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?

Question

Helena Mcgrath

Physics

15 August, 01:52

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?

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Answers (1)

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Yaka · Answer 1

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

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?

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?