$A$ particle moves according to the law $s=t^{3}-6t^{2}+9t+25$. Find the displacement of the particle when its velocity is zero. (in $\text{ units}$)

$A$ particle moves along a curve $y = \frac{2x^3 - 1}{3}$. The points on the curve at which the $y$-coordinate is changing $18$ times as fast as the $x$-coordinate are

Water is running into a hemispherical bowl of radius $180 \text{ cm}$ at the rate of $108 \text{ dm}^3/\text{min}$. How fast is the water level rising when the depth of the water in the bowl is $120 \text{ cm}$? $(1 \text{ dm} = 10 \text{ cm})$

$A$ stone thrown vertically upwards rises $s$ metres in $t$ seconds,where $s = 80t - 16t^2$. Then,the velocity after $2$ seconds is .......... $m/s$.

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

The displacement $x$ of a particle at time $t$ is given by $x = At^2 + Bt + C$, where $A, B, C$ are constants and $v$ is the velocity of the particle. Then the value of $4Ax - v^2$ is: