# 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 $a(x)= \frac{x}{s^2}$.

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It's not a homework but OK. – WindScar Oct 17 '11 at 0:50
The homework tag doesn't just apply to actual homework problems here, it applies to any questions of an educational nature. That being said, I think this might be general enough that it doesn't actually need the homework tag - especially if you were to remove the request for an example. – David Z Oct 17 '11 at 1:20

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

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this is the explicit solution... i guess david zaslanvasky didn't want to do your hw for you and gave you the answers to other similar problems – Timtam Oct 17 '11 at 4:26

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.

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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.

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