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Equation for jerk motion?

  1. Jan 28, 2010 #1
    What formula would one use to solve for delta d with regards to t, v1, a1, and constant J?
  2. jcsd
  3. Jan 28, 2010 #2


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    Staff: Mentor

    Please define your symbols, and provide the context of your question. Are you familiar with the concept of an impulse?


  4. Jan 28, 2010 #3
    Kinematics. As in,
    Code (Text):

    delta d = vt + (1/2)at[SUP]2[/SUP]
    except assuming that jerk motion is constant, rather than acceleration.
    Last edited: Jan 28, 2010
  5. Jan 28, 2010 #4
    Assuming that the mass is constant, if the force of the jerk J is constant, considering that F = ma so J = ma so a = J/m since J and m are constant A is constant.

    So if your implying there is an acceleration in place before the jerk, you would use vector addition to figure out the resultant acceleration. And use your formula to solve for d.
  6. Jan 28, 2010 #5
    *jerk motion (the derivative of acceleration). whoops.
  7. Jan 28, 2010 #6

    Char. Limit

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    Gold Member

    Well, going from the equation using velocity [tex]s=v t[/tex], to the equation using acceleration [tex]v t + \frac{1}{2}a t^2[/tex] I would assume that the equation using jerk is [tex]s=v t + \frac{1}{2}a t^2 + \frac{1}{6}j t^3[/tex].

    There's my guess.
  8. Jan 28, 2010 #7
    That's what I got. What confuses me was that I also got a different answer by substituting physics equations into one another. I ended up getting ΒΌ Jt^3 rather than 1/6 Jt^3. But integrating the equation for jerk motion twice gives the answer you gave, which I think is correct.
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