# Suvat equations in reverse direction

I have an object traveling in the positive direction with a velocity $v$ and I wish to apply a constant acceleration so that the final velocity is equal in magnitude but opposite in direction (i.e. $-v$).

I do not know the time, but know the distance over which I want to do this, the equation to use would be $v^2=u^2+2as$.

But then, $(-v)^2=v^2+2as$, so $v^2-v^2=2as$ and so I get an $a=0$ which is obviously incorrect. This stems from the fact that $s=\frac{(u+v)}{2}t$, so $s=\frac{v-v}{2}t$ and you can see where this is going.

What would be the best way to do this?

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Why the down vote? It's a basic question but a reasonable one, and Tray65 has at least had a go at solving it, which is more than can be said for many questions. – John Rennie Aug 4 '12 at 12:39
Comment to the question(v2): Note that $s$ in the suvat equations denotes a displacement (rather than a traveled distance), cf. e.g. this Wikipedia page. – Qmechanic Aug 4 '12 at 13:00

The equation is correctly telling you that $s = 0$.

Suppose you start at $x = 0$ and travelling in the positive $x$ direction with some velocity $v_0$. By the time you've accelerated to a halt (i.e. $v = 0$) you'll be at some positive $x$ value. You continue the acceleration to start moving in the opposite direction, and of course your position on the $x$ axis is now decreasing. By the time your speed is $-v_0$ you'll be back where you started i.e. $s = 0$. The equation is not telling you the total distance travelled, it's telling you the net distance travelled, and because you end up back where you started then net distance is zero.

To get the total distance travelled is easy. Just solve your equation for $u = v_0$ and $v = 0$, to find out how far you move before stopping, then double that distance if you want the total distance moved.

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Thanks! That cleared it up in my head. – Tray65 Aug 7 '12 at 11:25