The distance $x$ covered by a particle in time $t$ is given by $x = t^3 - 12t^2 + 6t + 8$. What is the velocity of the particle at the instant when its acceleration is zero?

The radius of a circular plate is increasing at the rate of $0.01 \text{ cm/sec}$. When the radius is $12 \text{ cm}$, the rate at which the area increases is (in $\text{cm}^2/\text{sec}$): (in $\pi$)

$A$ particle moves along the curve $6y = x^3 + 2$. Find the points on the curve at which the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate.

The distance $s$ (in meters) covered by a particle in $t$ seconds is given by $s = ae^t + \frac{b}{e^t}$. Then the acceleration of the particle at time $t$ is:

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

$A$ $2 \ m$ ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate of $25 \ cm/sec$,then the rate (in $cm/sec$) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is $1 \ m$ above the ground is: