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

$A$ uniform circular disc of radius $50\, cm$ at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of $2.0\, rad \,s^{-2}$. Its net acceleration in $m\,s^{-2}$ at the end of $2.0\,s$ is approximately (in $.0$)

$A$ particle is moving along a circular path of radius $R$ in such a way that at any instant the magnitude of radial acceleration and tangential acceleration are equal. If at $t = 0$ the velocity of the particle is $V_0$,the time period of the first revolution of the particle is:

$A$ 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 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:

$A$ 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: