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



  • $a$ = 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?

  • $\begingroup$ That's a quadratic equation in v $\endgroup$
    – Colin K
    Aug 13, 2012 at 16:40
  • $\begingroup$ I don't see how jerk is important here. In fact, if acceleration is constant then jerk is zero and your equation blows up. See if you can relate distance, acceleration, speed and time from what you know. $\endgroup$ Aug 13, 2012 at 18:53

1 Answer 1


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

$$\tag{2'} \Delta t ~\longrightarrow~ - \frac{v_0}{a_0}~>~0 \qquad \text{for} \qquad j~\longrightarrow ~0, $$


$$\tag{3'} s ~\longrightarrow~ - \frac{v^2_0}{2a_0}~>~0 \qquad \text{for} \qquad j~\longrightarrow ~0. $$

  • $\begingroup$ Worked perfectly! Thanx for the help $\endgroup$
    – stamp
    Aug 14, 2012 at 8:36

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