Rate of Change of Quantities Questions in English

(B) Given the equation $pv = 81$.
We can express $p$ in terms of $v$ as $p = \frac{81}{v} = 81v^{-1}$.
Now,differentiate $p$ with respect to $v$:
$\frac{dp}{dv} = \frac{d}{dv}(81v^{-1}) = 81(-1)v^{-2} = -\frac{81}{v^2}$.
To find the value at $v = 9$,substitute $v = 9$ into the derivative:
$\frac{dp}{dv} = -\frac{81}{(9)^2} = -\frac{81}{81} = -1$.
Thus,the correct option is $B$.

(A) Given the velocity equation: $v^2 = 2 - 3x$.
Differentiating both sides with respect to time $t$,we apply the chain rule:
$\frac{d}{dt}(v^2) = \frac{d}{dt}(2 - 3x)$
$2v \frac{dv}{dt} = -3 \frac{dx}{dt}$
Since acceleration $a = \frac{dv}{dt}$ and velocity $v = \frac{dx}{dt}$,we substitute these into the equation:
$2v \cdot a = -3v$
Assuming $v \neq 0$,we divide both sides by $2v$:
$a = -\frac{3}{2} \text{ m/s}^2$
Since the acceleration $a$ is a constant value $(-1.5 \text{ m/s}^2)$,the acceleration is uniform.

(A) Given the distance equation: $s = \frac{1}{2}g{t^2}$.
To find the velocity $v$,we differentiate $s$ with respect to $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(\frac{1}{2}g{t^2}) = \frac{1}{2}g(2t) = gt$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(gt) = g$.
Since $g$ (acceleration due to gravity) is a constant,the acceleration of the stone is uniform.

(C) The surface area $S$ of a sphere with radius $r$ is given by the formula $S = 4\pi r^2$.
To find the error in the surface area,we differentiate $S$ with respect to $r$:
$\frac{dS}{dr} = 8\pi r$.
Using the differential approximation,the error in surface area $\delta S$ is given by $\delta S \approx \frac{dS}{dr} \times \delta r$.
Given $r = 20 \, cm$ and $\delta r = 0.02 \, cm$,we substitute these values into the equation:
$\delta S = 8\pi \times 20 \times 0.02$.
$\delta S = 8\pi \times 0.4 = 3.2\pi$.
Using $\pi \approx 3.14159$,we get $\delta S \approx 3.2 \times 3.14159 = 10.053 \, sq \, cm$.
Rounding to two decimal places,the error is $10.05 \, sq \, cm$.

(C) Given the displacement function $s = 2t^2 - 3t + 1$.
The velocity $v$ is the first derivative of displacement with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(2t^2 - 3t + 1) = 4t - 3$.
The acceleration $a$ is the derivative of velocity with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(4t - 3) = 4$.
Thus,the acceleration of the particle is $4$ units.

(C) The velocity of the particle is given by the derivative of the displacement $s$ with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 9t^2 + 12t + 1) = 6t^2 - 18t + 12$.
For the particle to stop momentarily,its velocity must be zero:
$6t^2 - 18t + 12 = 0$.
Dividing the equation by $6$,we get:
$t^2 - 3t + 2 = 0$.
Factoring the quadratic equation:
$(t - 1)(t - 2) = 0$.
Thus,the particle stops at $t = 1 \, sec$ and $t = 2 \, sec$.

(D) Given the equation of motion: $s = t^2 - 2t$ $(i)$
The velocity $v$ is the rate of change of displacement with respect to time,so $v = \frac{ds}{dt} = 2t - 2$ $(ii)$
Given that the distance travelled $s = 15 \, km$,we substitute this into equation $(i)$:
$15 = t^2 - 2t$
$t^2 - 2t - 15 = 0$
Factoring the quadratic equation:
$(t - 5)(t + 3) = 0$
Since time $t$ cannot be negative,we have $t = 5 \, hours$.
Now,substitute $t = 5$ into equation $(ii)$ to find the velocity:
$v = 2(5) - 2 = 10 - 2 = 8 \, km/h$.
Therefore,the velocity of the car is $8 \, km/h$.

(B) Let the edge of the cube be $a$ and its volume be $V$.
Given that the rate of change of the edge is $\frac{da}{dt} = 5 \, cm/\sec$.
The volume of a cube is given by $V = a^3$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dV}{dt} = 3a^2 \frac{da}{dt}$.
Substitute the given values $a = 12 \, cm$ and $\frac{da}{dt} = 5 \, cm/\sec$ into the equation:
$\frac{dV}{dt} = 3 \times (12)^2 \times 5$
$\frac{dV}{dt} = 3 \times 144 \times 5$
$\frac{dV}{dt} = 432 \times 5 = 2160 \, cm^3/\sec$.
Thus,the volume of the cube is increasing at the rate of $2160 \, cm^3/\sec$.

(A) Given the equation of motion: $s = \frac{1}{2}vt$.
Multiplying by $2$,we get $2s = vt$.
Differentiating both sides with respect to $t$:
$2 \frac{ds}{dt} = v + t \frac{dv}{dt}$.
Since $v = \frac{ds}{dt}$,we substitute this into the equation:
$2v = v + t \frac{dv}{dt} \implies v = t \frac{dv}{dt}$.
Since acceleration $a = \frac{dv}{dt}$,we have $v = ta$.
Differentiating again with respect to $t$:
$\frac{dv}{dt} = a + t \frac{da}{dt}$.
Substituting $a = \frac{dv}{dt}$:
$a = a + t \frac{da}{dt} \implies t \frac{da}{dt} = 0$.
Since $t \neq 0$ for motion,we must have $\frac{da}{dt} = 0$.
This implies that the acceleration $a$ is constant.

(B) The position of the point is given by $s(t) = 15t - 2t^2$.
The average velocity over an interval $[t_1, t_2]$ is defined as the change in displacement divided by the change in time: $v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$.
Given $t_1 = 0$ and $t_2 = 3$:
$s(0) = 15(0) - 2(0)^2 = 0$.
$s(3) = 15(3) - 2(3)^2 = 45 - 18 = 27$.
Therefore,the average velocity is $v_{avg} = \frac{27 - 0}{3 - 0} = \frac{27}{3} = 9$.

(C) Given the distance function: $s = ae^t + \frac{b}{e^t}$.
To find the velocity $v$,we differentiate $s$ with respect to $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(ae^t + be^{-t}) = ae^t - be^{-t}$.
To find the acceleration $a_{acc}$,we differentiate the velocity $v$ with respect to $t$:
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(ae^t - be^{-t}) = ae^t - b(-e^{-t}) = ae^t + be^{-t}$.
Comparing this result with the original expression for $s$,we see that $a_{acc} = s$.
Therefore,the acceleration is equal to the distance $s$.

(B) Given the equation for velocity: ${v^2} = a + bx$.
To find the acceleration,we differentiate both sides with respect to time $t$:
$\frac{d}{dt}({v^2}) = \frac{d}{dt}(a + bx)$
Using the chain rule: $2v \frac{dv}{dt} = b \frac{dx}{dt}$.
Since velocity $v = \frac{dx}{dt}$,we substitute this into the equation:
$2v \frac{dv}{dt} = b v$.
Assuming $v \neq 0$,we divide both sides by $2v$:
$\frac{dv}{dt} = \frac{b}{2}$.
Since $b$ is a constant,the acceleration $\frac{dv}{dt} = \frac{b}{2}$ is also a constant.
Therefore,the acceleration is uniform.

(D) Given the relation between volume $V$ and depth $x$ is $V = 5x - \frac{x^2}{6}$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dV}{dt} = \frac{d}{dt}(5x - \frac{x^2}{6}) = 5\frac{dx}{dt} - \frac{2x}{6}\frac{dx}{dt} = (5 - \frac{x}{3})\frac{dx}{dt}$.
We are given that $\frac{dV}{dt} = 5 \, cm^3/sec$ and we need to find $\frac{dx}{dt}$ when $x = 2 \, cm$.
Substituting the values into the derivative equation:
$5 = (5 - \frac{2}{3})\frac{dx}{dt}$
$5 = (\frac{15 - 2}{3})\frac{dx}{dt}$
$5 = \frac{13}{3}\frac{dx}{dt}$
Solving for $\frac{dx}{dt}$:
$\frac{dx}{dt} = 5 \times \frac{3}{13} = \frac{15}{13} \, cm/sec$.
Since $\frac{15}{13} \, cm/sec$ is not among the given options $(A)$,$(B)$,or $(C)$,the correct answer is $(D)$.

(A) The equation of motion is $s = 10t - 3t^2$.
To find when the stone returns to the surface,we set the displacement $s = 0$.
$10t - 3t^2 = 0$
$t(10 - 3t) = 0$
Since $t=0$ is the time of projection,the stone returns to the surface when $10 - 3t = 0$.
$3t = 10$
$t = \frac{10}{3} \text{ sec}$.
Thus,the stone returns to the surface after $\frac{10}{3} \text{ sec}$.

(C) The velocity of the body is given by the formula $v = 1 + t^2$.
Acceleration $a$ is defined as the rate of change of velocity with respect to time,which is given by $a = \frac{dv}{dt}$.
Differentiating $v = 1 + t^2$ with respect to $t$,we get:
$a = \frac{d}{dt}(1 + t^2) = 0 + 2t = 2t$.
To find the acceleration after $3 \text{ s}$,we substitute $t = 3$ into the expression for acceleration:
$a = 2(3) = 6 \text{ cm/s}^2$.
Therefore,the correct option is $C$.

(A) Let the side of the square be $a$ and its area be $A$.
Given that the rate of change of the side length is $\frac{da}{dt} = 4 \, cm/sec$.
The area of the square is given by $A = a^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = \frac{d}{dt}(a^2) = 2a \frac{da}{dt}$.
We need to find the rate of change of the area when $a = 2 \, cm$.
Substituting the values $a = 2$ and $\frac{da}{dt} = 4$ into the equation:
$\frac{dA}{dt} = 2 \times 2 \times 4 = 16 \, cm^2/sec$.
Thus,the area of the sheet is increasing at the rate of $16 \, cm^2/sec$.

(B) Given the equation of motion: $10s = 10ut - 49t^2$.
Dividing by $10$,we get: $s = ut - 4.9t^2$.
To find the velocity $v$,we differentiate $s$ with respect to $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(ut - 4.9t^2) = u - 9.8t$.
At maximum height,the velocity $v$ becomes $0$.
Given that the maximum height is reached at $t = 5$ seconds,we substitute $v = 0$ and $t = 5$ into the velocity equation:
$0 = u - 9.8(5)$.
$u = 9.8 \times 5 = 49$.
Therefore,the value of $u$ is $49 \, m/sec$.

(A) Given the equation of motion: $s = 2t^3 - 9t^2 + 12t$.
To find the velocity $v$,we differentiate $s$ with respect to $t$: $v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 9t^2 + 12t) = 6t^2 - 18t + 12$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to $t$: $a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 18t + 12) = 12t - 18$.
For the acceleration to be zero,we set $a = 0$: $12t - 18 = 0$.
Solving for $t$: $12t = 18$,which gives $t = \frac{18}{12} = \frac{3}{2} \text{ } sec$.
Thus,the acceleration of the particle will be zero after $3/2 \text{ } sec$.

(B) The average velocity in the third second is defined as the displacement during the interval from $t = 2$ seconds to $t = 3$ seconds divided by the time interval $\Delta t = 1$ second.
First,calculate the position at $t = 3$ seconds:
$s(3) = (3)^2 + 8(3) + 12 = 9 + 24 + 12 = 45 \, m$.
Next,calculate the position at $t = 2$ seconds:
$s(2) = (2)^2 + 8(2) + 12 = 4 + 16 + 12 = 32 \, m$.
The displacement in the third second is:
$\Delta s = s(3) - s(2) = 45 - 32 = 13 \, m$.
Average velocity $= \frac{\Delta s}{\Delta t} = \frac{13 \, m}{1 \, s} = 13 \, m/s$.

(A) Let $OA = x$ and $OB = y$. Since the rod $AB$ has a constant length of $10 \, cm$,by the Pythagorean theorem,we have $x^2 + y^2 = 10^2 = 100$.
Differentiating both sides with respect to time $t$,we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$,which simplifies to $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$.
Given that $\frac{dx}{dt} = 2 \, cm/sec$ (as $A$ moves away from $O$,$x$ increases) and at the instant $x = 8 \, cm$,we find $y$ using $x^2 + y^2 = 100$: $8^2 + y^2 = 100 \Rightarrow 64 + y^2 = 100 \Rightarrow y^2 = 36 \Rightarrow y = 6 \, cm$.
Substituting these values into the differentiated equation: $8(2) + 6 \frac{dy}{dt} = 0 \Rightarrow 16 + 6 \frac{dy}{dt} = 0 \Rightarrow 6 \frac{dy}{dt} = -16 \Rightarrow \frac{dy}{dt} = -\frac{16}{6} = -\frac{8}{3} \, cm/sec$.
The negative sign indicates that the distance $y$ is decreasing,meaning the end $B$ is moving towards $O$ at a rate of $\frac{8}{3} \, cm/sec$.

(B) Given the position function $s(t) = at^2 + bt + 6$.
Velocity $v(t) = \frac{ds}{dt} = 2at + b$.
At $t = 4 \, s$,the particle comes to rest,so $v(4) = 0$.
$2a(4) + b = 0 \implies 8a + b = 0 \implies b = -8a$.
At $t = 0$,the position is $s(0) = a(0)^2 + b(0) + 6 = 6 \, m$.
At $t = 4 \, s$,the distance from the starting position is $16 \, m$,so $s(4) - s(0) = 16$.
$s(4) = 16 + 6 = 22 \, m$.
Substituting $t = 4$ into the position equation: $a(4)^2 + b(4) + 6 = 22$.
$16a + 4b = 16 \implies 4a + b = 4$.
Substitute $b = -8a$ into $4a + b = 4$:
$4a - 8a = 4 \implies -4a = 4 \implies a = -1$.
Then $b = -8(-1) = 8$.
The acceleration $acc = \frac{dv}{dt} = 2a = 2(-1) = -2 \, m/s^2$.
Retardation is the magnitude of negative acceleration,which is $2 \, m/s^2$.
Note: Based on the provided options and standard interpretation of such problems,if we re-evaluate the distance condition as $s(4) = 16$ (total position),then $16a + 4b + 6 = 16 \implies 16a + 4b = 10 \implies 8a + 2b = 5$.
Using $b = -8a$: $8a + 2(-8a) = 5 \implies -8a = 5 \implies a = -5/8$.
Acceleration $2a = -5/4 \, m/s^2$.
Retardation is $5/4 \, m/s^2$.

(C) The position of the particle is given by $s = 45t + 11t^2 - t^3$.
To find the velocity $v$,we differentiate $s$ with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(45t + 11t^2 - t^3) = 45 + 22t - 3t^2$.
When the particle comes to rest,its velocity $v$ must be $0$:
$45 + 22t - 3t^2 = 0$.
Multiplying by $-1$,we get $3t^2 - 22t - 45 = 0$.
Factoring the quadratic equation:
$3t^2 - 27t + 5t - 45 = 0$
$3t(t - 9) + 5(t - 9) = 0$
$(3t + 5)(t - 9) = 0$.
This gives $t = 9$ or $t = -\frac{5}{3}$.
Since time $t$ cannot be negative,the particle comes to rest at $t = 9 \text{ sec}$.

(C) The equation of motion is given by $s = ut - 4.9t^2$.
Velocity $v$ is the rate of change of displacement with respect to time,so $v = \frac{ds}{dt} = u - 9.8t$.
The ball returns to the ground after $6 \, s$,which means it reaches its maximum height at $t = \frac{6}{2} = 3 \, s$.
At the maximum height,the velocity $v = 0$.
Substituting $v = 0$ and $t = 3$ into the velocity equation:
$0 = u - 9.8(3)$
$u = 29.4 \, m/s$.
Thus,the initial velocity is $29.4 \, m/s$.

(A) Given that the rate of change of the radius is $\frac{dr}{dt} = 3 \, cm/s$.
The area of a circle is given by $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.
Substituting the given values $r = 10 \, cm$ and $\frac{dr}{dt} = 3 \, cm/s$:
$\frac{dA}{dt} = 2 \pi (10) (3) = 60 \pi \, cm^2/s$.
Thus,the rate of increase of the area is $60 \pi \, cm^2/s$.

(C) Given the displacement equation: $s = 13.8t - 4.9t^2$.
Velocity $v$ is defined as the rate of change of displacement with respect to time,given by $v = \frac{ds}{dt}$.
Differentiating $s$ with respect to $t$:
$v = \frac{d}{dt}(13.8t - 4.9t^2) = 13.8 - 9.8t$.
To find the velocity at $t = 1$ second,substitute $t = 1$ into the velocity equation:
$v = 13.8 - 9.8(1) = 4.0 \, m/s$.
Thus,the velocity at $t = 1$ second is $4 \, m/s$.

(A) The displacement of the particle is given by $s(t) = -4t^2 + 2t$.
The velocity $v(t)$ is the first derivative of displacement with respect to time $t$:
$v(t) = \frac{ds}{dt} = \frac{d}{dt}(-4t^2 + 2t) = -8t + 2$.
The acceleration $a(t)$ is the derivative of velocity with respect to time $t$:
$a(t) = \frac{dv}{dt} = \frac{d}{dt}(-8t + 2) = -8$.
At time $t = \frac{1}{2} \text{ s}$:
Velocity $v = -8(\frac{1}{2}) + 2 = -4 + 2 = -2$.
Acceleration $a = -8$.
Thus,the velocity and acceleration are $-2$ and $-8$ respectively.

(C) Given the distance $s = 180t - 16t^2$.
Velocity $v$ is the first derivative of distance with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(180t - 16t^2) = 180 - 32t$.
The rate of change in velocity is the acceleration $a$,which is the derivative of velocity with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(180 - 32t) = -32$.
Thus,the rate of change in velocity is $-32 \, unit$.

(B) Let $y$ be the distance of the man from the lamp post and $x$ be the length of his shadow.
Given that the man walks away from the lamp post at a speed of $5 \, m/h$,we have $\frac{dy}{dt} = 5 \, m/h$.
From the similar triangles in the figure,we have the relationship:
$\frac{x}{2} = \frac{x + y}{6}$
Multiplying both sides by $6$,we get:
$3x = x + y$
$2x = y$
$x = \frac{1}{2}y$
Differentiating both sides with respect to time $t$,we get:
$\frac{dx}{dt} = \frac{1}{2} \frac{dy}{dt}$
Substituting $\frac{dy}{dt} = 5 \, m/h$,we get:
$\frac{dx}{dt} = \frac{1}{2} \times 5 = \frac{5}{2} \, m/h$.
Thus,the rate at which the length of his shadow increases is $\frac{5}{2} \, m/h$.

(A) Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder from the ground.
Given that the length of the ladder is $5 \ m$,by Pythagoras theorem,we have $x^2 + y^2 = 5^2 = 25$ .....$(i)$
Differentiating equation $(i)$ with respect to time $t$,we get:
$2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$
$x \frac{dx}{dt} + y \frac{dy}{dt} = 0$ .....$(ii)$
Given that $\frac{dx}{dt} = 1.5 \ m/sec$ and at the instant $x = 4 \ m$:
From $(i)$,$4^2 + y^2 = 25 \Rightarrow 16 + y^2 = 25 \Rightarrow y^2 = 9 \Rightarrow y = 3 \ m$.
Substituting these values in $(ii)$:
$4(1.5) + 3 \frac{dy}{dt} = 0$
$6 + 3 \frac{dy}{dt} = 0$
$3 \frac{dy}{dt} = -6$
$\frac{dy}{dt} = -2 \ m/sec$.
The negative sign indicates that the height is decreasing. Thus,the height of the top of the ladder decreases at the rate of $2 \ m/sec$.

(A) Given that the rate of increase of the radius is $\frac{dr}{dt} = 3.5 \, cm/sec$ and the radius is $r = 10 \, cm$.
The area of the circular region is $A = \pi r^2$.
Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
Substituting the given values: $\frac{dA}{dt} = 2 \times \left( \frac{22}{7} \right) \times 10 \times 3.5$.
Since $3.5 = \frac{7}{2}$,we have $\frac{dA}{dt} = 2 \times \left( \frac{22}{7} \right) \times 10 \times \left( \frac{7}{2} \right)$.
Simplifying the expression: $\frac{dA}{dt} = 22 \times 10 = 220 \, cm^2/sec$.

(B) Let the ladder be represented by a line segment. The man is moving along the ladder at a speed of $v = 3 \, ft/sec$.
The angle between the ladder and the wall is $30^\circ$.
The component of the velocity vector along the horizontal direction (towards the wall) is given by $v \times \sin(\theta)$,where $\theta$ is the angle with the wall.
Alternatively,if we consider the angle with the ground,it is $60^\circ$. The horizontal velocity is $v \times \cos(60^\circ)$.
Rate of approaching the wall $= 3 \times \cos(60^\circ) = 3 \times \frac{1}{2} = \frac{3}{2} \, ft/sec$.

(B) Let $a$ be the edge of the cube and $V$ be its volume.
Given that the rate of change of the edge is $\frac{da}{dt} = 60 \, cm/s$.
The volume of a cube is given by $V = a^3$.
Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = 3a^2 \frac{da}{dt}$.
Substituting the given values $a = 90 \, cm$ and $\frac{da}{dt} = 60 \, cm/s$:
$\frac{dV}{dt} = 3 \times (90)^2 \times 60$
$\frac{dV}{dt} = 3 \times 8100 \times 60$
$\frac{dV}{dt} = 180 \times 8100 = 1458000 \, cm^3/s$.
Thus,the volume is increasing at the rate of $1458000 \, cm^3/s$.

(D) Given the distance function $s = a \sin t + b \cos 2t$.
The velocity $v$ is the first derivative of distance with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(a \sin t + b \cos 2t) = a \cos t - 2b \sin 2t$.
The acceleration $a_{acc}$ is the second derivative of distance with respect to time $t$:
$a_{acc} = \frac{d^2s}{dt^2} = \frac{d}{dt}(a \cos t - 2b \sin 2t) = -a \sin t - 4b \cos 2t$.
At $t = 0$,the acceleration is:
$a_{acc} = -a \sin(0) - 4b \cos(0) = -a(0) - 4b(1) = -4b$.
Thus,the acceleration at $t = 0$ is $-4b$.

(C) Given the position function $S = 6 + 48t - t^3$.
The velocity $v$ is the rate of change of position with respect to time $t$,given by $v = \frac{dS}{dt}$.
$v = \frac{d}{dt}(6 + 48t - t^3) = 48 - 3t^2$.
The direction of motion reverses when the velocity $v$ becomes zero.
Setting $v = 0$,we get $48 - 3t^2 = 0$.
$3t^2 = 48 \Rightarrow t^2 = 16 \Rightarrow t = 4$ (since time $t$ cannot be negative).
To find the distance moved when the direction reverses,substitute $t = 4$ into the position equation $S$:
$S(4) = 6 + 48(4) - (4)^3$.
$S(4) = 6 + 192 - 64 = 134$.
Thus,the particle reverses its direction after moving a distance of $134$ units.

(C) The volume $V$ of a spherical balloon is given by $V = \frac{4}{3}\pi r^3$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Given that $\frac{dV}{dt} = 900 \ cm^3/sec$ and $r = 15 \ cm$,we substitute these values into the equation:
$900 = 4\pi (15)^2 \frac{dr}{dt}$.
$900 = 4\pi (225) \frac{dr}{dt}$.
$900 = 900\pi \frac{dr}{dt}$.
$\frac{dr}{dt} = \frac{900}{900\pi} = \frac{1}{\pi} \ cm/sec$.
Since $\pi \approx \frac{22}{7}$,the value is $\frac{7}{22} \ cm/sec$.

(C) Given the position coordinates of the moving point are $x = at$ and $y = b \sin(at)$.
First,we find the velocity components by differentiating with respect to time $t$:
$v_x = \frac{dx}{dt} = a$
$v_y = \frac{dy}{dt} = ab \cos(at)$
Next,we find the acceleration components by differentiating the velocity components with respect to time $t$:
$a_x = \frac{d^2x}{dt^2} = 0$
$a_y = \frac{d^2y}{dt^2} = -ab^2 \sin(at)$
Since $y = b \sin(at)$,we can substitute this into the expression for $a_y$:
$a_y = -a^2(b \sin(at)) = -a^2y$
Thus,the acceleration vector is $\vec{a} = (0, -a^2y)$.
This shows that the acceleration is directed along the $y$-axis and its magnitude varies directly as the distance from the $x$-axis (which is the value of $y$).

(C) Let $P$ be the perimeter and $A$ be the area of the circle with radius $r$.
$P = 2\pi r$ and $A = \pi r^2$.
Differentiating both with respect to time $t$:
$\frac{dP}{dt} = 2\pi \frac{dr}{dt}$ and $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
From the first equation,$\frac{dr}{dt} = \frac{1}{2\pi} \frac{dP}{dt}$.
Substituting this into the expression for $\frac{dA}{dt}$:
$\frac{dA}{dt} = 2\pi r \left( \frac{1}{2\pi} \frac{dP}{dt} \right) = r \frac{dP}{dt}$.
Since the rate of increase of the perimeter $\frac{dP}{dt}$ is constant,let $\frac{dP}{dt} = k$ (where $k$ is a constant).
Then $\frac{dA}{dt} = k \cdot r$.
Therefore,the rate of increase of the area varies directly as the radius $r$.

(B) The displacement of the stone is given by the equation $s = 80t - 16t^2$.
To find the velocity $v$,we differentiate the displacement $s$ with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(80t - 16t^2)$
$v = 80 - 32t$
Now,we calculate the velocity at $t = 2$ seconds:
$v(2) = 80 - 32(2)$
$v(2) = 80 - 64$
$v(2) = 16 \, m/s$.
Therefore,the velocity after $2$ seconds is $16 \, m/s$.

(A) The distance covered by the particle is given by $s = t + 6t^2 - t^3$.
The velocity $v$ is the first derivative of distance with respect to time $t$:
$v = \frac{ds}{dt} = \frac{d}{dt}(t + 6t^2 - t^3) = 1 + 12t - 3t^2$.
The acceleration $a$ is the second derivative of distance with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(1 + 12t - 3t^2) = 12 - 6t$.
To find the time when the acceleration is zero,set $a = 0$:
$12 - 6t = 0$
$6t = 12$
$t = 2 \text{ seconds}$.
Thus,the acceleration is zero after $2$ seconds.

(C) The distance covered by the body is given by $s = 3t^2 - 8t + 5$.
Velocity $v$ is the rate of change of distance with respect to time,given by $v = \frac{ds}{dt}$.
$v = \frac{d}{dt}(3t^2 - 8t + 5) = 6t - 8$.
The body stops when its velocity becomes zero,i.e.,$v = 0$.
Setting $6t - 8 = 0$,we get $6t = 8$.
$t = \frac{8}{6} = \frac{4}{3} \, \text{sec}$.
Thus,the body will stop after $4/3 \, \text{sec}$.

(C) Let $y = \sqrt{x^2 + 16}$ and $z = \frac{x}{x - 1}$.
First,differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2\sqrt{x^2 + 16}} \cdot (2x) = \frac{x}{\sqrt{x^2 + 16}}$.
Next,differentiate $z$ with respect to $x$ using the quotient rule:
$\frac{dz}{dx} = \frac{(x - 1)(1) - x(1)}{(x - 1)^2} = \frac{-1}{(x - 1)^2}$.
Now,find the rate of change of $y$ with respect to $z$:
$\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{x}{\sqrt{x^2 + 16}} \cdot (-(x - 1)^2) = -\frac{x(x - 1)^2}{\sqrt{x^2 + 16}}$.
Substitute $x = 3$ into the expression:
$\left(\frac{dy}{dz}\right)_{x=3} = -\frac{3(3 - 1)^2}{\sqrt{3^2 + 16}} = -\frac{3(4)}{\sqrt{9 + 16}} = -\frac{12}{\sqrt{25}} = -\frac{12}{5}$.

(C) Given the equation of motion: $a + b{v^2} = {x^2}$.
Differentiating both sides with respect to time $t$:
$\frac{d}{dt}(a + b{v^2}) = \frac{d}{dt}({x^2})$
Since $a$ is a constant,its derivative is $0$. Using the chain rule:
$0 + b(2v \cdot \frac{dv}{dt}) = 2x \cdot \frac{dx}{dt}$
We know that acceleration $A = \frac{dv}{dt}$ and velocity $v = \frac{dx}{dt}$. Substituting these:
$2bv \cdot A = 2xv$
Assuming $v \neq 0$,we can divide both sides by $2v$:
$b \cdot A = x$
Therefore,the acceleration $A = \frac{x}{b}$.

(C) The position of the particle is given by $x = at^2 + bt + c$.
To find the velocity,we differentiate $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt}(at^2 + bt + c) = 2at + b$.
To find the acceleration,we differentiate the velocity with respect to time $t$:
$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(2at + b) = 2a$.
Since the acceleration $a_{acc} = 2a$ is a constant value (independent of time $t$),the particle is moving with uniform acceleration.

(C) Let $x$ be the length of each side of an equilateral triangle and $A$ be its area.
The area of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} x^2$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2x \cdot \frac{dx}{dt} = \frac{\sqrt{3}}{2} x \frac{dx}{dt}$.
Given that $\frac{dx}{dt} = 2 \, cm/sec$ and $x = 10 \, cm$.
Substituting these values into the derivative:
$\frac{dA}{dt} = \frac{\sqrt{3}}{2} \cdot 10 \cdot 2 = 10 \sqrt{3} \, cm^2/sec$.
Thus,the rate at which the area increases is $10 \sqrt{3} \, cm^2/sec$.

(C) Let the surface area of the sphere be $S = 4\pi r^2$.
Given that the rate of change of the radius is $\frac{dr}{dt} = 2 \text{ cm/sec}$.
Differentiating the surface area with respect to time $t$,we get:
$\frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) = 4\pi \times 2r \times \frac{dr}{dt}$.
Substituting the given value of $\frac{dr}{dt}$:
$\frac{dS}{dt} = 8\pi r \times 2 = 16\pi r$.
Since $16\pi$ is a constant,we have $\frac{dS}{dt} \propto r$.
Therefore,the rate of change of the surface area is proportional to $r$.

(C) The points encounter each other when their positions are equal:
$10 + 6t = 3 + t^2$
Rearranging the terms gives the quadratic equation:
$t^2 - 6t - 7 = 0$
$(t - 7)(t + 1) = 0$
Since time $t$ cannot be negative,we have $t = 7 \, sec$.
Now,we find the velocities of the two points by differentiating their position equations with respect to time $t$:
$v_1 = \frac{dx_1}{dt} = \frac{d}{dt}(10 + 6t) = 6 \, cm/sec$
$v_2 = \frac{dx_2}{dt} = \frac{d}{dt}(3 + t^2) = 2t$
At $t = 7 \, sec$,the velocity of the second point is:
$v_2 = 2 \times 7 = 14 \, cm/sec$
The relative speed at which they are receding from each other is the difference in their velocities:
$|v_2 - v_1| = |14 - 6| = 8 \, cm/sec$.

(A) Given that the rate of change of volume is $\frac{dV}{dt} = 30 \, ft^3/min$ and the radius is $r = 15 \, ft$.
The volume of a spherical balloon is given by $V = \frac{4}{3}\pi r^3$.
Differentiating both sides with respect to time $t$,we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.
Substituting the given values: $30 = 4 \times \pi \times (15)^2 \times \frac{dr}{dt}$.
$\frac{dr}{dt} = \frac{30}{4 \times \pi \times 225} = \frac{30}{900\pi} = \frac{1}{30\pi} \, ft/min$.

(C) Given the displacement function $s = t^3 - 3t^2$.
The velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{ds}{dt} = 3t^2 - 6t$.
The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = 6t - 6$.
Set the acceleration to zero to find the time $t$: $6t - 6 = 0 \implies t = 1 \text{ second}$.
Now,substitute $t = 1$ into the velocity equation to find the velocity at that instant:
$v = 3(1)^2 - 6(1) = 3 - 6 = -3 \text{ m/s}$.

(A) Given the displacement $s = \sqrt{t} = t^{1/2}$.
The velocity $v$ is the derivative of displacement with respect to time:
$v = \frac{ds}{dt} = \frac{1}{2} t^{-1/2} = \frac{1}{2\sqrt{t}}$.
From this,we can express $\sqrt{t}$ in terms of $v$:
$\sqrt{t} = \frac{1}{2v}$,which implies $t = \frac{1}{4v^2}$.
The acceleration $a$ is the derivative of velocity with respect to time:
$a = \frac{dv}{dt} = \frac{d}{dt} (\frac{1}{2} t^{-1/2}) = \frac{1}{2} \times (-\frac{1}{2}) t^{-3/2} = -\frac{1}{4} t^{-3/2}$.
Substitute $t = \frac{1}{4v^2}$ into the expression for $a$:
$a = -\frac{1}{4} (\frac{1}{4v^2})^{-3/2} = -\frac{1}{4} (4v^2)^{3/2} = -\frac{1}{4} (8v^3) = -2v^3$.
Therefore,the magnitude of acceleration is proportional to the cube of the velocity,i.e.,$a \propto v^3$.

(D) Let $x$ be the thickness of the ice layer. The radius of the sphere including the ice is $r = (10 + x) \, cm$.
The volume of the ice layer is $V = \frac{4}{3}\pi (10 + x)^3 - \frac{4}{3}\pi (10)^3$.
Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4\pi (10 + x)^2 \frac{dx}{dt}$.
Given that the ice melts at a rate of $50 \, cm^3/min$,we have $\frac{dV}{dt} = -50 \, cm^3/min$ (since volume is decreasing).
Substituting $x = 5$ and $\frac{dV}{dt} = -50$ into the equation:
$-50 = 4\pi (10 + 5)^2 \frac{dx}{dt}$
$-50 = 4\pi (15)^2 \frac{dx}{dt}$
$-50 = 4\pi (225) \frac{dx}{dt}$
$-50 = 900\pi \frac{dx}{dt}$
$\frac{dx}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \, cm/min$.
The rate at which the thickness decreases is $\frac{1}{18\pi} \, cm/min$.