Convert acceleration as a function of position to acceleration as a function of time? Suppose I have acceleration defined as a function of position, $a(x)$. How to convert it into a function of time $a(t)$? Please give an example for the case $a(x)= \frac{x}{s^2}$.
 A: Conserning the general case, there is a way to use potential and kinetic energy (every acceleration corresponds to a force on a point mass, and every force in 1 dimension has a potential energy V such that $F=\frac{-dV}{dx}$), but there is a more mathematical solution:
You know $a(x)$ (and, thus, $F(x)$), $v_0(x_0)$ and $t_0(x_0)$ as boundary conditions (otherwise the answer is ambiguous). You need to find $a(t)$.
The general definition:
$$a(x)=\frac{d v(x)}{d t}=\frac{d v(x)}{d x}\frac{d x}{d t}=\frac{d v(x)}{d x}v(x)$$
Recombine the differentials:
$$a(x)dx=v(x)dv(x)$$
Integrate both parts:
$$\int_{x_0}^x a(x)dx=\int_{v_0}^v v(x)dv(x) = \frac{v^2-v_0^2}{2}$$
Solve for $v(x)$:
$$v(x)=\sqrt{v_0^2+2\int_{x_0}^x a(x)dx}$$
Now, let's find $t(x)$. We will first find it`s derivative:
$$\frac{d t(x)}{d x}=\frac{1}{v(x)}$$
Again, split the differentials and integrate:
$$\int_{t_0}^t dt=\int_{x_0}^x \frac{dx}{v(x)}=\int_{x_0}^x \frac{dx}{\sqrt{v_0^2+2\int_{x_0}^x a(x)dx}}$$
Evaluate the leftmost integral:
$$t=t_0+\int_{x_0}^x \frac{dx}{\sqrt{v_0^2+2\int_{x_0}^x a(x)dx}}$$
Solving for x allows you substitute it and get $a(t)$.
This works perfectly when $x(t)$ is a 1-to-1 correspondence, but fails (to give the right answer), when the motion changes direction. In your case it seems that 1/s^2 is meant to be a positive real, so it should work fine.
A: rewrite as the function:
$ a_{(x)} = \frac {x}{s^2}   = \ddot {x} $
this differential equation can be rewritten as:
$ \ddot {x} - \frac{x}{s^2}   = 0 $
interestingly we need not go any further since obviously the 2nd derivitive must be equal to the function itself. THe obvious solution is some linear combination of hyperbolic functions
$a_{(t)}  =  A \sinh(st) + B \cosh(st)  $
etc.
if we are treating x as an independent variable, I think this is the only correct solution. this is not a SHO
A: The way to do this is to express position as a function of time, then for any time you can calculate the corresponding position and thus the acceleration.
$$a(t) = a(x(t))$$
So basically, you need to find the function $x(t)$. To do so, you need to solve the differential equation
$$a(x) = \ddot{x}$$
where $\ddot{x}$ denotes the second derivative of $x$ with respect to time.
In general, the solution to this equation is quite complicated. But there are certain special cases that are easily solvable. One that comes up very often in the solution of gravitationally bound systems (such as orbital motion) is
$$a(x) = -C x^{-2}$$
($C$ is a positive constant) which is discussed in this other answer I posted a while back. Another common case - in fact, probably even more common - is the simple harmonic oscillator,
$$a(x) = -C x$$
which has the solution
$$x(t) = A\cos(\omega t) + B\sin(\omega t)$$
where $\omega = \sqrt{C}$. You can also do the trivial case of constant acceleration,
$$a(x) = -C$$
which corresponds to such things as straight-line projectile motion.
