# What are the scalar equations displacement of acceleration obeys the inverse-square law?

In basic high school physics/calculus you learn that you can formulate equations for velocity and displacement under constant acceleration as:
$a\left(t\right)={a}_{0}$
$v\left(t\right)={a}_{0}t+{v}_{0}$
$x\left(t\right)=\frac{1}{2}{a}_{0}{t}^{2}+{v}_{0}t+{x}_{0}$
My question is how would you formulate a similar equation when acceleration is dependent on the inverse-square of distance from a point, such as Coulomb's Law or Isaac Newton's inverse-square law of universal gravitation?
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nutnhonyl8
Since $a=\frac{dv}{dt}=\frac{dx}{dt}\frac{dv}{dx}=v\frac{dv}{dx}=\frac{d}{dx}\left(\frac{1}{2}{v}^{2}\right)$, integration gives ${v}^{2}={v}_{\mathrm{\infty }}^{2}-\frac{2k}{x}$. Then $t=\int \frac{dx}{v}=\int \frac{dx}{\sqrt{{v}_{\mathrm{\infty }}^{2}-\frac{2k}{x}}}$. If you evaluate that integral (hint: substitute $x=\frac{2k}{{v}_{\mathrm{\infty }}^{2}}{\mathrm{csc}}^{2}\theta$), you'll have t as a function of x, not the other way round.