JackMeemai9ndsar7a
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A speeder passes a parked police car at a constant velocity of 30.0 m/s. The police car starts from rest with a uniform acceleration of 2.44 m/s^2.

a) How much time passes before the police car overtakes the speeder?

b) How far does the speeder get before being overtaken by the police car?

Respuesta :

meerkat18
The formula for the speeder should beV1 = d1/t1whereas the formula for the police car is d2  = vo t2 + 1/2 at2^2 where a is equal to 2.44 m/s2. when the overtaking takes place, d1 and d2 are the same, so as t1 and t2
30t = 1/2 *2.44*t2t is equal to 24.59 seconds.d then is equal to 737.70 meters

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