If the displacement of an object is not differentiable at some point, say $x(t)=t\sin(1/t)$ at $t=0$, how is its instant $v$ defined? If instant velocity at any given time $t_0$ is defined as the derivative of $x(t)$ at $t_0$, what if the derivative does not exist? How are we supposed to deal with $x(t)=|t|$ at $t=0$, or for more complex examples, how are we supposed to deal with the cases (i.e. define instant velocity at that 'bad' point) such as $x(t)=t\sin(1/t)$ at $t=0$ if we add $x(0)=0$?
 A: If $x(t)$ isn't differentiable at some $t_0$, then $v(t_0)\equiv x'(t_0)$ isn't defined.  That's what it means for a function not to be differentiable.
If you argue that the instantaneous velocity of a particle should always be well-defined, then you're saying that the position function should be everywhere differentiable.  Similarly, if you argue that the acceleration (and via Newton's 2nd law, the net force on the particle) should always be well-defined, then you're saying that the position function should be everywhere twice differentiable.
Both of these requirements are perfectly reasonable physical constraints to place on the trajectories of particles.  That being said, sometimes it's convenient to put them aside in favor of an unphysical - but simpler - model.  For example, your example $x(t) = v_0 |t|$ might be a model for a ball with speed $v_0$ undergoing a perfectly elastic, instantaneous collision with a wall at time $t=0$.  As long as you aren't interested in the precise details of the moment of collision, this is fine.  However, if you want a more physical picture, you might consider something like $x(t) = v_0 t \cdot \tanh(t/T)$, where $T$ encodes an effective collision time (an instantaneous collision would correspond to the limit $T\rightarrow 0$).
If you do this, you get trajectories which look like this:

                           

which correspond to velocities like this:

                           

As you can see, as $T\rightarrow 0$ the shift from $v=-v_0$ to $v=v_0$ happens more and more sharply, but as long as $T\neq 0$ the velocity (and acceleration, and indeed all other derivatives) are always well-defined.
Of course, this is just a model I made up.  If you'd like to actually calculate the (physical) trajectory for a squishy elastic collision with a wall, you'd need to model the ball as an elastic object, but that is a much more complex discussion.
A: For the $x(t)=|t|$ case, we, physicists, say that "there's no problem!"
There is no problem differentiating $x(t)=|t|$. It's just $v(t)=\dot{x}(t)=-1+2\mathrm{H}(t)$, where $\mathrm{H}(t)$ is the Heaviside step function! Even more, the acceleration is also well-defined: $a(t)=\ddot{x}(t)= 2\delta(t)$. No problem at all! (Sorry for mathematicians who complain about our sloppiness. Haha..) For physicists, nearly all "daily life" functions can be differentiated infinitely many times. (Different notion of "differentiability'' as compared to mathematicians.)
And here comes your interesting question: $x(t) = t \sin(1/t)$. In this case, an infinite number of oscillations occur within finite time! In my personal physics-oriented viewpoint, this will be related to the interesting physics of supertask. You may find it intriguing to have a look at this link: https://plato.stanford.edu/entries/spacetime-supertasks/
A: The answer to your question on how we deal with them is that we don't deal with the, they are unphysical since they cannot arise in any actual physical scenario. As other answers have noted, however, they may be useful to apply in intervals in which the position is defined.
It is also worth noting which force those position functions would they arise from Newton's second law. For example, your function $x(t)=t\sin(1/t)$ arises from the force $F=-Kx/t^{4}$ (constant added for dimensional reasons), which since energy is not conserved, the force is time dependent, may be useful to describe a specific type of open system, one which interacts with exterior matter, in the differentiable intervals.
