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Deriving terminal velocity

  1. Jun 20, 2013 #1
    Hello PF,
    I have once simple (well, not so simple for me) question.

    I'm trying to derive an equation for the velocity of a falling body with accordance to terminal velocity.

    The equation incorporates drag proportional to the speed.

    m*dv/dt=mg-bv

    and

    mg/b=terminal velocity vt

    So the steps I took were:

    m*dv/dt+bv=mg

    (m/b)*(dv/dt)+v=vt

    dv/dt=(b/m)(vt-v)

    dv/(vt-v)=(b/m)dt

    Integrating both sides would give
    ln[(vt-v)/(vt)]=(b/m)t

    But the textbook says that I'm supposed to get negative (b/m)t on the left side.

    Have I made a mistake on the integration part?

    Any help will be deeply appreciated.
     
  2. jcsd
  3. Jun 20, 2013 #2
    hmm. I seem to have gotten the answer if I just divided the entire equation by -b in the beginning without bringing the bv to the left side. Have I made a mistake on the integration part?
     
  4. Jun 20, 2013 #3
    You just missed the minus sign while integrating (chain rule)
     
  5. Jun 20, 2013 #4
    pshh. I can't believe I missed that. Thanks so much king vitamin!
     
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