Posted by **bill ** on Sunday, April 15, 2012 at 5:50pm.

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)?

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**Elena**, Sunday, April 15, 2012 at 5:51pm
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.

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**bill**, Sunday, April 15, 2012 at 6:01pm
i think their is a mistake, the answer is incorrect

- physics - college -
**Elena**, Sunday, April 15, 2012 at 6:07pm
The solution is right. The mistake may be in calculations. Calculate it carefully.

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