A speeder traveling at 30 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 2.4 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder?

Answer 1

driver goes 30 t
cop goes .5 a t^2 = 1.2 t^2

so
30 t = 1.2 t^2
t = 30/1.2 seconds

or of course they are also together at t = 0 :)

Answer 2

We want the distances to be equal, so

30t = 1/2 * 2.4 t^2
And now solve for t.