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Math Help - how can this be wrong?

  1. #1
    Senior Member furor celtica's Avatar
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    how can this be wrong?

    a particle is moving along a straight line with constant acceleration. in an interval of T seconds it moves D metres; in the next interval of 3T seconds it moves 9D metres. how far does it move in a further interval of T seconds?

    so I calculated the acceleration pretty simply, the velocity increased by 2D/T over 3T, so the constant acceleration is 2/3DT^-2. using this I calculated the distance travelled over the next interval of T seconds as being 0.5(3 + 3 + 2/3) = 20/6 D metres
    the book says 5D metres. where did i go wrong?
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  2. #2
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    Dear furor celtica,

    From S=ut+\frac{1}{2}at^2

    D=uT+\frac{1}{2}aT^2

    10D=4uT+\frac{1}{2}a(4T)^2

    10D=4(D-\frac{1}{2}aT^2)+\frac{1}{2}a(4T)^2

    By simplification, a=\frac{D}{T^2} and u=\frac{D}{2T}

    Therefore the distance traveled in the last T seconds,

    S=(\frac{D}{2T}\times{5T})+\frac{1}{2}\frac{D}{T^2  }(5T)^2-10T

    By simplification, S=5D

    So the book is correct. You haven't considered the fact that the particle could have an initial velocity which I had taken as u.
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