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April 23, 2014

April 23, 2014

Posted by **Michael** on Thursday, October 7, 2010 at 12:25am.

Precisely 1.00 sec. after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 1.90 m/s^2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

- Physics. -
**drwls**, Thursday, October 7, 2010 at 12:41amWrite two equations for distance travelled vs time after the speeder passes. One equation for each car.

Set the distances equal and solve for the time t.

V1 = 120 km/h = 33.33 m/s is the speeder's speed

V2 = 26.39 m/s + 1.90 (t-1)

is the policeman's speed,for t>1 s. For 0<t<1, it is 26.39 m/s

X1 (the speeder) = 33.33*t

X2 (the cop) = 26.39*t + 0.95(t-1)^2

Use X1 = X2 to solve for t

- Physics. -
**Anonymous**, Monday, June 27, 2011 at 12:08pm6.1 seconds

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