An edge of a variable cube is increasing at the rate of $3 \, cm/s$. How fast is the volume of the cube increasing when the edge is $10 \, cm$ long?

The equation of motion of a particle is $s = at^2 + bt + c$. If the displacement after $1 \text{ s}$ is $20 \text{ m}$,velocity after $2 \text{ s}$ is $30 \text{ m/s}$,and the acceleration is $10 \text{ m/s}^2$,then:

If a particle is moving in a straight line so that after $t$ seconds its distance $S$ (in $cm$) from a fixed point on the line is given by $S = f(t) = t^3 - 5t^2 + 8t$,then the acceleration of the particle at $t = 5 \text{ sec}$ is (in $cm/sec^2$).

If the displacement of a particle at a point is given by $s = 3t^{2} - 12t + 14$, then the displacement of the particle when its velocity becomes zero is (in $\text{ units}$)

The rate of change of the area of a sphere with respect to its radius, when the radius is $6 \text{ cm}$, is . . . . . . . (in $\pi$)

$A$ particle moves along a curve $y = \frac{2x^3 - 1}{3}$. The points on the curve at which the $y$-coordinate is changing $18$ times as fast as the $x$-coordinate are