If a velocity curve is a smooth one, does the acceleration curve need to be smooth as well? My physics professor says that an acceleration curve can be jagged.

Question

anonymous · Answer

that is more of a science question than math don't you think?

anonymous · Answer

I suppose it firstly depends on what you define as "smooth."

I would take it to mean that there are no "sharp turns" or cusps, like in the following two graphs:
|dw:1377792581207:dw|
In this case, you could describe smoothness of a curve as "continuity of the derivative:"
$$\lim_{t\to c^-}f'(t)=\lim_{t\to c^+}f'(t)$$
As you can see in the graph of $f$ above, the derivative to the left of $c$ is 0, while the derivative to the right is some positive number. So by this quick definition, the curve would not be considered smooth.

Now, if you wish to keep using this definition, it would be your task to find some function $v(t)$ where $\displaystyle\lim_{t\to c^-}v'(t)=\lim_{t\to c^+}v'(t)$, but $\displaystyle\lim_{t\to c^-}v''(t)\not=\lim_{t\to c^+}v''(t)$