Can we calculate centripetal acceleration by using this method $\frac{\mathbf v_2-\mathbf v_1}{T}$? If we know the angle between two velocity vectors $\mathbf v_1$ and $\mathbf v_2$, and if we know the time $T$ it takes for the velocity to change from $\mathbf v_1$ to $\mathbf v_2$,then is it possible to find the centripetal acceleration by using $$\frac{\mathbf v_2-\mathbf v_1}{T}$$
 A: Only if you take to the $t\to 0$ limit. Or perahps, in the case that acceleration is a constant vector --- in which case the figuring out ${\bf a}$ from $({\bf v}(t)- {\bf v}(0))/t$ requires the solution of the  vector differential equation
$$
\frac{d{\bf v}}{dt}={\bf a}.
$$
This last case does not apply for circular motion at a constant speed anyway as the acceleration vector is time dependent. In that case taking the limit is essential.
A: $$\frac{\mathbf v_2-\mathbf v_1}{T}$$
This expression gives the average acceleration of the body due to the resultant force acting on it.
Provided that there is no tangential acceleration,this expression will give you the centripetal acceleration iff $t\to 0$.
A: You can't do that unless you are in the limiting case. 
If you do it in the regular case then that would give you the average acceleration to move between two points. 
If you want to find the exact amount of acceleration that pulled the velocity vector to be tangent to the curve after some amount of time 't' then you can only do that under the limit. 
A: Thiknk of it in terms of calculus. See in a uniform circular motion, the velocity at any point in the motion remains constant. If your motion were to be non-uniform then you also account for the tangential component the equation being $$\vec{a}=\vec{a_c} +\vec{a_t} $$.So the answer to your question would be; As long as T approaches zero you will end up with instantaneous acceleration at a point.
