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27 June, 18:11

A train travels between two stations 1/2 mile apart in a minimum time of 41 sec. If the train accelerates and decelerates at 8 ft/sec^2, starting from the first station and coming to a stop at the second station, what is its maximum speed in mph? How long does it travel at this top speed?

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  1. 27 June, 21:55
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    It is important to notice firs of all that the train with no stop acceleration and them no stop acceleration would have a faster travel between stops. given this is safe to assume that there is a period during while the train is not accelerating nor decelerating. To find the time that is speed is constant

    let's assume non-stop accelerating and decelerating and the time lost (subtraction of both) is the time the train had constant speed:

    assuming same time accelerating a decelerating we have that the distance is

    x = voT + 1/2aT^2 (initiial speed = 0) (0.25 miles = 1320 feets)

    1320 = 1/2 (8) (T/2) ^2

    T/2 = 18.17 seg

    T = 36.34 seg

    time of constant speed

    T = 41 - 36.34 = 4.66 seg

    maximum speed

    v = v0 + aT (initial speed = 0)

    v = 8 (18.17) (time accelerating)

    v = 145.36 feets/sec

    distance at full speed

    x = v*t

    x = 145.36*4.66 = 677.38 feets
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