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)?
0
$\begingroup$
$\endgroup$
-
2$\begingroup$ 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
1
$\begingroup$
$\endgroup$
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 $a(t)$.
