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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!

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    $\begingroup$ Suppose you have a point moving along a circle. The point's distance from the center never changes, yet it's velocity is non-zero. What gives? $\endgroup$
    – David H
    Dec 31, 2013 at 13:55
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    $\begingroup$ The magnitude of $\vec{v}$ is constant, but not it's direction. $\endgroup$
    – jinawee
    Dec 31, 2013 at 14:04
  • $\begingroup$ There is much wrong with your question - namely the units don't work out for most of your equations except for the very first one. It's hard then to decipher what exactly you're asking. Jinawee makes a good point that you should keep in mind: Acceleration is defined by the $\frac{d \vec{v}}{dt}$ and since $\vec{v}$ is a vector, then if the direction of $\vec{v}$ changes in time (even with it's magnitude constant) then there is a non-zero acceleration. $\endgroup$
    – mcFreid
    Dec 31, 2013 at 15:00
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    $\begingroup$ See physics.stackexchange.com/q/38291 and physics.stackexchange.com/q/91797 $\endgroup$ Dec 31, 2013 at 15:55
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    $\begingroup$ I have no idea what you mean by "vector is vector." Magnitudes can have units. For example, the magnitude of the velocity is the speed and it has units of $\frac{m}{s}$. $\endgroup$
    – mcFreid
    Dec 31, 2013 at 16:37

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Physical meaning of $a$ is the length of an arc swept out by velocity vector.

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