$A$ square plate is contracting at the uniform rate $4 \,cm^2 / sec$. The rate at which the perimeter is decreasing, when the side of the square is $20 \,cm$, is

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 $\text{ cm}^2/\text{s}$)?

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 is moving on a straight line,where its position $s$ (in metre) is a function of time $t$ (in seconds) given by $s = at^2 + bt + 6, t \ge 0$. If it is known that the particle comes to rest after $4 \, s$ at a distance of $16 \, m$ from the starting position $(t = 0)$,then the retardation in its motion is

$A$ rod of length $41 \ ft$ with an end $A$ on the floor and another end $B$ on the wall perpendicular to the floor is sliding away horizontally from the wall at the rate of $3 \ ft/min$. When the end $B$ is at a height of $9 \ ft$ from the floor, the rate at which the area of the triangle formed by the rod with the wall and the floor changes at that instant is (in $ft^2/min$):

If the surface area of a cube is increasing at a rate of $3.6 \text{ cm}^2/\text{sec}$,while maintaining its shape,then the rate of change of its volume (in $\text{cm}^3/\text{sec}$),when the length of a side of the cube is $10 \text{ cm}$,is: