Answer:
a) It takes the police car 7.94 s to catch the speeder.
b) The two cars have traveled 142 m.
c) The velocity of the police car when it catches the speeder is 35.8 m/s
Explanation:
The equations for position and velocity of the police car are as follows:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where
x = position at time "t".
x0 = initial position.
t = time.
a = acceleration.
v = velocity.
For the speeder, the position equation will be:
x = x0 + v · t
a) When the police car catches the speeder, the position of both cars is the same. Then:
x speeder = x police car
x0 + v · t = x0 + v0 · t + 1/2 · a · t²
Let´s place the origin of the frame of reference at the point where the speeder passes the police car so that x0 = 0 for both. Since the police car starts from rest, v0 = 0, then:
v · t = 1/2 · a · t²
t = 2 · v /a
t = 2 · 17.9 m/s / 4.51 m/s²
t = 7.94 s
It takes the police car 7.94 s to catch the speeder.
b) Let´s calculate the position of the speeder at this time (the position of the police car will be the same):
x = v · t
x = 17.9 m/s · 7.94 s = 142 m
The two cars have traveled 142 m.
c) To calculate the velocity, we have to use the velocity equation:
v = v0 + a · t
v = 0 m/s + 4.51 m/s² · 7.94 s
v = 35.8 m/s
The velocity of the police car when it catches the speeder is 35.8 m/s
