# Can position be derived from acceleration in practice?

We know that acceleration is the derivative of velocity, and velocity is the derivative of position. But does that mean that we can find position from acceleration in practice (as opposed to in theory – say, in a basic physics textbook problem)? For instance, in a basic physics textbook problem, we might be given an acceleration of $$a \ \dfrac{\text{m}}{\text{s}^2}$$, and be asked to find the position. We then take the antiderivative of this to get velocity, and then the antiderivative of the velocity to get position. And although this resulting closed-form solution might be valid/sufficient for a textbook problem, it would still have the constant of integration, and so wouldn't actually be a position in practice, right? So, in the real world, when we are given some acceleration and want to use it to find the position, how does it actually work?

• The constants of integration are determined by initial conditions. Commented May 16, 2021 at 5:53
• @G.Smith Ahh, ok, that makes sense. But how would this work in practice? That is, what are these "initial conditions" in practice / how would these "initial conditions" be found? A simple, practical example would be nice (to better illustrate the idea behind this). Commented May 16, 2021 at 5:54
• For a single point particle, the initial conditions are its position and its velocity at some instant. Example: A pendulum is released from rest at an angle of 15 degrees from the vertical. Commented May 16, 2021 at 5:57
• @ThePointer take a simple practical example and try to work out how the integration would work, then you will see how to use the initial conditions, or others will help out where you are stuck Commented May 16, 2021 at 6:01
• @G.Smith Hmm, ok. But we obviously need to establish some "frame of reference" (I don't necessarily mean "frame of reference" as used in the relativity sense) / "coordinate frame" in order for "position" to make sense, right? This seems implicitly already done for us in a basic textbook example, but, in real life, one could imagine many circumstances where we need to establish this ourselves, right? (These are just my thoughts, not sure if this makes sense.) Commented May 16, 2021 at 6:01

If we know the initial position and velocity then yes double integration over the acceleration is enough to find your position after some time. Otherwise, commercial accelerometers such as linear accelerometers and gyroscopes, which are widely used in navigation would be pointless. An example, if our initial position is $$\textbf{r}_0$$ and initial velocity $$\textbf{v}_0$$, and we assume a constant acceleration $$\textbf{a}$$, then the position after some time $$t$$ is found by first finding the velocity by integrating
\begin{align}\textbf{v}(t)&=\int_0^t\textbf{a}dt+\textbf{v}_0\\ &=\textbf{a}t+\textbf{v}_0.\end{align}
\begin{align}\textbf{r}(t)&=\int_0^t\textbf{v}(t)dt+\textbf{r}_0\\ &=\textbf{r}_0+\textbf{v}_0t+\frac{1}{2}\textbf{a}t^2.\end{align}