A speeder traveling at 42 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 3.66 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder? Both located at the same place and time.

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I used the formula \[S=1/2(a)(t)^2+ Vi(t)+Si\]

Because of the same place and time I made cop=speeder

So the way I did it is (1/2)(3.66m/s^2)(t^2)=(42m/s)(t)

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