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Second Order Differential Equation to show s = ut +(1/2)at^2

  1. Dec 26, 2013 #1
    1. The problem statement, all variables and given/known data


    If d^2s/dt^2 = a, given that ds/dt = u and s = 0, when t = 0, where a, u are constants

    show that s = ut + .5at^2



    2. The attempt at a solution

    du/dt = a

    cross multiplying and then integrating and we get

    u = at

    ds/dt = at

    cross multiply and integrate

    s = .5at^2

    using limits when t = 0 then s = 0

    I can't seem to get out the constant u
     
  2. jcsd
  3. Dec 26, 2013 #2
    could you maybe use v = u + at ? and say v = u + at but v = ds/dt

    so ds/dt = u + at

    cross multiplying and integrating and you get

    s = ut + .5at^2
     
  4. Dec 26, 2013 #3

    CAF123

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    Gold Member

    Yes, but first you need to derive this from the starting equality d2s/dt2 = a.
     
  5. Dec 26, 2013 #4
    I'm not sure how though could I say dv/dt = a

    and then I'd get v = at but I just cant seem to get the u.
     
  6. Dec 26, 2013 #5

    rcgldr

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    Homework Helper

    After integration there's a constant. Say the initial velocity is u0, then

    u = at + u0

    The problem statement seems a bit off, if u = ds/dt, then there needs to be a constant for initial velocity in the equation such as u0:

    s = u0 t + 1/2 a t^2
     
  7. Dec 26, 2013 #6
    But would using the limits not eliminate the constant of integration?
     
  8. Dec 26, 2013 #7

    rcgldr

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    Homework Helper

    The goal here is to produce a quadratic equation, not an intergral with limits.
     
  9. Dec 26, 2013 #8
    I got now thanks.
     
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