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The given is the distance but the variable is the velocity $x(v)=k \,v^2$. I want to find the velocity $v(t)$.

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The given relation implies that at all times $t$, the function $x(t)$ obeys $$ x(t) = k \left( \frac{dx(t)}{dt} \right)^2 \quad \Rightarrow \quad \frac{dx(t)}{dt} = \pm \sqrt{\frac{x(t)}{k}} $$ This is just a differential equation for $x(t)$ and so can be solved for $x(t)$ using standard ODE techniques. (Note, in particular, that it is a separable equation.) Once this is done, the result can be differentiated to find $v(t)$.

  • This is only true if the LHS is meant to mean $x(v):=x(v(t)):=x(t)$, no?! – Tobias Fünke Sep 02 '22 at 13:41
  • @JasonFunderberker: I'm not quite sure what you mean there, but I've tried to clarify what I mean in the first sentence. If it doesn't address your concerns, please follow up. – Michael Seifert Sep 02 '22 at 13:44
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$x = kv^2 \\ \displaystyle \Rightarrow v = \frac {dx}{dt} = 2kv \frac {dv}{dt}$

One trivial solution is that $x=v=0$ for all $t$. However, if $v \ne 0$ then we have

$\displaystyle 2k \frac {dv}{dt} = 1 \\ \displaystyle \Rightarrow \frac {dv}{dt} = \frac 1 {2k}$

You can integrate this to find $v$ as a function of $t$. Integrating introduces a constant of integration, which you can set equal to the initial velocity $v|_{t=0}$.

gandalf61
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