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

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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