A speeder traveling at 38 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 2.03 m/s2. To the nearest tenth of a second how long does it take the policeman to catch the speeder?

The distance covered by speeder (uniform motion) is

s=vt
The same distance covered by policeman (accelerated motion from rest) is
s=(at^2)/2.

From vt=(at^2)/2
t=2v/a=2•38/2.03=37.4 s.