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If a particle is subject to a constant acceleration $a$, the position of the particle after a certain time is given by the formula $s(t)=\frac{1}{2}at^2$. This immediately results in the formula $v(t)=at$ for the velocity as function of time. However, it is also possible to specify the velocity as a function of the position, the formula for this is $v(s)=\sqrt{2as}$.

I am now wondering the reverse. If someone tells me that the velocity of a particle after travelling a distance $s$ is given by $v(s)=\sqrt{2as}$, can I conclude unambiguously that the velocity as a function of time is given by $s(t)=\frac{1}{2}at^2$? (I strongly suspect that this is the case, but I can't prove it.)

Samuel Renold
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2 Answers2

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The equation $v = \sqrt{2as}$ can be rewritten as $$ \frac{ds}{dt} = \sqrt{2as}. $$ This is a separable ODE, which means that we have $$ \int \frac{ds}{\sqrt{s}} = \sqrt{2a} \int dt \quad \Rightarrow \quad 2 \sqrt{s} = \sqrt{2 a} t + C $$ where $C$ is an arbitrary constant. Demanding that $s = 0$ when $t = 0$ implies that $C = 0$; and solving this equation for $s$ then yields $s = \frac12 a t^2$, as expected.

Michael Seifert
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Given the velocity profile $$ v = v(s)$$ you can extract the time needed to traverse a distance from $0$ to $s$ with the following integral

$$ t(s) = \int_{0}^{s} \tfrac{1}{v(s)}\,{\rm d}s =\int_{0}^s \tfrac{1}{\sqrt{2 a s}}\,{\rm d}s = \sqrt{\tfrac{2 s}{a}}$$

and you reverse this with

$$ s(t) = \tfrac{1}{2} a t^2 $$

Appendix 1

See this post (https://physics.stackexchange.com/a/797383/392) for the integral to use when variable acceleration is known.

John Alexiou
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