$A$ balloon,which always remains spherical,has a variable diameter $\frac{3}{2}(2x+1)$. Find the rate of change of its volume with respect to $x$.

The displacement of a particle at the time $t$ is given by $s = \sqrt{1+t}$. Then its acceleration $a$ is proportional to:

If the radius of a circle increases at the rate of $7 \text{ cm/sec}$,then the rate of increase of its area after $10 \text{ minutes}$ is:

$A$ particle is moving on a straight line. The distance $S$ travelled in time $t$ is given by $S = at^2 + bt + 6$. If the particle comes to rest after $4 \text{ s}$ at a distance of $16 \text{ m}$ from the starting point,then the acceleration of the particle is:

$A$ particle starts moving from rest from a fixed point in a fixed direction. The distance $s$ from the fixed point at a time $t$ is given by $s = t^{2} + at - b + 17$,where $a$ and $b$ are real numbers. If the particle comes to rest after $5 \ s$ at a distance of $s = 25$ units from the fixed point,then the values of $a$ and $b$ are respectively:

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 $g = 32 \ ft/sec^2$)