Physics 101
Homework 2
Due January 19
1. Can an object have a varying speed if its velocity is constant? If yes, give examples.
Solution:
By definition, if an object has a constant velocity, then both the object's speed and its
direction of motion are constant. Therefore the object CANNOT have a varying speed if
its velocity is constant.
2. 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.
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. A sprinter accelerates from rest to 15.0 m s in 3.00 s. What is his acceleration
(a) in m s 2 , and (b) in km h 2 ?
Solution:
The average acceleration of the sprinter is
v
a
(b)
a  5.00 m s 2 
t


15.0 m s  0.00 m s
(a)
3.00 s
 5.00 m s 2 .
2
1 km  3600 s 
4
2
  1000

  6.48  10 km h
m  1 h 

1
Physics 101
Homework 2
Due January 19
5. A small box moves from x1  30 cm to x2  40 cm during the time from t1  4.1s
to t 2  6.1 s. What is its average velocity?
Solution:
v 
The average velocity is given by
x
t

40 cm  30 cm
6.1s  4.1s

70 cm
2.0 s
 35 cm s .
6. A motorcycle is moving at constant speed; it travels 100 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:
v0 
The initial speed of the car before it accelerates is:
d
t

100 m
5.0 s
 20 m s
The final speed is v  0 , and the time to stop is 4.0 s. The acceleration is:
a
v v  v0 0  20 m s
 1g



 5.0 m s 2   5.0 m s 2  
 0.51 g 's
2 
t
t
4.0 s
 9.80 m s 
7. A car moves at 20 m s . It comes to rest in a distance of 80 m. What was its
acceleration, assumed constant?
Solution:
The acceleration can be found from the equation:
v 2  v02  2a  x  x0   a 
v 2  v02
2  x  x0 

2a x  x0   v 2  v02 :
0   20 m s 
2
 2.5 m s 2 .
2  80 m 
8. Find the length of a runway needed for an airplane to reach a speed of 40 m s if the
2
constant acceleration is 4.0 m s ? Assume that the plane starts from rest.
Solution:
The runway distance is found by solving equation
2a x  x0   v 2  v02 for x  x0 .
v  v  2a  x  x0   x  x0 
2
2
0
v 2  v02
2a

 40 m s 

2
0
2 4.0 m s
2

 2.0  102 m
2
Physics 101
Homework 2
Due January 19
9. 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 .

10. 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
v  v v 
2
x
2
y
4m / s 
2
From this follows:   37  .
y
2
vy
vx
x

v
3