What happens to the centripetal acceleration of a revolving body if you double the orbital speed $v$ and halve the angular velocity $\omega$?

$A$ car is moving on a circular path of radius $500 \ m$ with a speed of $30 \ m/s$. If the speed is increasing at a rate of $2 \ m/s^2$,the net acceleration will be ......... $m/s^2$.

The tangential acceleration of a particle moving in a circle of radius $1 \ m$ varies with time $t$ as shown in the graph (initial velocity of the particle is zero). The time after which the total acceleration of the particle makes an angle of $30^{\circ}$ with the radial acceleration is:

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 $m/s$?

$A$ body of mass $2 \,kg$ moves in a horizontal circular path of radius $5 \,m$. At an instant, its speed is $2 \sqrt{5} \,m/s$ and is increasing at the rate of $3 \,m/s^2$. The magnitude of the net force acting on the body at that instant is: (in $\,N$)

$A$ 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 $.0$)