A speeder passes a parked police car at a constant speed of 33.1 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.92 m/s 2 . How much time passes before the speeder is overtaken by the police car?

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A speeder passes a parked police car at a constant speed of 33.1 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.92 m/s 2 . How much time passes before the speeder is overtaken by the police car?

Physics
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v=u+at <-- Your equation
33.1/2.92 = 11.3 s
in 11.3 seconds, the speeder will have travelled 374.03m. but the cops will only have travelled 186.4m. 11.3 s is actually the time it takes for the cops to get to 33.1m/s. they will still be behind the speeder. one way to solve this is to compare the distances travelled by each in time t. for the speeder at constant velocity we can say \(x_s = v_s t\) for the cops we can say that \(x_c = u_c + \frac{1}{2}a_ct^2 \; \text{ but } \; u_c = 0 \; \text{ so} \; x_c = \frac{1}{2}a_ct^2\) \(x_s = x_c \implies v_st = \frac{1}{2}a_ct^2\) \[t = \frac{2 v_s}{a_c}\]

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