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

[RobotGymnast]

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

 

[berkeman]

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

.

 

[RobotGymnast]

Kinematics. As in,

delta d = vt + (1/2)at[SUP]2[/SUP]

except assuming that jerk motion is constant, rather than acceleration.

 

[Zula110100100]

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.

 

[RobotGymnast]

*jerk motion (the derivative of acceleration). whoops.

 

[Char. Limit]

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.

 

[RobotGymnast]

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.