$A$ particle is moving on a circular path with constant speed,then its acceleration will be

The centripetal acceleration is given by

$A$ particle is in uniform circular motion. The equation of its trajectory is given by $(x-2)^2+y^2=25$,where $x$ and $y$ are in meters. The speed of the particle is $2 \text{ m/s}$. When the particle attains the lowest $y$ coordinate,the acceleration of the particle is (in $\text{m/s}^2$):

$A$ particle moves along an arc of a circle of radius $R$. Its velocity depends on the distance covered as $v = a\sqrt{s}$,where $a$ is a constant. Then the angle $\alpha$ between the vector of the total acceleration and the vector of velocity as a function of $s$ will be:

In the given figure,$a = 15 \, m s^{-2}$ represents the total acceleration of a particle moving in the clockwise direction in a circle of radius $R = 2.5 \, m$ at a given instant of time. The speed of the particle is ........ $m/s$.

In circular motion,the centripetal acceleration is given by