Vp = 85,000m/3600s = 23.61 m/s.
Vs = 130,000/3600 = 36.11 m/s.
Ds = 36.11m/s * 3s = 108.3 m. Head start
Vo*t + 0.5a*t^2 = Vs*t + 108.3
23.61t + 0.5*2.30*t^2 = 36.11t + 108.3.
-12.5t + 1.15t^2 = 108.3.
1.15t^2 - 12.5t - 108.3 = 0.
Use Quadratic Formula.
t = 16.6 s.
An unmarked police car traveling a constant 85.0 km/h is passed by a speeder traveling 130 km/h .
Precisely 3.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.30 m/s2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?
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2 answers