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.