# Instataneous acceleration where velocity vs. time graph is not differentiable?

In the diagram below, what is the instantaneous acceleration at t=30s? There's a kink in the graph at that point; so in my opinion the instantaneous acceleration is undefined. Any thoughts appreciated.

• Welcome to physics.SE. This is a good first question. It would be even better if you made it more conceptual instead of asking in terms of a specific problem (of course it is not wrong to refer to a specific problem to illustrate your conceptual question). Jan 6, 2022 at 18:37
• Jan 6, 2022 at 22:40

The instantaneous acceleration is defined as $$a(t) = \frac{dv(t)}{dt}$$ therefore when $$v(t)$$ is not $$C^1$$ it can't be defined. What you can do in this case is take a finite interval enclosing your discontinuity and compute the average acceleration in that interval simply as $$a = \frac{\Delta v}{\Delta t}$$. If you are doing this from experimental data however you are dong this already, so you really don't have a problem.

In practice: you just compute an average acceleration for for every couple of points you have like this $$a_n = \frac{v_{n+1}-v_n}{t_{n+1}-t_n}$$ and that's well defined. This is approximating the real acceleration in the limit of infinite points.

• Nitpick: Smooth is usually be used to mean being an element of $C^\infty(\mathbb{R})$ while for $a(t)$ to exist pointwise $v \in C^1(\mathbb{R})$ is (more than) sufficient. Jan 6, 2022 at 19:32
• you are right I specified it :) Jan 6, 2022 at 19:39
• Thank you both for the thorough and insightful answers. Jan 7, 2022 at 19:11

You can see this question as a first step into the wonderland of weak derivatives and things like Sobolev spaces and distributions.

While there is no derivative at $$t = 30\,\mathrm{s}$$, this curve can still be seen as a solution of $$m\dot v = F(t).$$ for a certain time dependent force.

Intuitively, that the derivative is not defined in a single point does not really matter, since if we take the integral solution $$v(t) = v(0) + \int_0^t dt' \, \frac{F(t)}{m}$$ then this will happily give us solutions $$v(t)$$ for a much larger class of functions than those that fulfil the equation point-wise. (The value of $$a$$ at a single discrete point does not matter for the value of the integral.)

You can formalize this notion by introducing Sobolev spaces and weak derivatives. Then the differential equation will be fulfilled as an equation of elements in the Sobolev space, given that you interpret the derivative as the weak derivative in that space.

The weak derivate is still not defined at the point (since it is the element of an $$L^p$$-space so it is an equivalence class of functions that differ only on a set of measure zero), but it is a well-defined mathematical object. And the approach can be generalized to partial-differential equations.