Maximum velocity allowed for motion profile with constant jerk I´m trying to calculate what the maximum allowed velocity for a mechanical axis traveling towards a stop. The maximum velocity must be a secure speed so the axis have is able to stop in time.
I have:


*

*current position

*current distance to stop

*current velocity

*acceleration/deceleration constant

*jerk constant


I did found this formula
$$\frac{v^2}{2a}+\frac{va}{2j}=d$$
where


*

*$a$ = acceleration

*$v$ = velocity

*$j$ = jerk

*$d$ = stop distance


but I don't know how to calculate the $v$ instead.
Is this the correct formula to use and in that case, how should I use it?
 A: I) I'm not sure if this is relevant for what OP is asking, but imagine a 1 dimensional motion with constant (negative) jerk $j<0$, see e.g. the second half of this webpage. Let $v_0>0$ be an unknown positive initial velocity, and $a_0$ be a given initial acceleration. We interpret OP's question(v1) as 

For a given stopping distance $s>0$, what is the initial velocity $v_0$?  

Well, the final velocity is zero. This yields an equation 
$$\tag{1} 0~=~ v_f ~=~ v_0 + a_0 \Delta t + \frac{j}{2} \Delta t^2, $$
which is quadratic in time. There is one positive root
$$\tag{2} \Delta t ~=~ - \frac{\sqrt{a_0^2 -2j v_0}+a_0}{j} ~>~0. $$
The distance becomes
$$\tag{3}s ~=~ v_0 \Delta t+ \frac{a_0}{2} \Delta t^2+ \frac{j}{6} \Delta t^3 ~\stackrel{(2)}{=}~\frac{a_0(a_0^2-3jv_0)+ (a_0^2 -2j v_0)^{3/2}}{3j^2}.$$
For given $s>0$, $j<0$, and $a_0$, this equation (3) can be solved numerically for $v_0$. This ends our main answer.
II) As a consistency check of the above equations (2) and (3) in the case $a_0<0$, it is an instructive exercise to Taylor expand (or use l'Hopital's rule) in small $j$ to recover the well-known constant acceleration (suvat) formulas
$$\tag{2'} \Delta t ~\longrightarrow~ - \frac{v_0}{a_0}~>~0 \qquad \text{for} \qquad j~\longrightarrow ~0, $$
and
$$\tag{3'} s ~\longrightarrow~ - \frac{v^2_0}{2a_0}~>~0 \qquad \text{for} \qquad j~\longrightarrow ~0. $$
