Can I draw an acceleration vs. time graph from an acceleration vs. distance graph? An object with a known initial velocity, starting from the origin, moves along a line and its acceleration is graphed as a function of distance from the origin. I want to sketch $ x''(t) $ vs. $t$ given $ x''(t) $ = $ f(x(t))$. I will call these graphs $ a(t)$ and $ a(x)$ respectively. 
For instance, if graph $a(x)$ is linear with slope $k$, I would expect $a(t)$ to resemble a linear combination of $ke^{\pm\sqrt{k}t}$. This is an especially simple case, and I can't figure out a way to solve $x'' = f(x)$ in general. Is this possible? And if not, is there any other way I can transform the graphs (approximately), numerically or graphically or otherwise? $($I would prefer to not have to physically simulate the trajectory and measure :) $)$
Something I thought about is that the area under $a(x)$ represents something like kinetic energy over mass, and hence $\frac12v(x)^2$, but I don't know if this approach can get me anywhere.

$$x'' = f(x)$$
$$x'x'' = f(x)x'$$
Here we can integrate with respect to $t$. $F$ is an antiderivative of $f$.
$$\frac12 (x')^2 = F(x)+C_1$$
$$(x')^2 = 2F(x)+C_2$$
You could plug in initial values for position and velocity, to solve for $C_2$. Also, the sign of initial velocity determines the sign of this square root:
$$x' = \pm\sqrt{2F(x)+C_2}$$
$$\frac{x'}{\pm\sqrt{2F(x)+C_2}} = 1$$
$$\int\frac{dx}{\pm\sqrt{2F(x)+C_2}} = t+C_3$$
Find an antiderivative $g$ to simplify the left expression, and again use the initial position to solve for $C_3$.
$$g(x) = t+C_3$$
$$x = g^{-1}(t+C_3)$$
Now you can differentiate twice, if you want to find acceleration.
One possible problem is that $g$ may not be strictly invertible; $g^{-1}$ would be multi-valued. You could try, again, using initial position to find the correct branch of $g^{-1}$.
