Aspeeder doing 40.0 mi/h (about 17.9 m/s) in a 25 mi/h zone approaches a parked police car. the instant the speeder passes the police car, the police begin their pursuit. if the speeder maintains a constant velocity, and the police car accelerates with a constant acceleration of 4.51 m/s2, (a) how much time does it take for the police car to catch the speeder, (b) how far have the two cars traveled in this time, and (c) what is the velocity of the police car when it catches the speeder?
a) 7.96 s b) 143m c) 35.8 m/s

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