Question

An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 130km/h. Precisely 2.50s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.20m/s2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

Answer 1

2.50s after the speeder passes distance between police car and speeder equals:
S0 = t0*(Vs - Vp) = 2.5s*(35/3.6)m/s = 24.31 m
t0=2.5 s
Vs - speed of the speeder
Vp - speed of the police car
Distance from this point for speeder equals:
Ss = S0 + Vs*t
for police car:
Sp = Vp*t + a*t^2/2
a - the police car's acceleration
the police car overtakes the speeder then Ss = Sp
S0 + Vs*t = Vp*t + a*t^2/2
Or:
t^2 - 2(Vs-Vp)/a*t - 2S0/a = 0
t^2 - 2*35/2.2/3.6*t - 2*24.31/2.2 = 0
t^2 - 8.84*t - 22.1 = 0
Solving this equation:
t = 10.87 s
Finally, time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed):
T = t+t0 = 10.87+2.50 = 13.37 s
Answer: T = 13.37 s