Let t=0 when car passes police. We want the distance after that to be the same, so
49t = 19(t-.6) + 1/2 * 3 * (t-.6)^2
t = 21.54 s
A police car is traveling at a velocity of 19.0 m/s due north, when a car zooms by at a constant velocity of 49.0 m/s due north. After a reaction time 0.600 s the policeman begins to pursue the speeder with an acceleration of 6.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder?
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