physics - college
posted by bill on .
An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 14km/h .Precisely 2.50s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.30m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?
The time of police car motion is t,
the time of the speeder motion is (t+2.5).
The distance covered by
- the police car is v(p) •t+at^2/2,
- by speeder is v(s)•(t+2.5).
v(p) •t + at^2/2 = v(s)•(t+2.5).
v(p) = 95 km/h = 26.4 m/s
v(s) = 140 km/h = 38.9 m/s
v(p) •t + at^2/2 = v(s)• t + v(s)• 2.5,
at^2/2 + [v(p) – v(s)] •t -2.5•v(s) = 0,
t^2 –(2/a)•[v(s) – v(p)] •t – 2.5•2/a = 0,
t^2 – 10.9t - 84.6 = 0,
t = 5.24 s.
The total time is 5.24+2.5 = 7.74 s.