Question

An unmarked police car traveling a constant 85km/hr is passed by a speeder traveling 140km/hr. About 3.00s after the speeder passes, the police officer steps on the accelerator, if the police car's acceleration is 1.80m/s^2 , how much time passes before the police car ovetakes the speeder after the speeder passes?(assumed moving at constant speed)

Answer 1

This is a physics question involving kinematic equations. So far I've set up my problem as such: t = 3.00s, v-speeder140km/h, v-p: 85km/h. a-p: 1.80m/s^2. V stands for velocity of speeder and police car and a stands for acceleration.

Answer 2

I'm thinking that I would need to use this kinematics equation: d= vt+1/2at^2