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Math Help - Proving kinematics formula using Calc?

  1. #1
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    Proving kinematics formula using Calc?

    ok so say acceleration is constant.
    a=a

    the integral of a constant is
    \int a=at+C

    C in this case equals initial Velocity (V0)
    so we have
    v=at+V0

    But how do i show that
    v^2=(v0)^2+2a(x-x0)
    using this type of integration?

    Thanks for all the help in advance...
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  2. #2
    Member Abu-Khalil's Avatar
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    Just replace t from x-x_0=v_0t+\frac{at^2}{2}\Rightarrow t=\frac{-v_0\pm\sqrt{v_0^2+2a(x-x_0)}}{a}.
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  3. #3
    Member arpitagarwal82's Avatar
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    Use this
    a = dv/dt
    => a = (dv/dx) * (dx/dt)
    => a = v (dv/dt) since dx/dt = v

    Now proceed.
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  4. #4
    MHF Contributor
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    Quote Originally Posted by dandaman View Post
    ok so say acceleration is constant.
    a=a

    the integral of a constant is
    \int a=at+C

    C in this case equals initial Velocity (V0)
    so we have
    v=at+V0

    But how do i show that
    v^2=(v0)^2+2a(x-x0)
    using this type of integration?

    Thanks for all the help in advance...
    integrate velocity ...

    \Delta x = v_0t + \frac{1}{2}at^2

    using the velocity function ...

     <br />
v_f = v_0 + at<br />

    solve for t ...

    t = \frac{v_f - v_0}{a}

    substitute in the displacement equation to eliminate the parameter, t ...

    \Delta x = v_0 \left(\frac{v_f - v_0}{a}\right) + \frac{1}{2}a\left(\frac{v_f - v_0}{a}\right)^2

    \Delta x = \frac{2v_0(v_f - v_0)}{2a} + \frac{(v_f - v_0)^2}{2a}

    \Delta x = \frac{2v_0v_f - 2v_0^2 + v_f^2 - 2v_0v_f + v_0^2}{2a}

     <br />
\Delta x = \frac{v_f^2 - v_0^2}{2a}<br />

     <br />
v_0^2 + 2a\Delta x = v_f^2<br />
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  5. #5
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    ok now i see. Thanks to all.
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