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Thread: Solve for time

  1. #1
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    Solve for time

    Can somebody help me with this? How the hell do I solve for t:-

    x=vt+1/2at^2
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  2. #2
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    Use the quadratic formula:
    The solutions to $\displaystyle ax^2+ bx+ c= 0$ are given by
    $\displaystyle \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$

    Replace the "a", "b", and "c" with the coefficients from your equation"
    $\displaystyle \frac{1}{2}a t^2+ vt- x= 0$:
    So "a" is $\displaystyle \frac{1}{2}a$, "b" is $\displaystyle v$, and "c" is $\displaystyle -x$.
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  3. #3
    Senior Member Stroodle's Avatar
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    You could rearrange it (e.g. $\displaystyle at^2+2vt-2x=0$) and complete the square, or sub in the appropriate values into the quadratic formula.
    You ok with doing this?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Well, here I'd use the quadratic formula, or you can complete the square.

    Quadratic formula:
    Rearrange;
    $\displaystyle \frac{1}{2}at^2 + vt - x =0$

    $\displaystyle t = \frac{-v \+- \sqrt{v^2 - 4(\frac{a}{2})(-x)}}{2(\frac{a}{2})}$

    $\displaystyle t = \frac{-v \+- \sqrt{v^2 + 2ax}}{a}$

    But since time is positive, you take only the positive sign;

    $\displaystyle t = \frac{-v + \sqrt{v^2 + 2ax}}{a}$

    Completing the square;

    $\displaystyle \frac{1}{2}at^2 + vt - x =0$

    $\displaystyle \frac{a}{2}(t^2 + \frac{2v}{a}t) - x= 0$

    $\displaystyle \frac{a}{2}((t + \frac{v}{a})^2 - \frac{v^2}{a^2}) - x= 0$

    Rearrange;

    $\displaystyle (t + \frac{v}{a})^2 = \frac{2x}{a} + \frac{v^2}{a^2}$

    $\displaystyle t = \sqrt{\frac{2x}{a} + \frac{v^2}{a^2}} - \frac{v}{a}$

    EDIT: Oops, a little too late =S
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  5. #5
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    Thanks for that halls.
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  6. #6
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    I know about the quadratic formula and have used it before. For some reason I was trying to solve using other methods which were inapropriate in this case.
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