If $y = x - x^2$,then the rate of change of $y^2$ with respect to $x^2$ at $x = 2$ is

The displacement $S$ of a moving particle at a time $t$ is given by $S=5+\frac{48}{t}+t^3$. Then its acceleration when the velocity is zero,is

$A$ water tank has the shape of a right circular cone with its axis vertical and vertex downwards. Its semi-vertical angle is $\tan^{-1} \frac{3}{4}$. Water is poured into it at a constant rate of $6 \text{ m}^3/\text{hr}$. The rate (in $\text{m}^2/\text{hr}$) at which the wet curved surface area of the tank is increasing when the depth of water in the tank is $4 \text{ m}$ is:

$A$ particle is moving along a line according to the law $S = t^3 - 3t^2 + 4t - 2$,where $S$ is measured in meters and $t$ is measured in seconds. Then the velocity (in $m/s$) of the particle when its acceleration is zero is:

$A$ spherical snowball is forming such that its volume is increasing at the rate of $8 \text{ cm}^3/\text{sec}$. Find the rate of increase of its radius when the radius is $2 \text{ cm}$.

If the distance travelled by a point in time $t$ is $s = 180t - 16t^2$,then the rate of change in velocity is ......... $unit$.