We have this formula for centripetal acceleration - $a = v \frac{d\theta}{dt} = v\omega = \frac{v^2}{r}$
but in case of usual acceleration i know that speed in $t_1 = v_0 + a \cdot (t_1 - t_0)$
but in circular case i don't understande the nature of acceleration, speed is always the same if motion is uniform, but acceleration is not zero.
EDIT: Suppose T is 1. $t_0 = 0, t_1 = .25$, so $v$ was rotated for $\frac{\pi}{2}$. So $\frac{\bar{a}}{4} = \bar{v_1} - \bar{v_0}$ and $|a| = 4\sqrt{v^2 + v^2} = 4\sqrt{2} v$, but with previous formula magnitude of a is $\frac{v^2}{4r}$ .
ANSWER: physical meaning of $a$ is the length of an arc swept out by velocity vector!