$A$ 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:

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 ......... $m/s^2$.

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

$A$ 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 $.............$.