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I have an expression for one-dimensional acceleration as a function of position, $a(s)$, and I am hoping to achieve position as a function of time, $s(t)$. Once I have $s(t)$ I am hoping I will be able to find $v(t)$ and $a(t)$ by taking the first and second derivative of $s(t)$. I have been using the following resources for guidance:

youtube - http://bit.ly/2dE2kJs

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

Converting a function for "velocity vs. position", $v(x)$, to "position vs. time", $p(t)$

The first and second link seem to make sense to me. A few months ago I derived a result for $v(s)$ by following the video in the first link. Recently I have found the second link and the similarity between their result and mine leads me to believe that the work may be correct. I will show this work down below soon. The third link I do not understand but it seems to be what I'm after. $\\[10pt]$

My expression for $a(s)$ takes the form $$a(s)=k(\frac{1}{s^2}),$$ where $k$ is a constant. Since $a=\frac{dv}{dt}$ we have $$ks^{-2}=\frac{dv}{dt}.$$ Multiplying the right-hand side by $\frac{ds}{ds}$ and rearranging the expression we arrive at $$ks^{-2}ds=vdv.$$ Integrating gives us $$k\int_{s_0}^s s^{-2}ds =\int_{v_0}^v vdv.$$ When evaluated this becomes $$-k(\frac{1}{s}-\frac{1}{s_0})=\frac{1}{2}(v^2-v_0^2).$$ Isolating for $v$ we achieve $$v(s)=\pm\sqrt{-2k(\frac{1}{s}-\frac{1}{s_0})+v_0^2}.$$ This is the expression that seems to mimic the result in the second link. I have had this expression for some time now. It is what to do next that I am not sure.

So, here are my questions:

  • Is the work that I've done so far correct?
  • Is my basic strategy even possible? My plan is to convert $a(s)$ --> $v(s)$ -?-> $s(t)$. The third link leads me to believe that it is possible.
  • Is that strategy optimal? Is it possible to go from my initial expression $a(s)$ to $s(t)$ without the in between derivation of $v(s)$?
  • How does one actually convert $v(s)$ into $s(t)$? $\\[5pt]$

Part Two

That ends the official part of my post. The following is something that I worked out tonight and I am wondering if it makes any sense whatsoever.

The arguments are as follows:

$$v(s)=\frac{ds}{dt}$$

$$ds=v(s) dt.$$

Since $v(s)$ doesn't depend on time, I treat it as a constant with respect to time

$$\int_{s_0}^sds =v(s)\int_{t_0}^tdt$$ $$s-s_0=v(s)(t-t_0)$$ $$s=v(s)t-v(s)t_0+s_0$$ $$s=\pm\sqrt{-2k(\frac{1}{s}-\frac{1}{s_0})+v_0^2}t-\pm\sqrt{-2k(\frac{1}{s}-\frac{1}{s_0})+v_0^2}t_0+s_0.$$

  • Wouldn't this be $s(s,t)$? That doesn't sound quite right. I suspect the problem is with my statement that $v(s)$ doesn't depend on time. I know intuitively that the velocity does indeed change with time. But yet, the input for my function $v(s)$ only requires position.
  • I'm interested to know why I cannot treat $v(s)$ as a constant with respect to time in this case considering that it appears to depend only on $s$.

Thanks for reading. I haven't been taught this sort of thing and the whole exercise is just for fun.

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  • The result in the first part looks plausible. If you multiply both sides by the (unknown) mass, then after integrating the r.h.s. is the change in kinetic energy, while the l.h.s. is the work done on the body. – amd Sep 30 '16 at 06:18
  • You need to include the v(s) in the integration over s. i.e. ds/v(s). This gives you a nasty integral in s equalling t. – user121049 Sep 30 '16 at 06:30
  • Thank you for your reply, user121049. I also noticed the result that I believe you are referring to.$$v=\frac{ds}{dt}$$$$dt=v^{-1}ds$$$$dt=\pm(-2k(\frac{1}{s}-\frac{1}{s_0})+v_0^2)^{-1/2}ds$$$$\int_{t_0}^tdt=\int_{s_0}^s\pm(-2k(\frac{1}{s}-\frac{1}{s_0})+v_0^2)^{-1/2}ds$$ If anyone has insight on how to solve that one as well I would be interested to learn. – 1MTris-HClpH8point6 Sep 30 '16 at 21:08

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