Displacement =
1
m
1⎛
m ⎞
2
Vot + at 2 = -7.5 (1sec ) + ⎜ -9.8 2 ⎟ (1sec ) = −7.5m − 4.9m = −12.4m
2
s
2⎝
sec ⎠
Velocity = V0 + at = −7.5
m
m
m
+ −9.8 2 (1sec ) = −17.3
sec
sec
sec
After 1 sec., with an initial velocity of 7.5 m/s down, the ball is 12.4 meters below
the balcony and is moving with a velocity of 17.4 m/s downward.
2.30b. After 2 seconds with an initial velocity of 7.5 m/s down:
Displacement =
1
m
1⎛
m ⎞
2
Vot + at 2 = -7.5 (2 sec) + ⎜ -9.8 2 ⎟ ( 2 sec ) = −15 m − 19.6 m = −34.6 m
2
s
2⎝
sec ⎠
Velocity = V = Vo + at − 7.5
m
m
m
− 9.8 2 ( 2sec ) = −27.1
sec
sec
sec
After 2 sec., with an initial velocity of 7.5 m/s down, the ball is 34.6 meters below
the balcony and is moving with a velocity of 27.1 m/s downward.
Solutions to selected problems in ch.2
2.31. (I) Calculate the displacement and velocity of a rock that is dropped from the
Royal Gorge Bridge 1.0, 2.0, 3.0, and 4.0 sec after its release. Make graphs of the
displacement versus time and velocity versus time, drawing a line through the
calculated points, to represent the motion.
Solution: The initial velocity is 0 m/s and the acceleration is 9.8 m/sec2
down. If you take down as negative, you should get the following results: