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Thread: ODE help

  1. #1
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    ODE help

    Hey friends, I found this site searching for ode help and I think I might have found a home.. I do need help though and would appreciate any assistance you can give.

    I have an object that weighs 10kg and the drag force is proportional to the square of the velocity and the limiting velocity is 49 m/s. The equation of motion given is $\displaystyle m(dv/dt) = mg - 2v$
    or
    $\displaystyle dv/dt = 9.8 - v/5 $ after putting in the drag force constant, gravity and weight. So my objective is to prove how the equation of motion can be written as $\displaystyle dv/dt = [49^2 - v^2] / 245 $.

    If any of you happen to have Elementary Diff EQs and Boundry value problems its number 11 in section 1.2

    And this is the coolest site ever, I love math. T"hanks for your help.
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  2. #2
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    Quote Originally Posted by DinoJockey View Post
    I have an object that weighs 10kg and the drag force is proportional to the square of the velocity and the limiting velocity is 49 m/s. The equation of motion given is $\displaystyle m(dv/dt) = mg - 2v$
    or
    $\displaystyle dv/dt = 9.8 - v/5 $ after putting in the drag force constant, gravity and weight. So my objective is to prove how the equation of motion can be written as $\displaystyle dv/dt = [49^2 - v^2] / 245 $.
    your DE statement and your DE, $\displaystyle m(dv/dt) = mg - 2v$, do not match up.

    if the drag force is proportional to the square of velocity, then the DE should be

    $\displaystyle m\frac{dv}{dt} = mg - kv^2$
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  3. #3
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    Hmm I thought I missed something but that still leaves me at $\displaystyle \frac{dv}{dt} = 9.8 - \frac{v^2}{5}$ right?
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  4. #4
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    at limiting speed, $\displaystyle \frac{dv}{dt} = 0$

    $\displaystyle mg = kv^2$

    $\displaystyle \frac{mg}{v^2} = k$

    $\displaystyle \frac{98}{49^2} = k$

    $\displaystyle \frac{2}{49} = k$

    $\displaystyle \frac{dv}{dt} = g - \frac{k}{m} v^2$

    $\displaystyle \frac{dv}{dt} = 9.8 - \frac{v^2}{245}$
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  5. #5
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    ah ha! you are a scholar indeed. I appreciate your help.
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