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A train passes a control tower with a velocity u and has a constant acceleration.At 800m beyond the tower,the velocity is 48km/h;and at 1.6km beyond the tower,the velocity is 64km/h,what is the velocity u ?

  • physics -

    V2 = 48000/3600 = 13.33 m/s
    V3 = 64000/3600 = 17.78 m/s

    average speed between points 2 and 3 =(13.33+17.78)/2
    = 15.56 m/s

    distance between points 2 and 3 =1600 -800 = 800 m

    time between points 2 and 3 = 800/15.56 = 51.4 seconds

    acceleration = a = change in speed over time
    a = (17.78-13.33)/51.4 = .0866 m/s^2

    now between point 0 and point 1
    v = u + a t
    13.33 = u + .0866 t or u = (13.33-.0866t)
    x = 0 + ut +(1/2)at^2
    800 = u t +(1/2)(.0866) t^2
    800 = (13.33-.0866t)t +.0433 t^2
    solve quadratic for t, go back and get u

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