A speeder traveling at 28.8 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?

Answer 1

d1 = d2.
28.8t = 0.5*2.03*t^2
28.8t = 1.015t^2
1.015t^2 - 28.8t = 0
Use Quadratic Formula and get:
t = 28.4 s.