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)?
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Expert's answer
2013-03-08T05:24:53-0500
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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