The acceleration of a particle performing non-uniform circular motion is shown in the diagram at point $P$. If the radius of the circular path is $2 \ m$,then what is the velocity at point $P$ in $m/s$?

In a circular motion of a particle,the tangential acceleration of the particle is given by $a_t = 12 \ m/s^2$. The radius of the circle is $3 \ m$. The particle was initially at rest. The time after which the total acceleration of the particle makes an angle $45^{\circ}$ with the radial acceleration is

In non-uniform circular motion, the ratio of radial acceleration to tangential acceleration is (where $V$ is the velocity, $r$ is the radius, and $\alpha$ is the angular acceleration).

$A$ particle is moving in a circular path of radius $r$ under the action of a force $F$. If at an instant the velocity of the particle is $\vec{v}$,and the speed of the particle is increasing,then:

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 ........ $m/s$.

$A$ particle moves in the $xy$-plane with velocity $\vec{v} = x \hat{i} + yt \hat{j}$. At $t = \frac{x \sqrt{3}}{y}$,the tangential and normal accelerations are: