Question

A speeder who is doing 40 mi/hr (17.9m/s) in a 25 mi/hr zone approaches a parked police car. The instant the speeder passes the police car, the police car, the police begin their pursuit. If the speeder maintains a constant velocity and police car accelerates with a constant acceleration of 4.5 m/s^2, How long does it takes for the police car to catch the speeder ?

Answer 1

8.0 s

Step-by-step explanation:

For the speeder:

u = 17.9 m/s

a = 0 m/s²

Their position at time t is:

s = u t + ½ a t²

s = 17.9 t

For the police car:

u = 0 m/s

a = 4.5 m/s²

Their position at time t is:

s = u t + ½ a t²

s = 2.25 t²

When they are equal:

17.9 t = 2.25 t²

17.9 = 2.25 t

t = 7.96

Rounded to two significant figures, it takes 8.0 seconds for the police car to catch the speeder.