# Velocity as a function of position given acceleration as a function of position?

$v(x)$ given $a(x)$, where $v_0 = 0$ and $x_0 = 0$

I'm clueless. This is what I thought:

$dv = v, dx = x$

$a(x) = \frac{dv}{dt} \frac{dx}{dx} = \frac{dx}{dt} \frac{dv}{dx} = v\frac{dv}{dx} = \frac{v^2}{x}$

$v = \sqrt{a(x)x}$

But that makes no sense. I don't know why. It doesn't make any sense that I wouldn't have to integrate anything. But I don't get what's wrong with that algebra.

$$a(x)=\dfrac{dv}{dt}=\dfrac{dv}{dx}\dfrac{dx}{dt}=v \dfrac{dv}{dx}$$

$$a(x)dx=vdv$$

$$\int_{x_0=0}^{x} a(\bar{x}) d\bar{x}=\frac{1}{2}(v(x)^2-v_0^2)$$

$$\int_{0}^{x} a(\bar{x}) d\bar{x}=\frac{1}{2}v(x)^2$$

$$v(x)=\pm\sqrt{2\int_{0}^{x} a(\bar{x}) d\bar{x}}$$

• Nicely put. It may be worth noting that, if the equations are multiplied throughout by the mass $m$ of the particle, then this is basically just Newton's second law ($F_{net}=ma$) and the work-energy theorem. For instance, the third line becomes $$\int_{x_0=0}^x F_{net}(\bar{x})d\bar{x} = \frac12 m v(x)^2-\frac 12 m v_0^2 = K(x)-K_0$$ where we recognize the LHS as the work done on the particle and the RHS the change in kinetic energy. – Semiclassical Jun 9 '16 at 22:46
• What is $d\bar{x}$? I don't understand what the change in the mean of $x$ means. – user321514 Jun 9 '16 at 22:49
• @TheMostCuriousThing It's just a dummy variable for integration, since $x$ is already being used in the limit of integration. One could also use $x'$, $u$, etc. – Semiclassical Jun 9 '16 at 22:50