Honors Physics kinematics practice problems. – answers. Dr. Naumoff
1) A car accelerates from 5.0 m/s to 21 m/s at a rate of 3.0 m/s2. How far does it travel while
accelerating?
69 m
v 2f  vi2  2ad
d
v 2f  vi2
2a

441  25
m
 69.3
6
s
2) An airplane needs to reach a velocity of 203.0 km/h to take off. On a 2000 m runway, what is
the minimum acceleration necessary for the plane to take flight?
km   1000m   1h 
m

v f   203


  56.4
h   1km   3600s 
s

2
0.79 m/s2
m

56.4

v 2f  vi2 
m
s
v 2f  vi2  2ad
a

 0.795 2
2d
2  2000m
s
6) A car travels 95 km to the north at 70.0 km/h, then turns around and travels 21.9 km at 80.0
km/h. What is the difference between the average speed and the average velocity on this trip?
A) 27 km/h
t1 
95km
 1.36h
km
70
h
t2 
21.9km
 0.274h
km
80
h
t tot  1.36h  0.274h  1.634h
dis tan ce  95km  21.9km  116.9km
116.9km
km
average speed  v 
 71.5
1.634h
h
displacement  95km  21.9km  73.1km
73.1km
km
average velocity  v 
 44.7
North
1.634h
h
average speed  average velocity  71.5
km
km
km
 44.7
 26.8
h
h
h
9) A car is travelling north at 17.7 m/s. After 12 s its velocity is 14.1 m/s in the same direction.
Find the magnitude and direction of the car's average acceleration.
0.30 m/s2, South
m
m
v f  vi 14.1 s  17.7 s
m
m
aav 

 0.3 2 or 0.3 2 South
t
12s
s
s
15) A dragster travels 1/4 mi in 6.7 s. Assuming that acceleration is constant and the dragster is
initially at rest, what is its velocity when it crosses the finish line?
269 mi/h
vi  0 d= 0.25mi
t=6.7s
.25mi
mi
 0.0373
6.7 s
s
v+v
0+ v f
v
vaverage  i f 
 f
2
2
2
2  vaverage  v f
vaverage 
mi  
s
mi

 0.0373    3600   134.3
s  
h
h

mi
mi
 268.6
h
h
16) The average velocity of a car over a certain time interval is 37 mi/h. If the velocity of the car
was 65 mi/h at the end of this interval, what was its initial velocity? Assume that acceleration
was constant.
9 mi/h
mi
vi  65
vi  v f
mi
h
vav 
 37

2
h
2
mi 
mi
mi

2   37
 vi  9
  65
h 
h
h

2 134.3
20) A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket
engine provides constant upward acceleration during the burn phase. At the instant of engine
burnout, the rocket has risen to 64 m and acquired a velocity of 60 m/s. The rocket continues to
rise in unpowered flight, reaches maximum height, and falls back to the ground. The time
interval, during which the rocket engine provides upward acceleration, is closest to:
2.1 s
m
m
vi  0
s
s
2
2
v f  vi  2ad  0  2ad  2ad
d  64m
v 2f
2d
v f  60
a
2
 m
 60 
v
m
s
a=

 28.12 2
2d 2  64m
s
v f  vi  at  0  at  at
2
f
t
vf
a
m
s  2.133s
t

a 28.12 m
s2
vf
60