Provided some initial point $x(0)$, how do I convert the function for velocity vs. position, $v(x)$, into a function for position vs. time, $x(t)$, with time derivative $v(x(t))$?
Constant acceleration is not guaranteed. Surely this must always be possible?
Physics – Converting Velocity vs. Position Function $v(x)$ to Position vs. Time $p(t)$
physics
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Best Answer
In one dimension and assuming that the velocity is never zero, the velocity at time $t$ is $v(x(t))$ and also $\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}x(t)$, hence $\mathrm{d}t=\displaystyle\frac{\mathrm{d}x}{v(x)}$, which is solved by $$ t=\int_{x(0)}^{x(t)}\frac{\mathrm{d}z}{v(z)}. $$ This can be rewritten as follows: for every $q$, let $$ U(q)=\int_{0}^{q}\frac{\mathrm{d}z}{v(z)}, $$ then, for every nonnegative $t$, $t=U(x(t))-u_0$ with $u_0=U(x(0))$, hence $$ x(t)=U^{-1}(t+u_0). $$ Example If $v(x)=\mathrm{e}^{−x}$, then $U(q)=\mathrm{e}^{q}−1$ hence $U^{−1}(s)=\log(1+s)$ and $u_0=\mathrm{e}^{x(0)}−1$, which gives $x(t)=\log(\mathrm{e}^{x(0)}+t)$ for every nonnegative $t$.