Posted by Michael on .
An unmarked police car traveling a constant 95 km/h is passed by a speeder traveling 120 km/h.
Precisely 1.00 sec. after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 1.90 m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?
Write two equations for distance travelled vs time after the speeder passes. One equation for each car.
Set the distances equal and solve for the time t.
V1 = 120 km/h = 33.33 m/s is the speeder's speed
V2 = 26.39 m/s + 1.90 (t-1)
is the policeman's speed,for t>1 s. For 0<t<1, it is 26.39 m/s
X1 (the speeder) = 33.33*t
X2 (the cop) = 26.39*t + 0.95(t-1)^2
Use X1 = X2 to solve for t