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Answer to Question #25872 in Mechanics | Relativity for beren alper

Question #25872
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)?
Expert's answer
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

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