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 $m/s$?

$A$ particle moves along a circle of radius $3 \, m$ with its displacement given by $S = \frac{t^2}{2} + \frac{t^3}{3}$. The total acceleration at $t = 2 \, s$ is ....... $m/s^2$.

$A$ car is travelling with linear velocity '$V$' on a circular road of radius '$r$'. If its velocity is increasing at a rate of '$a$' $ms^{-2}$, then the resultant acceleration will be

$A$ particle moves in a circular path of radius $R$ with an angular velocity $\omega = a - bt$,where $a$ and $b$ are positive constants and $t$ is time. The magnitude of the acceleration of the particle after time $t = \frac{2a}{b}$ is:

$A$ car is moving on a circular path of radius $600\,m$ such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete the first quarter of a revolution,if it is moving with an initial speed of $54\,km/h$,is $t(1 - e^{-\pi/2})\,s$. The value of $t$ is $.............$.

For a particle in a uniformly accelerated circular motion: