Can I draw an acceleration vs. time graph from an acceleration vs. distance graph?

Question

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.

Answer 1

$$x'' = f(x)$$ $$x'x'' = f(x)x'$$ 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 $g^{-1}$.

Answer 2

You want to get ${\bf a}(t)$ from ${\bf a}\bigl({\bf x}(t)\bigr)$.

If you don’t know that $\bf a$ is constant, you want to go with the general

$${\bf x}(t) = {\bf x}(0) + {\bf v}(0)\,t + \frac{{\bf a}(0)\,t^2}{2} + \frac{{\bf a}’(0)\,t^3}{3!}+\cdots$$

Also, if you have $f(u)$, then you know all the derivatives $f^{(n)}(u)$. If you have just ${\bf a}(u)$, however, to complete your information about ${\bf x}(u)$ you need to know ${\bf v}(0)$ and ${\bf x}(0)$. (You can use any point to build your Maclaurin series, by the way.)

You said that the object moved with a known initial velocity from a known initial point. This tells us that you know everything about ${\bf x}(t)$. And if you know everything about ${\bf x}:\text{time}\mapsto\text{position}$, then you could hypothetically compute ${\bf x}^{-1}:\text{position}\mapsto\text{time}$.

Therefore, since you already know ${\bf a}(t)$, we know that you can create a function ${\bf x}\mapsto t \mapsto {\bf a}$.

Now, when you say “draw,” I hope you don’t mean freehand. Nevertheless, I suspect a powerful enough computer could handle this for you to a decent degree of accuracy and precision.