The equation of motion of a particle moving along a straight line is $s = 2t^{3} - 9t^{2} + 12t$,where the units of $s$ and $t$ are centimetre and second. The acceleration of the particle will be zero after: (in $s$)

The rate of change of the area of a circle with respect to its radius at $r = 3 \text{ cm}$ is . . . . . . $\text{cm}^2/\text{cm}$. (in $\pi$)

After $t$ seconds,the acceleration of a particle,which starts from rest and moves in a straight line is $(8-\frac{t}{5}) \text{ cm/s}^2$. The velocity of the particle at the instant when the acceleration is zero is: (in $\text{ cm/s}$)

The distance covered by a particle in $t$ seconds is given by $x = 3 + 8t - 4t^2$. After $1$ second, the velocity will be:

$A$ ball thrown vertically upwards falls back on the ground after $6 \, s$. Assuming that the equation of motion is of the form $s = ut - 4.9t^2$,where $s$ is in metres and $t$ is in seconds,find the initial velocity $u$ at $t = 0$ in $m/s$.

The volume of a cube is increasing at a rate of $9 \text{ cm}^3/\text{s}$. How fast is the surface area increasing when the length of an edge is $10 \text{ cm}$ (in $\text{ cm}^2/\text{s}$)?