The distance $s$ in meters covered by a particle in $t$ seconds is given by $s = 2 + 27t - t^3$. The particle will stop after covering a distance of:

Moving along the $x$-axis are two points with $x = 10 + 6t$ and $x = 3 + t^2$. The speed with which they are receding from each other at the time of encounter is ........... $cm/sec$. ($x$ is in $cm$ and $t$ is in seconds)

An object is moving in the clockwise direction around the unit circle $x^2+y^2=1$. As it passes through the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$,its $y$-coordinate is decreasing at the rate of $3 \text{ units/sec}$. The rate at which the $x$-coordinate changes at this point is

$A$ water tank has the shape of an inverted right circular cone whose semi-vertical angle is $\tan^{-1}\left(\frac{1}{2}\right)$. Water is poured into it at a constant rate of $5 \text{ m}^3/\text{min}$. The rate at which the level of water is rising in $\text{m/min}$ at the instant when the depth of water in the tank is $10 \text{ m}$ is:

The displacement of a particle at time $t$ is $s = t^{3} - 4t^{2} - 5t$. The velocity of the particle at $t = 2 \text{ sec}$ is:

What will be the rate of change of the volume of a sphere with radius $r$,with respect to its diameter?