A train which is traveling at 83 mi/hr applies its brakes as it reaches point A

Question

A train which is traveling at 83 mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased velocity is observed to be 63 mi/hr as it passes a point 1/2 mi beyond A. A car moving at 52 mi/hr passes point B at the same instant that the train reaches point A. In an unwise effort to beat the train to the crossing, the driver %u201Csteps on the gas.%u201D

a) Calculate the constant acceleration a that the car must have in order to beat the train to the crossing by 4.3 sec

b) find the velocity v of the car as it reaches the crossing.

Explanation / Answer

deaccelaration of the train = ( (80/3600)^2 - (63/3600)^2) / (2*0.5) = 1.86*10^-4 miles/sec^2

time taken by train be t

=> 1 = 0.022t - 9.3*10^-5 t^2

=> t = 61.38 s

=>time taken by car = 61.38 - 4.3 = 57.08

=> accelration of car = (1.3 - 0.793)/0.5*57.08^2 = 3.11*10^-4 miles/sec^2

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