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 $x$-coordinate changes on the curve $y=3x^5+15x-8$ at the rate of $\frac{1}{5} \text{ units/sec}$. If $A(x_1, y_1)$ and $B(x_2, y_2)$ are the points on the curve at which the $y$-coordinate changes at the rate of $6 \text{ units/sec}$, then the slope of $AB$ is:

$A$ particle moves along the curve $y=x^2+2x$. Then, the point on the curve such that the $x$ and $y$ coordinates of the particle change with the same rate is

$A$ spherical iron ball $10 \text{ cm}$ in radius is coated with a layer of ice of uniform thickness that melts at a rate of $50 \text{ cm}^3/\text{min}$. When the thickness of ice is $5 \text{ cm}$,then the rate at which the thickness of ice decreases is:

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

$A$ is a point on the circle with radius $8$ and centre at $O$. $A$ particle $P$ is moving on the circumference of the circle starting from $A$. $M$ is the foot of the perpendicular from $P$ on $OA$ and $\angle POM = \theta$. When $OM = 4$ and $\frac{d\theta}{dt} = 6 \text{ radians/sec}$,then the rate of change of $PM$ is (in units/sec)