A police car stopped at a set of lights has a speeder pass it at 100 km/h. If the police car can accelerate at 3.6 m/s^2, how long does it take to catch the speeder?

Answer 1

they travel the same distance, in the same time

change 100km/hr to 27.8m/s

ds=27.8*t
dp=1/2 3.6*t^2
set them equal

27.8 t= 1.8 t^2 solve for t.

Answer 2

why are you multiplying by 1/2

Answer 3

with constant acceleration a:

v = Vi + a t
x = Xi + Vi t + (1/2) a t^2

here Vi = Xi = 0
so
x = (1/2)at^2

Answer 4

how do you find the distance it has to travel before it catches speeder?