$A$ body moves according to the formula $v = 1 + t^2$,where $v$ is the velocity at time $t$. The acceleration after $3 \text{ s}$ will be .......... $\text{cm/s}^2$. ($v$ is in $\text{cm/s}$)

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

The distance $s$ in meters covered by a body in $t$ seconds is given by $s = 3t^2 - 8t + 5$. The body will stop after

$A$ particle moves along a straight line such that its distance $s$ in time $t$ seconds is given by $s = t + 6t^2 - t^3$. After what time is the acceleration zero?

If the displacement $S$ of a particle travelling along a straight line in $t$ seconds is given by $S = 2t^3 + 2t^2 - 2t - 3$,then the time taken (in seconds) by the particle to change its direction is

The rate of change of $\sqrt{x^2 + 16}$ with respect to $\frac{x}{x - 1}$ at $x = 3$ is