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Proving the equations of motion

  1. Jan 14, 2005 #1
    If I got [tex] v=u+at [/tex]

    i get, [tex] \int v = \int (u+at) dt [/tex]

    which is [tex] s=ut+\frac {1}{2} at^2 + C [/tex]

    how do i explain away the integration constant in this derivation?
     
  2. jcsd
  3. Jan 14, 2005 #2

    dextercioby

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    You can't.It's x(0),the initial position.You can set it to zero,if u want by chosing the origin of the coordinate system in that very point.

    Daniel.
     
  4. Jan 14, 2005 #3
    Why do you need to `explain away' the constant of integration. The constant [itex]C[/itex] corresponds to the position of the particle when [itex]t = 0[/itex]. Thus you have
    [itex]
    \begin{align*}
    x & = \frac{1}{2} a_xt^2 + u_xt + x_0 \\
    x - x_0 & = \frac{1}{2} a_xt^2 + u_xt \\
    s_x & = \frac{1}{2} a_x t^2 + u_xt.
    \end{align*}
    [/itex]
     
  5. Jan 14, 2005 #4
    that's the explanation i needed lol, thanks alot......it didn't occur to me then.
     
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