Recently I came across the concept of a flat function, which is a smooth function $f:\mathbb{R}\to\mathbb{R}$ all of whose derivatives vanish at a given point $x_0\in\mathbb{R}$, the canonical example being

$$ f(t) = \begin{cases} e^{-1/t^2},& t\neq 0\\ 0,& t=0. \end{cases} $$

Now I wonder if such a function can in any way describe the trajectory of a physical object where $t$ is time and $f(t)$ is its position. The reason why I am not sure is as follows: If I throw a ball in the air, its velocity becomes zero at the highest point. Naively I could reason that it must stay there, because it cannot move anymore once the velocity is zero. But we all know that this reasoning is flawed, because the 2n derivative, acceleration, is not zero at this point, which is why the ball starts picking up speed again and comes down to earth.

But what if the 2nd derivative, acceleration, would vanish too, and the third, and so on infinitely. Would this mean that the ball comes to a halt at this point. And, by generalizing from the ball to any particle, that the $f(t)$ shown above cannot be the trajectory of a physical object? Rather the function would have to be constant at least on one side of the flat point?


Surely this just depends on the potential well in which the particle finds itself. If the function you describe is the solution of the equation of motion, then that will indeed be the motion of the particle. The function you wrote will give rise to a path that "never ends" - the particle keeps slowing down, but never stops.

Of course in practice there are always going to be some forces that will stop this "eternal motion over infinitesimal distance" - because at some scale (of time / space) quantum effects will take over and dominate. But if you know the position of the object at any time, you know the velocity (derivative of position) and the acceleration (derivative of velocity). Given the mass, this tells you what the force is that needs to be applied.

Applying that force will cause that path...

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