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What is an equation for distance which relates the jerk and snap?

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Well I was perusing Wikipedia on physics and came across the "jerk" of an object, the rate of change of acceleration, as well as the "snap", the rate of change of the jerk. The units for these were [tex]\frac{m}{s^3}[/tex] and [tex]\frac{m}{s^4}[/tex] respectively.

    Is there some way to relate these to distance in the equation:

    [tex]d = x + v_{i}t + \frac{1}{2}at^2[/tex] ?


    2. Relevant equations


    [tex]d = x + v_{i}t + \frac{1}{2}at^2[/tex]

    3. The attempt at a solution

    The only good guess I have is that the coefficient before each variable is equal to [tex]\frac{1}{n!}[/tex] where [tex]n[/tex] is the power that [tex]t[/tex] is raised to.

    So according to my rule, the equation including both snap and jerk would be:

    [tex]d = \frac{1}{0!}xt^{0} + \frac{1}{1!}v_{i}t^{1} + \frac{1}{2!}at^2 + \frac{1}{3!}jt^3 + \frac{1}{4!}st^4[/tex]

    simplifying to:

    [tex]d = x + v_{i}t + \frac{1}{2}at^2 + \frac{1}{6}jt^3 + \frac{1}{24}st^4[/tex]

    Am I right?
     
    Last edited: Oct 3, 2009
  2. jcsd
  3. Oct 3, 2009 #2

    rock.freak667

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    Homework Helper

    As much as I can say about it is

    [tex]jerk= \frac{da}{dt} = \frac{d}{dt}(\frac{d^2x}{dt^2})= \frac{d^3x}{dt^3}[/tex]


    though I've never learned about jerk and snap .
     
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