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Acceleration as a function of displacement

  1. Nov 13, 2014 #1
    In one my classes my lecturer showed us the following derivation of acceleration as a function of displacement

    dv/dt = v(dv/dx) = d(.5v^2)/dx

    I understand how to get from the first to the second part. But I'm not sure how he got from the second part to the third. Its almost like he integrated v dv on the right hand side?

    Any help would be appreciated.
  2. jcsd
  3. Nov 13, 2014 #2


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    [itex]\frac{dv}{dt}=v\frac{dv}{dx}=\frac{1}{2}2v\frac{dv}{dx}=\frac{d}{dx}(\frac{1}{2} v^2)[/itex]
  4. Nov 13, 2014 #3
    Oh so you're really just reversing the chain rule
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