90km/hr = 25m/s
128km/hr = 35.56m/s
Let t=0 be the moment the speeder passes
the cop. Then we want t when
35.56t = 25(t-1) + 1.40(t-1)^2
t = 11
An unmarked police car traveling a constant 90 km/h is passed by a speeder traveling 128 km/h. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.80 m/s2 , how much time passes after the police car is passed by a speeder and before the police car overtakes the speeder (assumed moving at constant speed)?
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