An unmarked police car traveling a constant 80 km/h is passed by a speeder traveling 125 km/h. Precisely 3.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.40 m/s^2, how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

Answer 1

Xf(police)=Xf(car)

Xf(police)= (22.22)(3) + (22.22)t + 0.5(2.40)t^2..(Xf=Xi+Vt+0.5at^2)

Xf(car)= (34.72)(3) + 34.72t..(Xf=Xi+Vt)-constant velocity