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Derivative of Acceleration

  1. Jun 12, 2005 #1
    If the derivative of displacement if velocity, and the derivative of displacement is acceleration, does the derivative of acceleration give you anything? We were trying to think of something in class today but couldnt.
  2. jcsd
  3. Jun 12, 2005 #2


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    The time derivative of acceleration is called 'jerk.'

    From wikipedia (http://en.wikipedia.org/wiki/Jerk):

    - Warren
  4. Jun 12, 2005 #3
    There are infinite time derivatives of position, but only the first six I think are actually named. The position function for constant jerk becomes

    [tex] x(t) = x_0 + v_0 t + \frac{1}{2} at^2 + \frac{1}{6} j t^3 [/tex]

    Also, the derivative of velocity is acceleration, you seem to have a little mix-up there.
    Last edited: Jun 12, 2005
  5. Jun 13, 2005 #4


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    The first six? I knew "jerk" but what are the other three?
  6. Jun 13, 2005 #5


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  7. Jun 15, 2005 #6
    ya i meant the derivative of velocity is acceleration. i didnt even notice i wrote it wrong though.
  8. Jun 15, 2005 #7


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    You know,more time derivatives on the position vector are not welcome for various reasons.Take the Lorentz-Dirac reaction force.It has the third time derivative.It poses problems with the causality.Newtonian physics,however,seems to accomodate the time varying acceleration.Thankfully,in quantum physics the problems generated by more than 2 time derivatives are absent.There's no such thing as force,nor acceleration.

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