The Concept of Instantaneous Velocity The concept of instantaneous velocity really becomes counter-intuitive to me when I really think deeply about it. Instantaneous velocity is the velocity of something at an instant of time; however, at the very next instant the velocity changes. In general, speed tells us how quickly something is changing its position with respect to some point in space. Then by instantaneous velocity we mean that the object at some instant is changing its position with respect to time, but may change to a velocity much faster. But if it is changing its position with that rate then the time passes by and that instant as well and at another instant the velocity is different. Did it even move with that velocity for a moment? I’m sorry that I’m not exactly able to tell you what my problem is; however, it is something like I've described. 
 A: 
Instantaneous speed is the speed of something at an instant of time...

It may not be productive to think of instantaneous speed in this way.  The concept of instantaneous speed should be no more difficult to grasp than the concept of instantaneous slope which is to say that it is to be understood in the context of limits.
Recall that, in words, the instantaneous slope of a curve at a point is the slope of the line tangent to the curve at that point.  This slope is just the limit of the average slope of the curve as the displacement from that point goes to zero.
Now, if we measure a particle's position at one instant and at another later instant, the average speed is just the ratio of the change in position to the change in time:
$$\bar v = \dfrac{\Delta x}{\Delta t}$$
Now, imagine taking the position measurements closer and closer together in time.  Assuming the limit exists, the instantaneous speed is simply the limit of the average speed as the time between position measurements goes to zero.
A: Maybe it helps to change your perspective. Consider a record spining on a record table with constant rotational speed, and you note that at some time the part of the record under the needle has speed $\vec{v}_A$ where A is the location of the needle. Now some time later the same part of the record has different speed because it is located elsewhere, but under the needle the speed is still $\vec{v}_A$ (with a different part of the record being observed).
So when the body (record) goes through constant motion the spatial velocity (velocity at any fixed point in space) is constant also. The instantaneous velocity reflects the velocity of whatever part you are observing (in our case point A). 
Now the non-constant speed case. Again at any instant the spatial velocity is some value of time $\vec{v}_A(t)$. Since the point of observation is fixed in space (and time) the instantaneous velocity is well defined. The time derivative $\dot{\vec{v}_A}$ is called the spatial acceleration which is different from the material acceleration (the acceleration of a point on the moving body). When the rotational acceleration is zero so is the spatial acceleration. The instananeous velocity is constant in this case. In general the instanteneous velocity although varies, it always answers the question what is the speed of rigid body under an observation point.
As a side note, at any instant, if you observe the instanteneous velocity to be $\vec{v}_A$ under a point located at $\vec{r}_A$, and the body is rotating with $\vec{\omega}$ then the instant center of rotation is located at
$$\vec{r}_{cen} = \vec{r}_A + \frac{\vec{\omega} \times \vec{v}_A}{|\vec{v}_A|^2} $$
where $\times$ is the vector cross product, and $|\cdot|$ is the magnitude of a vector. The instant center of rotation is the point in space (actually a line) with either zero velocity, or velocity that is only parallel to the rotation axis.
Example
A record is rotating with $\vec{\omega} = 10 \hat{k}$, and I observe the speed under the needle located at $\vec{r}_A = (4.5, 1.2, 0)$ as $\vec{v}_A = (8.4,4.7,0)$ where is the center of rotation of the record (or frisby!)
$$\vec{r}_{cen} = (4.5,1.2,0) + \frac{ (0,0,10) \times (8.4,4.7,0) }{|(0,0,10)|^2} 
\\ = (4.5,1.2,0) + \frac{ (-47,84,0) }{100} = (4.03, 2.04, 0) $$
Verification
In the problem above check the instantenous velocity at A.
$$\vec{v}_A = \vec{\omega} \times (\vec{r}_A - \vec{r}_{cen}) \\ 
 = (0,0,10) \times (0.47, -0.84, 0) = (8.4, 4.7, 0) \; \checkmark$$
A: Anything that changes in speed will do so as a result of momentum transfer into the object. This requires some interaction with other forces. Force fields are the source of momentum transfer, this transfer takes some time, thus we get acceleration.
