The equation of motion of a stone thrown vertically upward from the surface of a planet is given by $s = 10t - 3t^2$,where the units of $s$ and $t$ are $cm$ and $sec$ respectively. The stone will return to the surface of the planet after:

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

$A$ rod of length $41 \ ft$ with an end $A$ on the floor and another end $B$ on the wall perpendicular to the floor is sliding away horizontally from the wall at the rate of $3 \ ft/min$. When the end $B$ is at a height of $9 \ ft$ from the floor, the rate at which the area of the triangle formed by the rod with the wall and the floor changes at that instant is (in $ft^2/min$):

If the rate of change of $x$ is more than the rate of change of $y$ on the curve $x^3 = 12y$ for $x > 0$,then $x$ lies in the interval:

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:

Find the rate of change of the area of a circle with respect to its radius $r$ when $r=4 \, cm$. (in $\pi \, cm$)