$A$ 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

For a particle in a non-uniform accelerated circular motion:

$A$ car is moving at a speed of $40\,m/s$ on a circular track of radius $400\,m.$ This speed is increasing at the rate of $3\,m/s^2.$ The acceleration of the car is ........ $m/s^2$.

$A$ particle is moving in a circle of radius $50 \ cm$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t=0$ is $4 \ m/s$,the time taken to complete the first revolution will be $\frac{1}{\alpha}[1-e^{-2 \pi}] \ s$,where $\alpha=$ . . . . . . .

$A$ car is moving with a speed of $30 \ m/s$ on a circular path of radius $500 \ m$. Its speed is increasing at the rate of $2 \ m/s^2$. What is the acceleration of the car in $m/s^2$?

$A$ 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 $\,N$)