$A$ 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:

$A$ particle moves in a circular path of radius $r$ with speed $v.$ It then increases its speed to $2v$ while travelling along the same circular path. The centripetal acceleration of the particle has changed by a factor of

The centripetal acceleration is given by

$A$ cyclist is riding with a speed of $27 \; km/h$. As he approaches a circular turn on the road of radius $80 \; m$,he applies brakes and reduces his speed at the constant rate of $0.50 \; m/s^2$. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

$A$ particle moves in a circle with speed $v$ varying with time as $v(t) = 2t$. The total acceleration of the particle after it completes $2$ rounds of the cycle is:

If the radius of curvature of the path of two particles of same mass are in the ratio $3:4,$ then in order to have constant centripetal force,their velocities will be in the ratio of: