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Math Help - acceleration with respect to time

  1. #1
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    acceleration with respect to time

    Could anyone help with this?

    a(t)= -g/b^2(v(t)^2 - b^2)

    i need to find t in terms of v, b, g, vo,

    I think I need to separate the variables but not sure how to progress.

    Thanks.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by thermalwarrior View Post
    Could anyone help with this?

    a(t)= -g/b^2(v(t)^2 - b^2)

    i need to find t in terms of v, b, g, vo,

    I think I need to separate the variables but not sure how to progress.

    Thanks.
    Please use parenthesis! I assume this is the following:
    \frac{dv}{dt} = -\frac{g}{b^2} \cdot (v^2 - b^2)

    \frac{dv}{v^2 - b^2} = -\frac{g}{b^2}~dt

    \int_{v_0}^v \frac{dv}{v^2 - b^2} = -\frac{g}{b^2}\int_0^t dt

    The integral on the right is trivial. The integral on the left can be done by a partial fraction expansion:
    \int \frac{1}{v^2 - b^2} ~dv = \frac{1}{2b} \int \left ( \frac{1}{v + b} - \frac{1}{v - b} \right ) ~dv

    = \frac{1}{2b} \cdot ln \left | \frac{v + b}{v - b} \right | + C

    So going back to the original problem we have:
    \frac{1}{2b} \cdot ln \left | \frac{v + b}{v - b} \cdot \frac{v_0 + b}{v_0 - b} \right | = \frac{g}{b^2} \cdot t

    So finally:
    t = -\frac{b}{2g} \cdot ln \left | \frac{v + b}{v - b} \cdot \frac{v_0 + b}{v_0 - b} \right |

    -Dan
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  3. #3
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    Smile

    thats spot on, thanks!

    one question- can you explain what you meant by the integral in trivial?

    thanks again
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by thermalwarrior View Post
    thats spot on, thanks!

    one question- can you explain what you meant by the integral in trivial?

    thanks again
    \int dt = t + C
    That's about the simplest integral there is. Thus "trivial" as in, everyone knows it off the top of their heads kind of thing.

    -Dan
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