How could I find the acceleration of a body knowing only that its position and velocity satisfy $v^2 = f(x)$, where $f(x)$ is a known function of $x$ (position)?

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    $\begingroup$ Hi Maxxtack, and welcome to Physics Stack Exchange! I edited your question to make it more generally useful. $\endgroup$ – David Z Nov 9 '11 at 0:45

Let's say you're looking for acceleration as a function of position. Starting from

$$v^2 = f(x)$$

you can differentiate the equation to get

$$2v\frac{\mathrm{d}v}{\mathrm{d}t} = f'(x)\frac{\mathrm{d}x}{\mathrm{d}t} = f'(x)v$$

Then you can solve for acceleration as

$$a(x) = \frac{1}{2}f'(x)$$

For your particular case, since you have a condition for $v^2$, another way to do it is to express your equation as

$$\frac{1}{2}mv^2 - \frac{m}{2}f(x) = 0$$

This corresponds to a particle moving under the influence of a potential energy function $U(x) = -\frac{m}{2}f(x)$. You can then determine the force on the particle as

$$F = -\frac{\mathrm{d}U}{\mathrm{d}x} = \frac{m}{2}f'(x)$$

and the acceleration as

$$a(x) = \frac{F}{m} = \frac{1}{2}f'(x)$$

If you need acceleration as a function of time, then you can use the techniques described in this other question to convert $a(x)$ to $a(t)$.


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