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A police car is traveling at a velocity of 19.0 m/s due north, when a car zooms

ID: 2160725 • Letter: A

Question

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 41.0 m/s due north. After a reaction time 0.800 s the policeman begins to pursue the speeder with an acceleration of 3.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder?

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 41.0 m/s due north. After a reaction time 0.800 s the policeman begins to pursue the speeder with an acceleration of 3.00 m/s^2. Including the reaction time, how long does it take for the police car to catch up with the speeder?

Explanation / Answer

Let t = time it takes for the police car to catch up with the speeder t = 0.700 s + t1, where t1 = time after reaction time for police car to catch up with the speeder When the police car catches up with the speeder, the distances they traveled from the point the speeder passed the police car are Equal to each other. Dp = Ds where Dp=distance traveled by police car and Ds = distance traveled by speeder Vi(t1) + (1/2)(a)(t1)^2 = (48.0 m/s)(t1) (18.0 m/s)(t1) + (1/2)(5.00 m/s^2)(t1)^2 = (48.0 m/s)(t1) 2.5(t1)^2 - 30(t1) = 0 t1(2.5t1 - 30) = 0 t1 = 0 2.5t1 - 30 = 0 t1 = 30/2.5 t1 = 12 s Therefore, t = 0.700 s + 12 s t = 12.700 s answer

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