Applications of the
Kinematic Equations
h
vo
-H
AP Physics B
Lecture Notes
g
One Dimensional Motion
H = -50 m
What is the maximum height
the ball reaches?
vo =30 m/s
g = -9.8 m/s2
h
v2f = v2o + 2a Dx
vo
0=
H
g
v2
o
+ 2(g)h
vf = 0
a=g
h = Dx
-h =v2o/2g
-h 
30 m/s 2
2
2 -9.8 m/s 


= 45.9 m
2
One Dimensional Motion
H = -50 m
vo =30 m/s
How much time to reach
the maximum height?
g = -9.8 m/s2
h
vf = vo + at
vo
vf = 0
xo  0
a=g
0 = vo + (g)tr
H
g
t = tr
tr  vo
g
-30 m/s
t 
r
2
-9.8 m/s
=
3.06 s
3
One Dimensional Motion
H = -50 m
What is the speed of the ball
at xo = 0?
h
g = -9.8 m/s2
v2f = v2o + 2a Dx
vo
xo  0
H
vo =30 m/s
g
v2f = v2o + 2a 0
a=g
Dx = 0
v2f = v2o
vf =
-30.0 m/s
4
One Dimensional Motion
H = -50 m
vo =30 m/s
How much time to reach
the xo = 0?
h
vo
g = -9.8 m/s2
vf = vo + at
vf = -30 m/s
a=g
xo  0
vf = vo + at1
H
g
t = t1
-30 m/s = 30 m/s + (-9.8 m/s2)t1
t
1

-60 m/s
2
=
6.12 s
-9.8 m/s
5
One Dimensional Motion
H = -50 m
What is the final speed of the ball?
vo =30 m/s
g = -9.8 m/s2
h
v2f = v2o + 2a Dx
vo
xo  0
H
g
v2f = v2o + 2a -50m
a=g
Dx = H
v2f = (30 m/s)2 + 2(-9.8 m/s2 -50m
vf =
-43.36 m/s
6
One Dimensional Motion
H = -50 m
How much total time in the air?
vo =30 m/s
g = -9.8 m/s2
h
vo
vf = vo + at
vf = -43.36 m/s
a=g
xo  0
vf = vo + atT
H
g
-43.36 m/s = 30 m/s + (-9.8 m/s2)tT
t
-vf
t = tT
1

-73.36 m/s
2
=
7.49 s
-9.8 m/s
7
One Dimensional Motion
A motorcycle is moving at 30 m/s when the rider applies
the brakes, giving the motorcycle a constant deceleration.
During the 3.0 s interval immediately after braking begins,
the speed decreases to 15 m/s. What distance does the
motorcycle travel from the instant braking begins until the
2
2
motorcycle stops?
v  v  2ax
f
vf = 0
o
v - vo
a
Dt
 v - vo 
2
0  v o  2
x
 Dt 
2
x-
x-
v o Dt
2 v - v o 
30 m/s 2 3s 
215 m/s - 30 m/s 
=
90 m
8
One Dimensional Motion
A sprinter has a top speed of 11.0 m/s. If the sprinter starts
from rest and accelerates at a constant rate, he is able to
reach his top speed in a distance of 12.0 m. He is then able to
maintain this top speed for the remainder of a 100 m race.
What is his time for the 100 m race?
11
88 m
12 m
t1
t2
t
9
11
88 m
12 m
t1
0
vo  vf
v avg 

2
x 2  vf t 2
t
t2
x1
t1
212 m 
t1 
2x1
vf
11 m/s
t2 
x2
88 m
vf


 2.2 s
 8.0 s
11 m/s
t  t1  t 2
 10.2 s
10
One Dimensional Motion
On a dry road a car with good tires is able to brake
with a constant deceleration of 4.92 m/s2. If the car is
initially traveling at 24.6 m/s,
(a) how much time is required to stop?
0
v f  v o  at
v o  - 24.6 m/s
t2
- 4.92 m/s
a

 5s

(b) how far does it travel in this time?
02
2
v f  v o  2ax
2
x-
vo
2a
-
24.6 m/s 2

2 - 4.92 m/s
2

 61.5 m
11
One Dimensional Motion
The maximum acceleration of a subway train is 1.34 m/s2,
and subway stations are located 806 m apart,
(a) what is the maximum speed a subway train can attain
between stations?
(m/s)
a = 1.34 m/s2
vmax
403 m
403 m
t
(s)
12
(m/s)
a = 1.34 m/s2
vmax
x1 = 403 m
x2 = 403 m
t
(s)
0
2
2
v max  v o  2ax1
v max 
2ax1 

2 1.34 m/s
2
403 m
v max  32.9 m/s
13
(m/s)
a = 1.34 m/s2
vmax
v max  32.9 m/s
x1 = 403 m
x2 = 403 m
t1
t2
(b) what is the travel time between stations?
(s)
0
t  t1  t 2  2t1
v max
2 v max  232.9 m/s 
t
2
1.34 m/s
a
v max  v o  at1
t1 
t
a
t  49.1 s
14
One Dimensional Motion
END