A train which is traveling at 73mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased ve

Question

A train which is traveling at 73mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased velocity is observed to be 56mi/hr as it passes a point 1/2mi beyond A. A car moving at 45mi/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 “steps on the gas.” Calculate the constant acceleration a that the car must have in order to beat the train to the crossing by 3.9sec and find the velocity v of the car as it reaches the crossing.

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Thanh Hà 3 years 2021-08-01T06:54:50+00:00 1 Answers 120 views 0

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    2021-08-01T06:56:22+00:00

    Answer:

    First to find deceleration of the train we use

    v²= u²+ 2as

    56²= 73²+ 2(0.5)a

    a= -2193mi/hr²

    Then we find time in which the train does the intersection

    Using

    S= ut+ 1/2 at²

    1= 73t-1/2(1293)t²

    t =68.5s

    But since the train is to intersect in 3.9s the time will be the difference which is

    65.68s

    So finding acceleration

    S= ut + 1/2at²

    1.3mi= 45/3600mi/s(65.58s)+ 1/2a(65.5)²

    So a= 1.179ft/s²

    To find velocity we use

    V= u + at

    = 45/3600mi/s + (-2.33E-4mis²)(65.58s)

    V= 0.0271mi/s

    = 97.6ft/s

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