Physics 198
Homework 2
1. What is the difference between a longitudinal and a transverse wave? Give an example of
each.
Solution:
The difference is that in longitudinal waves the motion of the medium is parallel to the direction
of wave travel, while in transverse waves the motion of the medium is perpendicular to the
direction of wave travel. A sound wave in air is longitudinal wave. A wave on a rope shaken up
and down at one end is transverse wave.
2. A slope of a graph of distance versus time is equal to what quantity?
Solution:
The slope of a graph of position versus time is equal to the speed.
3. Can an object have zero velocity and nonzero acceleration at the same time? Give examples.
Solution:
If an object is at the instant of reversing direction (like an object thrown upward, at the top of its
path), it instantaneously has a zero velocity and a non-zero acceleration at the same time. A
person at the exact bottom of a “bungee” cord plunge also has an instantaneous velocity of zero
but a non-zero (upward) acceleration at the same time.
4. Which one of these motions is not at constant acceleration: a rock falling from a cliff, an
elevator moving from the second floor to the fifth floor making stops along the way, a dish
resting on a table?
Solution:
The elevator moving from the second floor to the fifth floor is NOT an example of constant
acceleration. The elevator accelerates upward each time it starts to move, and it accelerates
downward each time it stops.
Ignoring air resistance, a rock falling from a cliff would have a constant acceleration. (If air
resistance is included, then the acceleration will be decreasing as the rock falls.) A dish resting
on a table has an acceleration of 0, so the acceleration is constant.
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Physics 198
Homework 2
5. A sprinter accelerates from rest to 10 .0 m s in 1.35 s. What is his acceleration
(a) in m s 2 , and (b) in km h 2 ? (c) Compare this to the acceleration of on object in free fall.
Solution:
The average acceleration of the sprinter is
v
a
(b)
a  7.41m s 2 
(c)
The acceleration of an object in free fall is: g  9.8m / s 2
t

So,

10.0 m s  0.0 m s
(a)
1.35 s
 7.41m s 2 .
2
1 km  3600 s 
4
2
  1000

  9.60  10 km h
m  1 h 

a 7.41m / s 2

 0.76
g 9.8m / s 2
6. A rolling ball moves from x1  3.4 cm to x 2  4.2 cm during the time from t1  3.0 s to
t 2  6.1 s. What is its average velocity?
Solution:
The average velocity is given by
v 
x
t

4.2 cm  3.4 cm
6.1s  3.0 s

7.6 cm
3.1s
 2.5 cm s .
7. A sports car moving at constant speed travels 110 m in 5.0 s. If it then brakes and comes to a
stop in 4.0 s, what is its acceleration in m s 2 ? Express the answer in terms of “g’s,” where
1.00 g  9.80 m s .
2
Solution
The initial speed of the car before it accelerates is:
v
d
t

110 m
5.0 s
 22 m s  v0
The final speed is v  0 , and the time to stop is 4.0 s. The acceleration is:
a
v v  v0 0  22 m s
 1g


 5.5 m s 2   5.5 m s 2  
2
t
t
4.0 s
 9.80 m s

  0.56 g 's

2
Physics 198
Homework 2
8. Figure below shows the velocity of a train as a function of time. (a) At what time was its
velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what
periods, if any, was the acceleration constant? (d) When was the magnitude of the
acceleration greatest?
Solution:
Slightly different answers may be obtained since the data comes from reading the graph.
(a) The greatest velocity is found at the highest point on the graph, which is at t  48 s .
(b) The indication of a constant velocity on a velocity-time graph is a slope of 0, which occurs
from t  90 s to t  108 s .
(c) The indication of a constant acceleration on a velocity-time graph is a constant slope, which
occurs from t  0 s to t  38 s , again from t  65 s to t  83 s , and again from
t  90 s to t  108 s .
(d) The magnitude of the acceleration is greatest when the magnitude of the slope is greatest,
which occurs from t  65 s to t  83 s .

9. If v x  4m / s and v y  3m / s , determine the magnitude and direction of v .
Solution:
The magnitude is
  3m / s   5m / s .

If  is the angle between v and x-direction then
v y  3m / s
tan  

 0.75 .
vx
4m / s
From this follows:   37  .
v  v v 
2
x
2
y
4m / s 
2
y
2
vy
vx
x

v
3