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  1. #1
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    Red face application

    hey everyone please help me figue out the distance of the train , i need A and b c and d too , this is my last project so i was hoping i can get some help on it
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  2. #2
    Bar0n janvdl's Avatar
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    b)  v^2 = u^2 + 2as
     1296 = + (2)(2)(s)
     s = 324

     s = ut + \frac{1}{2} a t^2
     s = ut \ (There \ was \ no \ acceleration)
     s = (36)(90)
     s = 3240

     v^2 = u^2 + 2as
     0 = 36^2 + (2)(-3)(s)
     s = 216

    Add all those distances together and you get: 3780 metres
    Last edited by janvdl; May 28th 2007 at 12:06 PM.
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  3. #3
    Bar0n janvdl's Avatar
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    c) T2 takes 150s to complete the journey.
    I calculated T1 to take only 118s.

    But they travelled the same distance and route. Im guessing that Train 1 was therefore faster. Maybe it had a bigger average speed.

    d) For T2
    s = 3870
    t = 150
    u = 0
    v = 0
    a = ?

     s = ut + \frac{1}{2} a t^2
     3870 = \frac{1}{2} a 150^2
     a = \frac{43}{125}

     v = u + at
     v = 0 + \frac{43}{125} (150)
     v = 51,6 ms^-1
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