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My question is very similar to others asking about how to calculate the time to a given distance given a specific acceleration value but I have a little more complexity and thus the reason for asking an additional question.

most of the questions (and answers) wrt acceleration work with a simplified model of acceleration using constant acceleration and follow the formula y=1/2at^2+vt where a is the acceleration, t is time and v is initial velocity. The issue is that in the real world when accelerating or decelerating a large load its difficult (or impossible) to accelerate it at constant acceleration, likewise, once you reach a target velocity there is a lot of momentum in the system that will attempt to continue its acceleration... abruptly stopping this constant acceleration into the system will result in a not insignificant jolt (or jerk as we should call it) into the system. so there is a third order formula for acceleration that allows for curving the acceleration curve (this third element is formally called jerk). The formula I've found to determine the displacement/distance relative to time is: s=1/6jt^3+1/2at^2+vt with the 1/6jt^3 being a new term introduced to account for the jerk smoothing of the acceleration curve. j is the jerk.

My need is to determine the amount of time necessary to cover a given distance given the acceleration, jerk and distance - in other words I just need to refactor the equation to solve for t instead of s.

other similar questions have pointed readers to use the quadratic equation for acceleration questions but If I'm not mistaken its just a bit of a fluke that the normal (constant) acceleration curve fit the ax^2+bx+c formula that works with the quadratic equation and I'm not aware of a way to use a third order equation such as mine into quadratic... or maybe so - i'm not sure - anyway any guidance here would be appreciated.

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  • $\begingroup$ There is a formula to solve a cubic, and you can find it here: math.vanderbilt.edu/~schectex/courses/cubic It is a bit more complicated. There is also one for a fourth-order equation, more complicated still. It is an interesting fact of mathematics that there cannot be a general formula for orders 5 and higher. $\endgroup$ – Eric Auld Aug 2 '13 at 15:41
  • $\begingroup$ oh you totally rock - i've been looking for that for 2 days $\endgroup$ – wenger Aug 2 '13 at 15:53
  • $\begingroup$ Eric - your answer is more than satisfactory for my need but I don't see how to mark it as a correct/accepted answer. What do I need to do to get you appropriate credit for your response? $\endgroup$ – wenger Aug 2 '13 at 17:14
  • $\begingroup$ Thanks for asking! I didn't know that would solve your problem, or else I'd have posted it as an answer... $\endgroup$ – Eric Auld Aug 2 '13 at 17:42
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There is a formula to solve a cubic, and you can find it here:

http://www.math.vanderbilt.edu/~schectex/courses/cubic/

It is a bit more complicated. There is also one for a fourth-order equation, more complicated still. It is an interesting fact of mathematics that there cannot be a general formula for orders 5 and higher.

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