Can you solve for delta d in jerk motion using this formula?

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SUMMARY

The discussion focuses on calculating delta d in jerk motion using the formula delta d = vt + (1/2)at² + (1/6)Jt³, where J represents constant jerk. Participants clarify that jerk is the derivative of acceleration and emphasize the importance of vector addition when considering pre-existing acceleration. The confusion arises from differing results when substituting physics equations, with one participant noting a discrepancy between ¼ Jt³ and 1/6 Jt³. Ultimately, integrating the equation for jerk motion twice confirms the latter as the correct formulation.

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What formula would one use to solve for delta d with regards to t, v1, a1, and constant J?
 
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RobotGymnast said:
What formula would one use to solve for delta d with regards to t, v1, a1, and constant J?

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

http://en.wikipedia.org/wiki/Impulse

.
 
Kinematics. As in,
Code:
delta d = vt + (1/2)at[SUP]2[/SUP]

except assuming that jerk motion is constant, rather than acceleration.
 
Last edited:
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.
 
*jerk motion (the derivative of acceleration). whoops.
 
Well, going from the equation using velocity s=v t, to the equation using acceleration v t + \frac{1}{2}a t^2 I would assume that the equation using jerk is s=v t + \frac{1}{2}a t^2 + \frac{1}{6}j t^3.

There's my guess.
 
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 ¼ 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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