A speeder traveling at 37 m/s passes a motorcycle policeman at rest at the side of the road. The policeman accelerates at 2.62 m/s2, To the nearest tenth of a second how long does it take the policeman to catch the speeder?
When the policeman catches up, the
distance traveled by each will be equal:
d1 = d2.
V*t = 0.5at^2,
37*t = 0.5*2.62*t^2,
37*t = 1.31*t^2,
1.31t^2 - 37t = 0,
Factor out t:
t(1.31t - 37) = 0,
t = 0,
1.31t - 37 = 0,
1.31t = 37,
t = 37 / 1.31 = 28.244 s.