Posted by Michael on Thursday, October 7, 2010 at 12:25am.
An unmarked police car traveling a constant 95 km/h is passed by a speeder traveling 120 km/h.
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 12:41am
Write 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 (t1)
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(t1)^2
Use X1 = X2 to solve for t

Physics.  Anonymous, Monday, June 27, 2011 at 12:08pm
6.1 seconds

Physics.  Anonymous, Tuesday, January 19, 2016 at 5:58pm
6.1
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