Given an initial velocity $v_0$, initial acceleration $a_0$, and a maximum (positive or negative) jerk $j$, what will be the final resting position $p$ (in relation to the initial position) when coming to a complete stop (velocity and acceleration both 0) in the shortest amount of time?
Velocity and acceleration are both unbounded.
For simplicity please assume velocity is a positive value.
I believe I've partially solved the problem.
Given the situation where acceleration is zero the shortest stopping time should be achieved with negative jerk $j$ for $t$ time (at which point velocity should equal $\frac{v_0}2$) and then positive $j$ for $t$. At this point velocity and acceleration will both be zero.
$t$ can be determined with: $\frac{v_0}2 = v_0 - \frac{1}2jt^2$
$t = \frac{\sqrt{v_0}}{\sqrt{j}}$
Distance traveled during this "first" jerk is then: $d_0 = v_0 t - \frac{j t^3}6$
$d_0 = \frac{5 v_0^{\frac{3}2}}{6\sqrt{j}}$
Acceleration after this "first" jerk: $a_1 = -j t$
Distance traveled during the "second" jerk should be: $d_1 = \frac{1}2v_0 t + \frac{1}2 a_1 t^2 + \frac{1}6 j t^3$
$d_1 = \frac{v_0^\frac{3}2}{6\sqrt{j}}$
Stopping distance in relation to initial position is then of course: $p = d_0 + d_1$
That's just when initial acceleration $a_0$ is zero.
If $a_0$ is positive (accelerating) the problem is almost as easy. Just apply negative jerk $j$ until acceleration is zero, get the distance traveled and new velocity (easy), solve per above now that acceleration is zero, and add all the distances together.
But I am not sure how to solve the problem when $a_0$ is negative. (starting in a state of deceleration.)
