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    Hi,

    I am really having trouble with a maths question from Advancing Maths for AQA 2nd edition. M1

    In chapter 2 kinematics there is the following question.

    "Telegraph poles, 40m apart stand alongside a straight railway line. The times taken for a locomotive to pass the gaps between three consecutive poles are 2.5 seconds and 2.3 seconds, respectively. Assume that the acceleration of the train is constant. Calculate the Acceleration of the train and the speed past the first post."

    I tried using suvat, but we only have 3 of the 4 required variables.
    E.g.

    S = 40m
    U
    V
    A = X
    T = 2.5

    S = 40
    U
    V
    A = X
    T = 2.3

    I tried making V = Y and tried so solve it using suvat and simultaneous equations, but I just failed. I also tried the classic V = D / T but that returned the wrong awnser. Supposedly A = 0.58ms-2 and V = 15.3ms-1 but I cannot understand how.
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    I'd work with the first interval, and the two intervals combined, that way you have the same u and a for each set of equations.

    Then using s=ut+\frac{1}{2}at^2 and plug in your figures.

    It does work out (but the figures are messy), post some working if you're still stuck.
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    Working

    1. 40 = 2.5u + 3.125a
    2. 80 = 4.8u + 2.645a

    4.8 / 2.5 = 1.92

    1. 76.8 = 4.8u + 6a
    2. 80 = 4.8u + 2.645a

    a = 76.8 / 3.355
    a = 22.89

    ??????
    • Study Helper
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    (Original post by fernandez56)
    2. 80 = 4.8u + 2.645a
    The bit in red is not correct. Should be \frac{1}{2}\times 4.8^2

    Edit: And in your final couple of lines when you've subtracted one equation from the other, you've not subtracted one constant from the other on the left hand side.
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    Don't worry, thanks for your time, I managed to get it without using simultaneous equations.
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    (Original post by fernandez56)
    Don't worry, thanks for your time, I managed to get it without using simultaneous equations.
    If you've found a better way, can you post your method for the benefit of others, or drop me a PM if you wish.
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    hmm I still want to try and solve it with simultaneous equations though lol for future referance so....

    1. 40 =2.5u + 3.125a
    2. 80 = 4.8u + 11.52a

    1 = (x 1.92)
    76.8 = 4.8u + 6a
    80 = 4.8u + 11.52a

    3.2 = 5.52a
    3.2/5.52 = a
    a = 0.5797101449

    Thankyou for all your help ghostwalker
    Now I can do it 2 ways XD.

    Just out of curiosity how did you know to calculate suvat for the 1st interval and the combined interval as opposed to just the 1st and 2nd?
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    other method:

    U = 40/2.5 = 16
    V = 40/2.3 = 17.39136435

    V = U + AT

    Sub into suvat for both t = 2.3 and t = 2.5
    Calculate the mean result.

    Note: I had help from another from a friend
    • Study Helper
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    (Original post by fernandez56)
    ...
    Looking at using simultaneous equations, you know s and t, and want to find a and u, so you need both those quantities present, and the only way you can do that is by taking the intervals as we did.

    Which is really me trying to explain something that I learnt long ago, then forgot, and then "rememebered".... So although the explanation is valid, it's more an attempt at how one might go about considering it.

    Edit: Cheers for the other method.
 
 
 
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