Over on twitter, @MegaManSE asked

does anyone know the equation to find the acceleration to stop a moving object in a constant distance given some random starting velocity?

I didn’t, at the time, know … but I do know how to work it out from first principles, and it makes a decent little classical mechanics exercise, and also an excuse to figure out how to get MathJax hooked up on this blog, which might be useful in the future. So here’s how it’s done.

The first step in solving one of these problems is to rewrite the question as formally as possible:

At time $t=0$ an object is at position $x=0$ and moving with velocity $\nu=v$. Find the constant acceleration $a$ such that at some future time $t=T$, when the object is at position $x=d$, its velocity will be zero.

Now how do we do that? It’s time for just a little bit of integral calculus. Velocity is the rate at which a moving object’s position changes, as a function of time, and acceleration is the rate at which a moving object’s *velocity* changes, also as a function of time. The calculus was invented to answer the question, if I know what one of these is, what are the other two? It has a somewhat-deserved reputation for being confusing, but mostly that’s because it’s hard to explain *how you come up with its rules*. If you know the rules, they’re pretty easy to apply. The acceleration in this problem is constant, $a$, and we know at time $0$ the velocity is $v$ and the position is $0$. Therefore, the velocity at time $t$ is

$\nu(t) = v + \int_0^t a\; \text{d}t = v + at$

and the position is

$x(t) = 0 + \int_0^t v + at\; \text{d}t = 0 + vt + \frac{at^2}{2}$

These are both functions of *time*, but we want to solve for acceleration as a function of *distance* and starting velocity. But that’s just a matter of algebra. We want $\nu(T) = 0$, so we plug that into the first of these equations and solve for $T$:

$0 = v + aT \quad\rightarrow\quad T = \frac{-v}{a}$

And we want $x(T) = d$, so we plug both that and the formula for $T$ into the second equation:

$d = v\frac{-v}{a} + \frac{a}{2}\left(\frac{-v}{a}\right)^2$

Now all we have to do is solve for $a$:

$d = \frac{-v^2}{a} + \frac{v^2}{2a}$

$d = \frac{-2v^2 + v^2}{2a}$

$2ad = -v^2$

$a = \frac{-v^2}{2d}$

Wait, the acceleration comes out to be *negative*?! Yes. That’s how you know the object is slowing down rather than speeding up. (If the object weren’t moving in a straight line, its position, velocity, and acceleration would all have to be treated as 2- or 3-dimensional vectors, but the calculations would wind up being very nearly the same, only with more boldface. Also, if the *velocity* were negative, it would mean the object was moving backward. This is, in fact, the difference between velocity and speed: speed is the *magnitude* of velocity, without the direction, so it can never be negative.)