Posted by ally on Sunday, April 15, 2012 at 4:38pm.
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.
im still a bit confused, i still dont know the answer to the question
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