Can a particle have no instantaneous velocity at all points of the path taken but a finite average velocity? I have a question on kinematics.
Say the path traced by a particle is given by a Koch curve or Koch snowflake. 

Now consider the particle starts from some arbitrary point $A$ on the curve and continues moving with some acceleration. It moves a finite distance on the curve and reaches another point $B$ which is different from $A$ and the particle has not crossed the same point twice.
So there is a net finite displacement covered in a finite time. Hence the particle has a finite average velocity.
But the curve is not differentiable at any point, by definition of the curve. So the particle has no instantaneous velocity at all points of the path taken.
QUESTION: Can a particle have no instantaneous velocity at all points of the path taken but still a finite average velocity?
Is this possible? Can anyone explain this?
 A: No, it's not possible, because one of the underlying assumptions of kinematics is that all paths are at least twice differentiable. Before you complain about this requirement, remember that physics is about building models that can be used to describe and predict measurements. Measurements always have some amount of uncertainty, and even if you suppose that it is possible for a particle to travel along a nondifferentiable path $x(t)$, it is still always possible to construct a twice-differentiable path that matches $x(t)$ to any desired level of precision. That twice-differentiable path is what you use for the model.
Even beyond that, make sure not to mix up "no instantaneous velocity" with "zero instantaneous velocity". Usually we use these terms interchangeably in physics, but we have the luxury of doing so because (we normally assume) paths are always differentiable and thus there is not really any such thing as, literally, having no instantaneous velocity. If you want to work with nondifferentiable paths, then you have to be more careful. It's conceivable that in such a model, a particle could have a perfectly well defined average velocity between any two points in time and yet never have an instantaneous velocity. This is still fine (if useless) because no physical process actually measures instantaneous velocity. The closest you get is an exceedingly short-time average, e.g. over roughly a period of oscillation of an EM wave when using the Doppler effect.
A: The length along any segment of the Koch snowflake is infinite. It has finite area but infinite perimeter. So, for a particle to move from one place on the snowflake to another it would have to travel an infinite distance. This is why differentiability is important.
A: This is an excellent question. It is pedantic to whine about non differentiability ,however there is in fact a point to be made on that topic . We seem to be conflating the derivative dy/dx with the time derivative dx/dt(x is the position here). One may have a differentiable path but still the instantaneous velocity can remain undefined. Whether the converse is also false I do not know. First of all I am not mathematically adept enough to obtain an expression for the koch snowflake or any non differentiable continuous function for that matter. But we do know a simple continuous non differentiable path . Suppose particle A moves with velocity u and suddenly changes direction. At this point the particle certainly has infinite acceleration. But what about the velocity? You seem to think it is undefined. 
A: It is possible if the instantaneous velocity is regularized in appropriate mathematical sense. Take $ \sqrt{x} $ for example. It has no derivative at the origin, yet in any finite interval about 0
$$
\frac{\Delta \sqrt{x}}{\Delta x}
$$
is finite. 
