Review Interactive LearningWare 2.2 in preparation for this problem. A police ca

Question

Review Interactive LearningWare 2.2 in preparation for this problem. A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 41.0 m/s due north. After a reaction time 0.800 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? YP vos North Police car catches up with speeder vos --Police car begins to accelerate YOP - Speeder passes police car

Explanation / Answer

for this question you use the kinematic equation:

distance(S) = [initial velocity(U) x time(t)] + 1/2[acceleration(a) x time(t)^2]

S = Ut + 1/2at^2

when the police car catches the speeding car, their 'distances' are said to be equal

so [cars distance] = [police distance]
Ut + 1/2at^2 = Ut + 1/2at^2

for the car: for the police:
U = 41m/s U = 18m/s
a = 0 a = 6m/s
t = ? t = ?

so, plugging those values in we get

41xt + 1/2x0xt^2 = 18xt + 1/2x6xt^2 - sub in values
41t = 18t +3t^2 - work through
41 = 18 + 3t - divide all by t
t = (41 - 18) / 3 - re-arrange to get t =
t = 7.666

Therefore, t = 0.800 s + 7.666 s
t = 8.466 s answer

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