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:

If the volume of a spherical balloon is increasing at the rate of $900 \ cm^3/sec$,then find the rate of change of the radius of the balloon at the instant when the radius is $15 \ cm$ (in $cm/sec$).

From a balloon rising vertically with a uniform velocity of $v \ ft/sec$, a stone is dropped. If the stone reaches the ground after $4 \ sec$, what is the height of the balloon above the ground at that moment (in $ft$)? (Take $g = 32 \ ft/sec^2$)

$A$ particle moves along a straight line according to the law $s=16-2t+3t^{3}$,where $s$ metres is the distance of the particle from a fixed point at the end of $t$ seconds. The acceleration of the particle at the end of $2 \ s$ is

$A$ balloon which always remains spherical is being inflated by pumping in $10 \text{ cm}^3$ of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is $15 \text{ cm}$.

$A$ particle moves along a straight line according to the law $s = \frac{1}{3} t^3 - 3 t^2 + 9 t + 17$,where $s$ is in metre and $t$ is in second. Its velocity decreases in