posted by Matthew .
An unmarked police car traveling a constant 80km/h is passed by a speeder traveling 135km/h.
Precisely 3.00s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 1.50m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?
80km/hr = 22.22 m/s
135km/hr = 37.5 m/s
so, at time t when the police overtake the scoff-law,
37.5t = 22.22t + 0.75 (t-3)^2