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

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?

anonymous · Answer

Consider t is the time when police car catches speeder
$$30 t = 0.5* 2.44*t^2$$
t = 24.59 sec

s = 30*24.59 = 737.70 m

anonymous · Answer

DUUUDE! THANK YOU SO MUCH!! Makes more sense!

anonymous · Answer

how do you put it on da calculator???