I am currently using the equation:

$$ \vec p=\frac12\vec at^2+\vec vt+\vec x$$ to find the acceleration I need to move from one point to another.


  • $\vec p$ = target position (point 2d)
  • $\vec a$ = acceleration (vector 2d)
  • $t$ = time (scalar)
  • $\vec v$ = initial velocity (vector 2d)
  • $\vec x$ = initial displacement (point 2d)

I am able to rearrange this equation to find the required acceleration to reach the target in time $t$. But I need to limit the acceleration applied, and the time is irrelevant (it can be any value) to my problem.

Can I rearrange this to find $t$ and $\vec a$ where $\left|\vec a\right |$ is equal to some value $n$?

  • $\begingroup$ What is $\mathrm{mag}$? $\mathrm{mag}\,\vec a =|\vec a|$? $\endgroup$ – Michael Galuza Jul 10 '15 at 8:58

From $$ \frac12 t^2 \vec a=\vec p-\vec x-t\vec v$$ we obtain $$ \frac14t^4n^2=(\vec p-\vec x)^2-2t\langle \vec p-\vec x,\vec v\rangle+t^2\vec v^2$$ where all but $t$ is given. If we solve this for $t$ (numerically, I suggest), you can obtain your desired $\vec a$ from this as $\vec a=\frac{2(\vec p-\vec x-t\vec v)}{t^2}$.

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