Do there exist motions in which the second derivative of position is not well defined? After some thought today, I realized that all the power of Newton's laws is fundamentally rooted in the fact that $\frac{d^2 x}{dt^2}$ is a sensible thing to write i.e: there exists a function  $x$ modelling the position of the particle such that the second derivative of $x$ is governed by the following equation:
$$ m\frac{d^2 x}{dt^2}  = \sum_i F_i$$
where $m$ is the mass of the particle and $\sum_i F_i$ is the net force on the particle
Does there exist edge cases where the second derivative of the particle/rigid body's position function does not exist? If so, could we modify the derivative operator in regular newton law's to be applicable in those cases as well?
 A: As physicists we usually don't get lost into existential questions about it. If some case would arise where this thing is not differentiable, it would likely be so bizarre that it would be obvious that the classical approach has some problems. But I can guess you are more a mathematician at heart, and I will try to bring some elements along those lines
In classical mechanics:
Well, the realm of classical mechanics is pretty much defined as those systems that can be described by Newton’s equations (I think). So, well, by definition, the acceleration has to be well defined in any classical mechanics situation.
 
But we can look at what it means. I think we can agree on the fact that the speed of a particle (in the generic sense) is a continuous function of time, as it could not make sense on an energetic point of view to jump instantly from a speed to another. That does not make it differentiable just yet; a continuous function could still have some nasty behavior at specific points.
 
Let’s assume that the speed is differentiable on at least some interval $[t_1,t_2[$, but some nasty thing makes it non-differentiable at $t_2$ (for a speed to never be differentiable on any interval at all, it would have to be so insanely discontinuous that I do not take that case into account in a reasonable physical framework). We can then define the acceleration $a(t)$ for times t in $[t_1,t_2[$. The non-differentiability in $t_2$ implies that the following condition is not met:
 
$$\lim_{\epsilon \to 0^+} \frac{v(t_2+\epsilon)-v(t_2)}{\epsilon}=\lim_{\epsilon \to 0^-} \frac{v(t_2+\epsilon)-v(t_2)}{\epsilon}=a(t_2)$$
 
Well, let’s consider the two reasons why the above condition could not be met. Let’s first imagine that at this specific time $t_2$, this limit takes an infinite value. That means I can pick an arbitrarily small $\epsilon>0$ such that $a(t_2-\epsilon)$ is arbitrarily large. That would require through newton’s law (which is still defined at $t_2-\epsilon$), an arbitrarily large force. I’m sure you can see how that would be unphysical to have an infinitely large force.
 
In the second case, where $a(t_2^-)$ and  $a(t_2^+)$ are finite but are not equal (think about the absolute value function in 0), that means that acceleration is defined on both sides, but is basically discontinuous in $t_2$ (and so not defined at that point). That implies that your force on your system has been been changing infinitely fast at $t_2$. Which means that some information has travelled faster than light. And that would piss off Einstein. And we kinda like the guy so we avoid doing that.
 
So it would be quite unphysical to have a time where you can not define the acceleration.
 
In other formalisms:
Well, the law you are referring to simply does not hold in many contexts. In quantum mechanics for example, position itself is ill-defined, so that’s an issue right from the start.
In fact, the notion of momentum is more general than the idea of a second derivative of position. In different formalisms, momentum may take different forms. In quantum mechanics, you can define the momentum of a monochromatic plane wave as $p=h/\lambda$. In a relativistic framework, it looks more like
$$p=\frac{m_0v}{\sqrt{1-v^2/c^2}}$$
So that already makes quite a difference from the quantity in Newton’s law
The equations of motions are then generalized using a Lagrangian formalism, rather than Newtonian, so the equation you mention becomes quite irrelevant in those fields.
 
A: Consider a particle on a table. Let it have an initial velocity towards the edge of the table. At some point, the particle leaves the table. At this point, the acceleration is undefined. The vertical component of acceleration experiences a jump from 0 to -1 g. This is because the normal force from the table suddenly vanishes. Discontinuity in normal force implies discontinuity in acceleration.
This is a case of infinite jerk. The jerk vector is the time derivative of the acceleration vector.
Another example is to consider a roller coaster design where you have a straight track going into a perfect circle. You can imagine the straight portion of the track as being tangent to the circle. When the roller coaster transitions into the circular portion of the track, it experiences a discontinuity in acceleration, leading to infinite jerk. In order for the coaster to stay on the track, a centripetal acceleration must suddenly appear. This type of roller coaster design was tried in the early days and resulted in death, neck, and back injuries.
https://www.youtube.com/watch?v=7C5kxkBPhpE See 10:33.
A: In Brownian motion particles bounce around randomly because of jostling from the surrounding medium. Mathematically this can be modelled as a random process that does not have any derivatives. Statistical mechanics generalizes Newtonian mechanics for this kind of motion with random processes.
In practice particles move in a straight line until they get deflected after a collision; if one regards the collision interaction as instantaneous (which may make sense in particle physics if not in normal fluids) then the second derivative is always zero except for at random points where it is undefined. Depending on how strict you want to be this may mean a lack of a second derivative function, or just that it is not always defined.
