Kinematics-1D

To determine the nature of the motion, calculate the average velocity for each 1-second interval:
For time interval $t = 0 \text{ s}$ to $t = 1 \text{ s}$:
$v_1 = \frac{0 - (-2)}{1 - 0} = 2 \text{ m/s}$
For time interval $t = 1 \text{ s}$ to $t = 2 \text{ s}$:
$v_2 = \frac{6 - 0}{2 - 1} = 6 \text{ m/s}$
For time interval $t = 2 \text{ s}$ to $t = 3 \text{ s}$:
$v_3 = \frac{16 - 6}{3 - 2} = 10 \text{ m/s}$
Now, calculate the rate of change of velocity (acceleration) between consecutive time intervals:
$a_1 = \frac{v_2 - v_1}{\Delta t} = \frac{6 - 2}{1} = 4 \text{ m/s}^2$
$a_2 = \frac{v_3 - v_2}{\Delta t} = \frac{10 - 6}{1} = 4 \text{ m/s}^2$
Since the acceleration remains constant at $4 \text{ m/s}^2$, the motion is uniform, accelerated.

For uniform motion, the speed of the body remains constant with time.
The relationship between distance $s$ and time $t$ for uniform motion is given by:
$s = v \cdot t$
Since velocity $v = \frac{ds}{dt}$ represents the slope of the $s-t$ graph, a constant velocity implies a constant slope.
A straight line inclined at a constant angle to the time axis represents a constant slope, and therefore, it correctly represents uniform motion.

The velocity of a particle from a displacement-time graph is given by the slope of the line:
$v = \tan\theta$
For particle $A$:
$v_A = \tan 30^\circ = \frac{1}{\sqrt{3}}$
For particle $B$:
$v_B = \tan 60^\circ = \sqrt{3}$
The ratio of their velocities $v_A : v_B$ is:
$\frac{v_A}{v_B} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{\sqrt{3} \times \sqrt{3}} = \frac{1}{3}$
Thus, the ratio $v_A : v_B$ is $1 : 3$.

In the given graph, time is plotted on the vertical axis and displacement is plotted on the horizontal axis.
The angle made by the line with the displacement axis is $30^\circ$.
Therefore, the angle made by the line with the time axis is:
$\theta = 90^\circ - 30^\circ = 60^\circ$
Velocity is given by the slope of the graph with respect to the time axis:
$v = \tan\theta = \tan 60^\circ = \sqrt{3} \text{ m/s}$

From the given displacement-time graph:
At initial time $t = 0 \text{ s}$, the initial position is $x(0) = 0 \text{ m}$.
At final time $t = 5 \text{ s}$, the final position is $x(5) = -10 \text{ m}$.
Total displacement $\Delta x = x(5) - x(0) = -10 - 0 = -10 \text{ m}$.
Total time interval $\Delta t = 5 - 0 = 5 \text{ s}$.
Average velocity is given by:
$v_{\text{av}} = \frac{\Delta x}{\Delta t} = \frac{-10}{5} = -2 \text{ ms}^{-1}$

From the given displacement-time graph:
1. Speed in the first second ($t = 0$ to $t = 1 \text{ s}$):
Displacement changes from $0 \text{ m}$ to $30 \text{ m}$ in $1 \text{ s}$.
$v_1 = \frac{30 - 0}{1 - 0} = 30 \text{ m/s}$
2. Speed in the next two seconds ($t = 1$ to $t = 3 \text{ s}$):
Displacement changes from $30 \text{ m}$ to $0 \text{ m}$ in $2 \text{ s}$.
$v_2 = \left\vert{} \frac{0 - 30}{3 - 1} \right\vert{} = \left\vert{} \frac{-30}{2} \right\vert{} = 15 \text{ m/s}$
3. Ratio of the speeds:
$\frac{v_1}{v_2} = \frac{30}{15} = \frac{2}{1}$
Thus, the ratio of the speed in the first second to that in the next two seconds is $2 : 1$.

During the upward motion, both gravity and air resistance act downwards, so the net retardation is $(g + a)$. Consequently, the speed decreases to zero rapidly with a larger slope magnitude.
During the downward motion, gravity acts downwards and air resistance acts upwards, so the net downward acceleration is $(g - a)$. Consequently, the speed increases from zero at a slower rate with a smaller slope magnitude.
Graph (c) correctly shows a steeper linear decrease in speed during the upward journey followed by a less steep linear increase in speed during the downward journey.

Maximum acceleration corresponds to the maximum positive slope in the speed-time graph.
Looking at the different segments of the graph:
1. From $t = 0 \text{ h}$ to $t = 0.25 \text{ h}$:
$a_1 = \frac{20 - 0}{0.25 - 0} = 80 \text{ km h}^{-2}$
2. From $t = 0.25 \text{ h}$ to $t = 0.75 \text{ h}$:
$a_2 = \frac{20 - 20}{0.75 - 0.25} = 0 \text{ km h}^{-2}$
3. From $t = 0.75 \text{ h}$ to $t = 1.00 \text{ h}$:
$a_3 = \frac{20 - 20}{1.00 - 0.75} = 0 \text{ km h}^{-2}$
4. From $t = 1.00 \text{ h}$ to $t = 1.25 \text{ h}$ (segment CD):
$a_{\text{max}} = \frac{60 - 20}{1.25 - 1.00} = \frac{40}{0.25} = 160 \text{ km h}^{-2}$
Therefore, the maximum acceleration during the journey is $160 \text{ km h}^{-2}$.

At $t = 0$, the tangent to the displacement-time curve has a positive slope, meaning the initial velocity is positive ($v > 0$). As time progresses towards the peak of the displacement curve, the slope becomes less steep, indicating that velocity decreases until it reaches zero ($v = 0$) at the maximum displacement. Past the peak, the slope becomes negative and increasingly steep, showing that the velocity becomes negative ($v < 0$). Since the rate of change of the slope is constant for a parabola, the velocity decreases linearly with time. Graph (a) correctly depicts a straight line starting from a positive velocity, crossing the time axis at $v = 0$, and continuing into negative velocity values.

Given displacement $S = t^3 - 6t^2 + 3t + 4$
Velocity $v = \frac{dS}{dt} = 3t^2 - 12t + 3$
Acceleration $a = \frac{dv}{dt} = 6t - 12$
To find the time when acceleration is zero, set $a = 0$:
$6t - 12 = 0$
$6t = 12$
$t = 2 \text{ s}$
Now, substitute $t = 2 \text{ s}$ into the velocity equation:
$v = 3(2)^2 - 12(2) + 3$
$v = 3(4) - 24 + 3$
$v = 12 - 24 + 3$
$v = -9 \text{ ms}^{-1}$

Given displacement equation:
$x = at + bt^2 - ct^3$
Differentiating $x$ with respect to $t$ to find velocity $v$:
$v = \frac{dx}{dt} = a + 2bt - 3ct^2$
Differentiating velocity $v$ with respect to $t$ to find acceleration:
$\text{Acceleration} = \frac{dv}{dt} = 2b - 6ct$

Given displacement:
$x = 2t^2 + t + 5$
Velocity is the first derivative of $x$:
$v = \frac{dx}{dt} = 4t + 1$
Acceleration is the derivative of velocity:
$a = \frac{dv}{dt} = 4\text{ m/s}^2$
Since the acceleration is constant ($4\text{ m/s}^2$), its value at $t = 2\text{ s}$ remains $4\text{ m/s}^2$.
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Given velocity:
$v = 20 + 0.1t^2$
Differentiating with respect to time $t$:
$a = \frac{dv}{dt} = 0.2t$
Since the acceleration depends on $t$, it increases linearly with time and is therefore non-uniform acceleration.
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Given $x = kt^3$, where $k$ is a positive constant.
Velocity $v = \frac{dx}{dt} = 3kt^2$
Acceleration $a = \frac{dv}{dt} = 6kt$
Since acceleration $a$ is directly proportional to time $t$, its magnitude increases with time.
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When a body is thrown vertically upwards, at the highest point its velocity becomes zero momentarily, but its acceleration due to gravity remains $g$ downwards. Hence, zero velocity does not imply zero acceleration.
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Given $s = t^3 - 3t^2 + 2$
Velocity $v = \frac{ds}{dt} = 3t^2 - 6t$
Acceleration $a = \frac{dv}{dt} = 6t - 6$
Setting acceleration to zero:
$6t - 6 = 0$
$t = 1\text{ s}$
Substituting $t = 1\text{ s}$ into the displacement equation:
$s = (1)^3 - 3(1)^2 + 2 = 1 - 3 + 2 = 0\text{ meter}$
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During retarded motion, the acceleration vector opposes the velocity vector and displacement vector. Since they act in opposite directions along the same line, the angle between instantaneous displacement and acceleration is $180^\circ$ or $\pi\text{ radians}$.

Given acceleration equation:
$A = \frac{dv}{dt} = -a\omega^2 \sin\omega t$
Integrating acceleration with respect to time $t$ to find velocity $v$:
$v = \int -a\omega^2 \sin\omega t \, dt$
$v = a\omega \cos\omega t + C_1$
Since the particle starts from rest, $v = 0$ at $t = 0$:
$0 = a\omega \cos(0) + C_1$
$0 = a\omega + C_1$
$C_1 = -a\omega$
Substituting $C_1$ back into the velocity expression:
$v = \frac{dx}{dt} = a\omega \cos\omega t - a\omega$
Integrating velocity with respect to time $t$ to find displacement $x$:
$x = \int (a\omega \cos\omega t - a\omega) \, dt$
$x = a \sin\omega t - a\omega t + C_2$
Assuming the displacement at origin $x(0) = 0$ at $t = 0$:
$0 = a \sin(0) - 0 + C_2 \implies C_2 = 0$
Thus, the harmonic oscillatory term representing displacement amplitude is:
$x = a \sin\omega t$

Given velocity function:
$v(t) = 10 + 2t^2$
Calculate velocity at $t_1 = 2\text{ s}$:
$v(2) = 10 + 2(2)^2 = 10 + 2(4) = 10 + 8 = 18 \text{ m/s}$
Calculate velocity at $t_2 = 5\text{ s}$:
$v(5) = 10 + 2(5)^2 = 10 + 2(25) = 10 + 50 = 60 \text{ m/s}$
Calculate the change in velocity:
$\Delta v = v(5) - v(2) = 60 - 18 = 42 \text{ m/s}$
Calculate the time interval:
$\Delta t = 5 - 2 = 3\text{ s}$
Calculate average acceleration:
$a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{42}{3} = 14 \text{ m/s}^2$

When no force acts on a body, its acceleration is zero ($a = 0$).
Zero acceleration means the velocity of the body is constant.
In a displacement-time graph, velocity is equal to the slope of the graph.
A constant velocity is shown by a straight line region where the slope does not change.
In the given graph, region $bc$ is a flat horizontal straight line, indicating constant velocity (zero velocity) and zero acceleration.
Therefore, no force is acting on the body during region $bc$.

Deceleration means that the speed or magnitude of velocity decreases with time.
In a displacement-time graph, the slope at any point corresponds to the velocity of the object.
For deceleration, the slope of the displacement-time graph must continuously decrease over time.
Graph (a) shows an increasing slope with time, which represents acceleration.
Graph (b) shows a constant positive slope, which represents constant velocity with zero acceleration.
Graph (c) shows a constant negative slope, which also represents constant velocity with zero acceleration.
Graph (d) shows a curve whose slope flattens out and decreases over time, representing constant deceleration.

Distance is a scalar quantity that measures the total path length covered by an object.
Unlike displacement, total distance covered can never decrease over time.
Graph (a) shows distance increasing with time, which is physically possible.
Graph (b) shows distance continuously decreasing with time, which is impossible because distance cannot decrease.
Graph (c) shows distance remaining constant and then increasing, which is physically possible.
Graph (d) shows distance remaining constant over time (body at rest), which is physically possible.
Hence, graph (b) cannot represent a distance-time graph.

In a valid physical motion, time must always continuously increase.
A single object cannot have multiple displacements at a single instant of time.
Graph (b), (c), and (d) all represent single-valued functions of time where time moves forward.
Graph (a) contains regions where the line loops backwards along the time axis.
This backward loop implies that time is moving backwards and that the particle possesses multiple position values at the same instant of time.
Since time cannot move backwards and a particle cannot be in multiple places at once, graph (a) is physically impossible.

In the interval $OA$, the graph is concave downwards, meaning the slope $\frac{dx}{dt}$ decreases with time, so the acceleration is negative ($-$).
In the interval $AB$, the graph is a straight line, meaning the slope is constant, so the acceleration is zero ($0$).
In the interval $BC$, the graph is concave upwards, meaning the slope $\frac{dx}{dt}$ increases with time, so the acceleration is positive ($+$).
In the interval $CD$, the graph is a straight line, meaning the slope is constant, so the acceleration is zero ($0$).
Thus, the acceleration signs for intervals $OA$, $AB$, $BC$, and $CD$ are $-$, $0$, $+$, $0$.

From $t = 0$ to $t = t_1$, the $x-t$ graph is a straight line with a constant positive slope.
A constant slope in a displacement-time graph indicates that the object is moving at a constant speed.
At time $t = t_1$, the graph becomes a horizontal line parallel to the time axis.
A horizontal line means that the displacement does not change with time, indicating that the body has come to rest.
Therefore, the graph shows that the body travels with constant speed up to time $t_1$ and then stops.

Given equation relating displacement and velocity:
$x = \sqrt{v+1}$
Squaring both sides gives:
$x^2 = v + 1$
Rearranging to express velocity in terms of displacement:
$v = x^2 - 1$
Differentiating velocity with respect to time to find acceleration $a$:
$a = \frac{dv}{dt} = \frac{d}{dt}(x^2 - 1)$
Using the chain rule $\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$:
$a = 2x \cdot \frac{dx}{dt}$
Since $\frac{dx}{dt} = v$:
$a = 2x \cdot v$
Substituting $v = x^2 - 1$ into the acceleration expression:
$a = 2x(x^2 - 1)$
Now, evaluating acceleration at $x = 5\text{ m}$:
$a = 2(5)(5^2 - 1)$
$a = 10(25 - 1)$
$a = 10(24)$
$a = 240\text{ m/s}^2$

Given velocity relationship:
$v = \alpha \sqrt{x}$
Substitute $v = \frac{dx}{dt}$:
$\frac{dx}{dt} = \alpha \sqrt{x}$
Separating variables to integrate:
$\frac{dx}{\sqrt{x}} = \alpha dt$
Integrating both sides from $x = 0$ at $t = 0$ to displacement $x$ at time $t$:
$\int_0^x x^{-1/2} dx = \int_0^t \alpha dt$
$2 \sqrt{x} = \alpha t$
Squaring both sides:
$4x = \alpha^2 t^2$
$x = \frac{1}{4} \alpha^2 t^2$
Differentiating displacement with respect to time to get velocity:
$v = \frac{dx}{dt} = \frac{1}{2} \alpha^2 t$
Differentiating velocity with respect to time to get acceleration:
$a = \frac{dv}{dt} = \frac{d^2 x}{dt^2} = \frac{1}{2} \alpha^2$

(a) Distance travelled by Ram = 100 m
Distance travelled by Shyam = $\pi (50\text{ m}) = 50\pi\text{ m}$
(b) Displacement of Ram = 100 m
Displacement of Shyam = 100 m
Step Solution:
Step 1: Identify the path taken by Ram. Ram moves along path 1, which is a straight line of length 100 m from point P to point Q. Therefore, the distance travelled by Ram is 100 m.
Step 2: Identify the path taken by Shyam. Shyam moves along path 2, which is a semicircular arc with a diameter of 100 m (radius $r = 50\text{ m}$). The length of a semicircular arc is given by $\pi r$. Thus, the distance travelled by Shyam is $\pi \times 50\text{ m} = 50\pi\text{ m}$.
Step 3: Determine the displacement for both Ram and Shyam. Displacement is defined as the shortest straight-line distance from the initial position P to the final position Q. Since both Ram and Shyam start at point P and end at point Q, their displacement is identical and equal to the straight-line distance between P and Q, which is 100 m.

Since $t = \sqrt{x} + 3$
$\Rightarrow x = (t - 3)^2$
So, we get $x$ as a function of $t$. Let us draw a table for $x$ and $t$ from $t = 0$ to $t = 6$. Then we get
$t = 0 \Rightarrow x = 9$
$t = 1 \Rightarrow x = 4$
$t = 2 \Rightarrow x = 1$
$t = 3 \Rightarrow x = 0$
$t = 4 \Rightarrow x = 1$
$t = 5 \Rightarrow x = 4$
$t = 6 \Rightarrow x = 9$
This table gives us the answers to the three parts.
So, for (a), we have $\Delta x_1 = -9\text{ m}$
For (b), we have $\Delta x_2 = +9\text{ m}$
For (c), distance travelled $= \vert{}-9\vert{} + \vert{}+9\vert{} = 18\text{ m}$ and displacement $= \Delta x_1 + \Delta x_2 = -9 + 9 = 0\text{ m}$
Step Solution:
Step 1: Express position $x$ as an explicit function of time $t$. Rearranging $t = \sqrt{x} + 3$ gives $\sqrt{x} = t - 3$, which yields $x(t) = (t - 3)^2$.
Step 2: Calculate the positions at key time intervals: at $t = 0\text{ s}$, $x(0) = (0 - 3)^2 = 9\text{ m}$; at $t = 3\text{ s}$, $x(3) = (3 - 3)^2 = 0\text{ m}$; and at $t = 6\text{ s}$, $x(6) = (6 - 3)^2 = 9\text{ m}$.
Step 3: Determine displacement from $t = 0$ to $t = 3\text{ s}$. $\Delta x_1 = x(3) - x(0) = 0 - 9 = -9\text{ m}$.
Step 4: Determine displacement from $t = 3$ to $t = 6\text{ s}$. $\Delta x_2 = x(6) - x(3) = 9 - 0 = +9\text{ m}$.
Step 5: Determine total displacement and total distance from $t = 0$ to $t = 6\text{ s}$. Total displacement is $\Delta x = x(6) - x(0) = 9 - 9 = 0\text{ m}$. Total distance is the sum of magnitudes of individual path segments: $\vert{}\Delta x_1\vert{} + \vert{}\Delta x_2\vert{} = \vert{}-9\vert{} + \vert{}+9\vert{} = 18\text{ m}$.

Average Speed = $\frac{\text{Total Distance Travelled}}{\text{Total Time Taken}}$
CASE-1: $v_{av} = \frac{2 v_1 v_2}{v_1 + v_2} = \frac{(2)(40)(60)}{40 + 60} = 48\text{ kmh}^{-1}$
Average Velocity = $\vec{v}_{av} = \vec{0}$
{$\because$ train returns to its station}
Step Solution:
Step 1: Identify the given speeds for the two equal distance legs of the journey: $v_1 = 40\text{ kmh}^{-1}$ and $v_2 = 60\text{ kmh}^{-1}$.
Step 2: Apply the harmonic mean formula for average speed when distances are equal: $v_{av} = \frac{2 v_1 v_2}{v_1 + v_2} = \frac{2 \times 40 \times 60}{40 + 60} = \frac{4800}{100} = 48\text{ kmh}^{-1}$.
Step 3: Determine the average velocity by evaluating total displacement. Because the train finishes its journey at the initial station, the total displacement is $0$, making the average velocity $0\text{ kmh}^{-1}$.

$v_{av} = \frac{v_1 + v_2}{2} = \frac{1 + 3}{2} = 2\text{ ms}^{-1}$
Average Velocity = $2\text{ ms}^{-1}$
Step Solution:
Step 1: Note the given speeds $v_1 = 1\text{ ms}^{-1}$ and $v_2 = 3\text{ ms}^{-1}$ and equal time intervals $t_1 = t_2 = 1\text{ min} = 60\text{ s}$.
Step 2: Calculate average speed using the formula for equal time intervals: $v_{av} = \frac{v_1 + v_2}{2} = \frac{1 + 3}{2} = 2\text{ ms}^{-1}$.
Step 3: Since the motion takes place continuously along a straight path without reversing direction, total displacement equals total distance. Hence, average velocity is also $2\text{ ms}^{-1}$.

Total Distance $= 720 + 4 \times (5 \times 60) + 1 \times (3 \times 60) = 720 + 1200 + 180 = 2100\text{ m}$
Total Time $= \frac{720}{2} + (5 \times 60) + (3 \times 60) = 360 + 300 + 180 = 840\text{ s}$
$v_{av} = \frac{2100}{840} = 2.5\text{ ms}^{-1}$
Average Velocity = $2.5\text{ ms}^{-1}$
Step Solution:
Step 1: Calculate distance and time for each segment.
Segment 1: $d_1 = 720\text{ m}$, $v_1 = 2\text{ ms}^{-1} \Rightarrow t_1 = \frac{720}{2} = 360\text{ s}$.
Segment 2: $v_2 = 4\text{ ms}^{-1}$, $t_2 = 5\text{ min} = 300\text{ s} \Rightarrow d_2 = 4 \times 300 = 1200\text{ m}$.
Segment 3: $v_3 = 1\text{ ms}^{-1}$, $t_3 = 3\text{ min} = 180\text{ s} \Rightarrow d_3 = 1 \times 180 = 180\text{ m}$.
Step 2: Find total distance travelled: $D = 720 + 1200 + 180 = 2100\text{ m}$.
Step 3: Find total time elapsed: $T = 360 + 300 + 180 = 840\text{ s}$.
Step 4: Compute average speed: $v_{av} = \frac{2100}{840} = 2.5\text{ ms}^{-1}$.
Step 5: As the motion is unidirectional in a straight line, total displacement equals total distance, so average velocity is $2.5\text{ ms}^{-1}$.

Let us first draw a pictorial representation to the problem.
$s_1 = \frac{l}{3}$, $t_1 = \frac{l}{3 v_0}$
$s_2 = v_1 \frac{t_2}{2} + v_2 \frac{t_2}{2} = \frac{2l}{3}$
$\Rightarrow t_2 = \frac{4l}{3(v_1 + v_2)}$
$v_{av} = \frac{l}{t_1 + t_2} = \frac{l}{\frac{l}{3 v_0} + \frac{4l}{3(v_1 + v_2)}}$
$\Rightarrow v_{av} = \frac{3 v_0 (v_1 + v_2)}{4 v_0 + v_1 + v_2}$
Step Solution:
Step 1: Let the total distance of the journey be $l$. The first one-third of the distance is $s_1 = \frac{l}{3}$, which is covered with constant velocity $v_0$. The time taken for this segment is $t_1 = \frac{s_1}{v_0} = \frac{l}{3 v_0}$.
Step 2: The remaining distance is $s_2 = \frac{2l}{3}$. Let $t_2$ be the total time spent covering this remaining portion. For the first half of this time interval ($\frac{t_2}{2}$), the particle moves with velocity $v_1$, and for the second half ($\frac{t_2}{2}$), it moves with velocity $v_2$.
Step 3: Express the distance $s_2$ in terms of $t_2$: $s_2 = v_1 \left(\frac{t_2}{2}\right) + v_2 \left(\frac{t_2}{2}\right) = \frac{(v_1 + v_2) t_2}{2}$. Equating this to $\frac{2l}{3}$, we get $\frac{(v_1 + v_2) t_2}{2} = \frac{2l}{3} \Rightarrow t_2 = \frac{4l}{3(v_1 + v_2)}$.
Step 4: Determine the total time of motion: $t_{total} = t_1 + t_2 = \frac{l}{3 v_0} + \frac{4l}{3(v_1 + v_2)} = \frac{l}{3} \left[ \frac{1}{v_0} + \frac{4}{v_1 + v_2} \right] = \frac{l}{3} \left[ \frac{v_1 + v_2 + 4 v_0}{v_0 (v_1 + v_2)} \right]$.
Step 5: Compute the mean velocity averaged over the whole time: $v_{av} = \frac{l}{t_{total}} = \frac{l}{\frac{l (4 v_0 + v_1 + v_2)}{3 v_0 (v_1 + v_2)}} = \frac{3 v_0 (v_1 + v_2)}{4 v_0 + v_1 + v_2}$.

$x = 8t^2 - 3t^3$ ...(1)
$\Rightarrow v = \frac{dx}{dt} = 16t - 9t^2$ ...(2)
$\Rightarrow a = \frac{dv}{dt} = 16 - 18t$ ...(3)
(a) At $t = 1\text{ s}$, $v = 16(1) - 9(1)^2 = 7\text{ ms}^{-1}$
(b) Body is at rest when $v = 0$
$\Rightarrow 16t - 9t^2 = 0 \Rightarrow t(16 - 9t) = 0$
$\Rightarrow t = 0\text{ s}$ or $t = \frac{16}{9}\text{ s}$
Position at $t = \frac{16}{9}\text{ s}$:
$x = 8\left(\frac{16}{9}\right)^2 - 3\left(\frac{16}{9}\right)^3 = \frac{2048}{243}\text{ m}$
(c) Acceleration at $t = 4\text{ s}$:
$a = 16 - 18(4) = 16 - 72 = -56\text{ ms}^{-2}$
Step Solution:
Step 1: Differentiate position $x(t) = 8t^2 - 3t^3$ to obtain velocity $v(t) = 16t - 9t^2$.
Step 2: Differentiate velocity $v(t)$ to obtain acceleration $a(t) = 16 - 18t$.
Step 3: Evaluate instantaneous velocity at $t = 1\text{ s}$: $v(1) = 16(1) - 9(1)^2 = 7\text{ ms}^{-1}$.
Step 4: Find time of rest by setting $v(t) = 0$, giving $t = \frac{16}{9}\text{ s}$. Substitute $t = \frac{16}{9}\text{ s}$ into $x(t)$ to get position $x = \frac{2048}{243}\text{ m}$.
Step 5: Evaluate acceleration at $t = 4\text{ s}$: $a(4) = 16 - 18(4) = -56\text{ ms}^{-2}$.

At $t = 0$, $x_0 = 0$
At $t = 4\text{ s}$, $x_4 = 8(4)^2 - 3(4)^3 = 128 - 192 = -64\text{ m}$
Displacement $\Delta x = x_4 - x_0 = -64 - 0 = -64\text{ m}$
Average velocity $= \frac{-64}{4} = -16\text{ ms}^{-1}$
Body reverses direction at $t = \frac{16}{9}\text{ s}$ where $x = \frac{2048}{243}\text{ m}$.
Distance from $t = 0$ to $t = \frac{16}{9}\text{ s}$ is $d_1 = \frac{2048}{243}\text{ m}$.
Distance from $t = \frac{16}{9}\text{ s}$ to $t = 4\text{ s}$ is $d_2 = \left\vert{}-64 - \frac{2048}{243}\right\vert{} = \frac{17600}{243}\text{ m}$.
Total distance $= d_1 + d_2 = \frac{2048}{243} + \frac{17600}{243} = \frac{19648}{243}\text{ m}$.
Average speed $= \frac{\frac{19648}{243}}{4} = \frac{4912}{243} = \frac{544}{27}\text{ ms}^{-1}$
Step Solution:
Step 1: Calculate position at initial time $t = 0\text{ s}$, $x(0) = 0\text{ m}$, and final time $t = 4\text{ s}$, $x(4) = 8(4)^2 - 3(4)^3 = -64\text{ m}$.
Step 2: Find total displacement $\Delta x = x(4) - x(0) = -64\text{ m}$. Compute average velocity $= \frac{\Delta x}{\Delta t} = \frac{-64}{4} = -16\text{ ms}^{-1}$.
Step 3: Determine turning point by setting velocity $v = 16t - 9t^2 = 0$, which yields $t = \frac{16}{9}\text{ s}$ (within $0 \le t \le 4$).
Step 4: Find positions at key times: $x(0) = 0\text{ m}$, $x\left(\frac{16}{9}\right) = \frac{2048}{243}\text{ m}$, and $x(4) = -64\text{ m}$.
Step 5: Compute distance for first segment $d_1 = \frac{2048}{243}\text{ m}$ and second segment $d_2 = \left\vert{}-64 - \frac{2048}{243}\right\vert{} = \frac{17600}{243}\text{ m}$. Sum to get total distance $= \frac{19648}{243}\text{ m}$.
Step 6: Compute average speed $= \frac{\text{Total Distance}}{\Delta t} = \frac{\frac{19648}{243}}{4} = \frac{544}{27}\text{ ms}^{-1}$.

(a) $v = 12 - 3t^2$ ...(1)
$\Rightarrow a = \left.\frac{dv}{dt} = -6t\right\vert{}_{t=4} = -24\text{ ms}^{-2}$
(b) Further $v = \frac{dx}{dt}$
$\Rightarrow \int_{-10}^{x} dx = \int_{1}^{t} v \, dt = \int_{1}^{t} (12 - 3t^2) \, dt$
$\Rightarrow x - (-10) = \left[ 12t - t^3 \right]_1^t$
$\Rightarrow x + 10 = 12t - t^3 - (12 - 1)$
$\Rightarrow x = 12t - t^3 - 21$ ...(2)
Displacement from $t = 0$ to $t = 10\text{ s}$:
At $t = 0\text{ s}$, $x(0) = -21\text{ m}$
At $t = 10\text{ s}$, $x(10) = 12(10) - (10)^3 - 21 = 120 - 1000 - 21 = -901\text{ m}$
Displacement $= x(10) - x(0) = -901 - (-21) = -880\text{ m}$
(c) For distance, $v = 0 \Rightarrow 12 - 3t^2 = 0 \Rightarrow t = 2\text{ s}$
Position at $t = 2\text{ s}$: $x(2) = 12(2) - (2)^3 - 21 = 24 - 8 - 21 = -5\text{ m}$
Distance from $t = 0$ to $t = 2\text{ s}$ is $d_1 = \vert{}x(2) - x(0)\vert{} = \vert{}-5 - (-21)\vert{} = 16\text{ m}$
Distance from $t = 2$ to $t = 10\text{ s}$ is $d_2 = \vert{}x(10) - x(2)\vert{} = \vert{}-901 - (-5)\vert{} = 896\text{ m}$
Total distance $= d_1 + d_2 = 16 + 896 = 912\text{ m}$
Step Solution:
Step 1: Calculate acceleration by differentiating velocity $v(t) = 12 - 3t^2$ with respect to time $t$. This gives $a(t) = \frac{dv}{dt} = -6t$. Evaluating at $t = 4\text{ s}$ gives $a(4) = -6(4) = -24\text{ ms}^{-2}$.
Step 2: Determine position function $x(t)$ using the given initial condition $x(1) = -10\text{ m}$. Integrate $\int_{-10}^{x} dx = \int_{1}^{t} (12 - 3t^2) \, dt$, which yields $x(t) = 12t - t^3 - 21$.
Step 3: Calculate position at $t = 0\text{ s}$ to be $x(0) = -21\text{ m}$ and position at $t = 10\text{ s}$ to be $x(10) = 12(10) - (10)^3 - 21 = -901\text{ m}$.
Step 4: Compute displacement from $t = 0$ to $t = 10\text{ s}$: $\Delta x = x(10) - x(0) = -901 - (-21) = -880\text{ m}$.
Step 5: Find the turning point of motion by setting velocity $v(t) = 0 \Rightarrow 12 - 3t^2 = 0 \Rightarrow t = 2\text{ s}$ (since $t \ge 0$). Position at $t = 2\text{ s}$ is $x(2) = 12(2) - (2)^3 - 21 = -5\text{ m}$.
Step 6: Compute total distance by summing absolute displacements of the two sub-intervals: $d_1 = \vert{}x(2) - x(0)\vert{} = \vert{}-5 - (-21)\vert{} = 16\text{ m}$ and $d_2 = \vert{}x(10) - x(2)\vert{} = \vert{}-901 - (-5)\vert{} = 896\text{ m}$. Total distance $= 16\text{ m} + 896\text{ m} = 912\text{ m}$.

(a) Let $v_m$ be the velocity at $C$, the midpoint of $A$ and $B$.
Using $v^2 = u^2 + 2as$:
For path $AC$: $v_m^2 - u^2 = 2a\left(\frac{l}{2}\right) = al$
For path $CB$: $v^2 - v_m^2 = 2a\left(\frac{l}{2}\right) = al$
Equating both equations:
$v_m^2 - u^2 = v^2 - v_m^2$
$2v_m^2 = u^2 + v^2$
$v_m = \sqrt{\frac{u^2 + v^2}{2}}$
(b) Let $t_1$ be the time taken from $A$ to $C$ and $t_2$ be the time taken from $C$ to $B$. Given $t_1 = 2t_2$.
Displacement for path $AC$: $\frac{l}{2} = \left(\frac{u + v_m}{2}\right) 2t_2$
Displacement for path $CB$: $\frac{l}{2} = \left(\frac{v_m + v}{2}\right) t_2$
Equating both expressions gives $u + v_m = \frac{v_m + v}{2} \implies v_m = v - 2u$.
Squaring both sides gives $v_m^2 = v^2 + 4u^2 - 4uv$.
Substituting $v_m^2 = \frac{u^2 + v^2}{2}$:
$\frac{u^2 + v^2}{2} = v^2 + 4u^2 - 4uv$
$u^2 + v^2 = 2v^2 + 8u^2 - 8uv$
$v^2 - 8uv + 7u^2 = 0$
$(v - u)(v - 7u) = 0$
Since motion is accelerated, $v - u \neq 0$, so $v = 7u \implies \frac{v}{u} = 7$.
Step Solution:
Step 1: Calculate the velocity $v_m$ at the midpoint $C$.
Let $l$ be the distance $AB$. Then $AC = CB = \frac{l}{2}$.
Using $v^2 = u^2 + 2as$:
For path $AC$: $v_m^2 = u^2 + 2a\left(\frac{l}{2}\right) \implies v_m^2 - u^2 = al$
For path $CB$: $v^2 = v_m^2 + 2a\left(\frac{l}{2}\right) \implies v^2 - v_m^2 = al$
Equating the two expressions for $al$:
$v_m^2 - u^2 = v^2 - v_m^2$
$2v_m^2 = u^2 + v^2$
$v_m = \sqrt{\frac{u^2 + v^2}{2}}$
Step 2: Relate the time taken for segments $AC$ and $CB$.
Let time for $CB$ be $t$. Since time for $AC$ is twice as much, time for $AC$ is $2t$.
Using $s = \left(\frac{u_{\text{initial}} + u_{\text{final}}}{2}\right) t$:
For $AC$: $\frac{l}{2} = \left(\frac{u + v_m}{2}\right) 2t = (u + v_m)t$
For $CB$: $\frac{l}{2} = \left(\frac{v_m + v}{2}\right) t$
Step 3: Solve for $v_m$ in terms of $u$ and $v$.
Equating the displacement equations:
$(u + v_m)t = \left(\frac{v_m + v}{2}\right) t$
$u + v_m = \frac{v_m + v}{2}$
$2u + 2v_m = v_m + v$
$v_m = v - 2u$
Step 4: Determine the ratio $\frac{v}{u}$.
Square $v_m = v - 2u$:
$v_m^2 = (v - 2u)^2 = v^2 + 4u^2 - 4uv$
Substitute $v_m^2 = \frac{u^2 + v^2}{2}$:
$\frac{u^2 + v^2}{2} = v^2 + 4u^2 - 4uv$
$u^2 + v^2 = 2v^2 + 8u^2 - 8uv$
$v^2 - 8uv + 7u^2 = 0$
$(v - 7u)(v - u) = 0$
As motion is accelerated, $v \neq u$, hence $v - 7u = 0 \implies v = 7u \implies \frac{v}{u} = 7$.

For $OA(= l)$:
$(50)^2 - (10)^2 = 2al \implies al = 1200$
Now, when the particle just attains the velocity of $50\text{ ms}^{-1}$ at $A$, acceleration is reversed in direction i.e., it starts acting as retardation for $50\text{ ms}^{-1}$ which will first be reduced to zero after travelling a distance $AB$ and then the particle will again be accelerated from $B$ to $A$ with the same acceleration. So, velocity of the particle again at $A$ will be $50\text{ ms}^{-1}$, but in opposite direction.
Now, for path $AO$:
$v^2 = (-50)^2 + 2a(l)$
$v^2 = 2500 + 2(1200) = 2500 + 2400 = 4900$
$v = \sqrt{4900} = 70\text{ ms}^{-1}$
Since the direction is towards the origin $O$ (opposite to initial motion), the velocity of the particle as it reaches the starting point is $-70\text{ ms}^{-1}$.
Step Solution:
Step 1: Apply equation of motion for the initial segment $OA$ of distance $l$.
Using $v^2 = u^2 + 2al$:
$(50)^2 = (10)^2 + 2al$
$2500 = 100 + 2al$
$2al = 2400 \implies al = 1200$
Step 2: Analyze the motion after point $A$.
When acceleration reverses direction at point $A$, the particle slows down to rest at point $B$ and accelerates back towards $A$ under acceleration of magnitude $a$.
By symmetry of motion between $A$ and $B$, the magnitude of velocity of the particle when it returns to point $A$ heading towards $O$ is $50\text{ ms}^{-1}$ directed towards $O$.
Step 3: Apply equation of motion for the journey from $A$ back to origin $O$.
Initial speed at $A$ heading left is $50\text{ ms}^{-1}$.
During motion from $A$ to $O$, the acceleration $a$ acts in the same direction of motion (towards $O$).
Using $v^2 = u^2 + 2al$:
$v^2 = (50)^2 + 2al$
$v^2 = 2500 + 2(1200) = 2500 + 2400 = 4900$
$v = 70\text{ ms}^{-1}$
Step 4: Determine final velocity taking direction into account.
Since the particle is moving in the direction opposite to its initial motion (towards the left), its velocity upon reaching the starting point $O$ is $-70\text{ ms}^{-1}$.

Let the particle $P$ reach $C$ (the midpoint of $AB$) in time $t$, then the particle $Q$ must reach $C$ also in time $t$.
Let $AB = l$, so $AC = CB = \frac{l}{2}$.
For particle $P$: $\frac{l}{2} = u_1 t + \frac{1}{2} a_1 t^2$
For particle $Q$: $\frac{l}{2} = u_2 t + \frac{1}{2} a_2 t^2$
Equating both expressions for $\frac{l}{2}$:
$u_1 t + \frac{1}{2} a_1 t^2 = u_2 t + \frac{1}{2} a_2 t^2$
$\frac{1}{2}(a_1 - a_2) t^2 = (u_2 - u_1) t \implies t = 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right)$
Substituting $t$ back into displacement equation:
$\frac{l}{2} = u_1 \left[ 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \right] + \frac{1}{2} a_1 \left[ 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \right]^2$
$l = \frac{4(u_2 - u_1)}{(a_1 - a_2)^2} (a_1 u_2 - a_2 u_1)$
Also, $P$ reaches $B$ and $Q$ reaches $A$ with equal velocities $v$.
For $P$: $v^2 - u_1^2 = 2 a_1 l$
For $Q$: $v^2 - u_2^2 = 2 a_2 l$
Subtracting the two equations gives $u_2^2 - u_1^2 = 2(a_1 - a_2)l$.
Substituting the expression for $l$:
$(u_2 - u_1)(u_2 + u_1) = 2(a_1 - a_2) \left[ \frac{4(u_2 - u_1)}{(a_1 - a_2)^2} (a_1 u_2 - a_2 u_1) \right]$
$(u_1 + u_2)(a_1 - a_2) = 8(a_1 u_2 - a_2 u_1)$
Step Solution:
Step 1: Set up the displacement equations for both particles to reach midpoint $C$.
Let length of $AB = l$. Midpoint distance $AC = CB = \frac{l}{2}$.
Since $P$ and $Q$ meet at midpoint $C$, time taken by both is $t$.
For particle $P$: $\frac{l}{2} = u_1 t + \frac{1}{2} a_1 t^2$
For particle $Q$: $\frac{l}{2} = u_2 t + \frac{1}{2} a_2 t^2$
Step 2: Solve for time $t$ in terms of initial velocities and accelerations.
$u_1 t + \frac{1}{2} a_1 t^2 = u_2 t + \frac{1}{2} a_2 t^2$
$\frac{1}{2} (a_1 - a_2) t^2 = (u_2 - u_1) t$
$t = 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right)$
Step 3: Express total length $l$ in terms of $u_1, u_2, a_1, a_2$.
Substitute $t$ into $\frac{l}{2} = u_1 t + \frac{1}{2} a_1 t^2$:
$\frac{1}{2} l = u_1 \left[ 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \right] + \frac{1}{2} a_1 \left[ 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \right]^2$
$\frac{1}{2} l = 2 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \left[ u_1 + a_1 \left( \frac{u_2 - u_1}{a_1 - a_2} \right) \right]$
$l = \frac{4(u_2 - u_1)}{(a_1 - a_2)^2} (a_1 u_2 - a_2 u_1)$
Step 4: Use third equation of motion for final velocities at the opposite ends.
Both particles reach the opposite ends with equal velocity $v$.
For $P$ reaching $B$: $v^2 - u_1^2 = 2 a_1 l$
For $Q$ reaching $A$: $v^2 - u_2^2 = 2 a_2 l$
Subtracting the two equations:
$u_2^2 - u_1^2 = 2(a_1 - a_2)l$
Step 5: Combine equations to obtain the final relationship.
Substitute $l$ into $u_2^2 - u_1^2 = 2(a_1 - a_2)l$:
$(u_2 - u_1)(u_1 + u_2) = 2(a_1 - a_2) \left[ \frac{4(u_2 - u_1)}{(a_1 - a_2)^2} (a_1 u_2 - a_2 u_1) \right]$
$(u_2 - u_1)(u_1 + u_2) = \frac{8(u_2 - u_1)}{a_1 - a_2} (a_1 u_2 - a_2 u_1)$
$(u_1 + u_2)(a_1 - a_2) = 8(a_1 u_2 - a_2 u_1)$

Let the man catch the bus at time $t$.
For bus:
$x = x_0 + ut + \frac{1}{2}at^2$
$\implies x = 40 + 0(t) + \frac{1}{2}(1)t^2$
$\implies x = 40 + \frac{t^2}{2}$
For man:
$x = 9t$
Equating both expressions for position:
$40 + \frac{t^2}{2} = 9t$
$\implies t^2 - 18t + 80 = 0$
$\implies t^2 - 10t - 8t + 80 = 0$
$\implies (t - 8)(t - 10) = 0$
$\implies t = 8\text{ s}$ or $t = 10\text{ s}$
Step Solution:
Step 1: Set up the equation of motion for the bus taking the man's initial position as origin $x = 0$.
At $t = 0$, position of bus is $x_0 = 40\text{ m}$ and its initial speed $u = 0\text{ ms}^{-1}$ with acceleration $a = 1\text{ ms}^{-2}$.
$x_{\text{bus}} = x_0 + ut + \frac{1}{2}at^2 = 40 + 0 + \frac{1}{2}(1)t^2 = 40 + \frac{t^2}{2}$
Step 2: Set up the equation of motion for the man moving with constant speed $v = 9\text{ ms}^{-1}$.
$x_{\text{man}} = 9t$
Step 3: Equate positions of the man and the bus to solve for time $t$.
$x_{\text{man}} = x_{\text{bus}}$
$9t = 40 + \frac{t^2}{2}$
$18t = 80 + t^2$
$t^2 - 18t + 80 = 0$
Step 4: Factorize and solve the quadratic equation for $t$.
$t^2 - 10t - 8t + 80 = 0$
$t(t - 10) - 8(t - 10) = 0$
$(t - 8)(t - 10) = 0$
$t = 8\text{ s}$ or $t = 10\text{ s}$

Furthermore, the average speed
$v_{av} = 72\text{ kmh}^{-1} = 20\text{ ms}^{-1}$
$\implies v_{av} = \frac{\text{Total Distance Travelled}}{\text{Total Time Taken}} = \frac{l_1 + l_2 + l_1}{25}$
$\implies 2l_1 + l_2 = (20)(25) = 500$
$\implies 2l_1 + l_2 = 500$
where $l_1 = \frac{5}{2} t_1^2$
Also, for $OA$, we have
$v = 5t_1$
For $AB$, we have
$l_2 = v t_2 = 5 t_1 t_2$
$\implies 2\left(\frac{5}{2} t_1^2\right) + 5 t_1 t_2 = 500$ {substituting in (2)}
$\implies 5 t_1^2 + 5 t_1 (25 - 2t_1) = 500$
$\implies 5 t_1^2 + 125 t_1 - 10 t_1^2 = 500$
$\implies 5 t_1^2 - 125 t_1 + 500 = 0$
$\implies t_1^2 - 25 t_1 + 100 = 0$
$\implies (t_1 - 5)(t_1 - 20) = 0$
$\implies t_1 = 5\text{ s}$ (since $t_1 = 20\text{ s}$ makes $t_2 < 0$)
$\implies t_2 = 25 - 2(5) = 15\text{ s}$
Step Solution:
Step 1: Convert average speed to $\text{ms}^{-1}$ and express total distance.
$v_{av} = 72 \times \frac{5}{18} = 20\text{ ms}^{-1}$
Total distance $S = v_{av} \times t_{\text{total}} = 20 \times 25 = 500\text{ m}$.
Step 2: Express distance traveled in terms of time intervals $t_1$ and $t_2$.
Let $t_1$ be acceleration time, $t_2$ be uniform motion time, and $t_1$ be deceleration time.
$2t_1 + t_2 = 25 \implies t_2 = 25 - 2t_1$
Distance during acceleration: $l_1 = \frac{1}{2} (5) t_1^2 = 2.5 t_1^2$
Maximum speed reached: $v = a t_1 = 5 t_1$
Distance during uniform motion: $l_2 = v t_2 = (5 t_1) t_2$
Total distance $2l_1 + l_2 = 500 \implies 2(2.5 t_1^2) + 5 t_1 t_2 = 500 \implies 5 t_1^2 + 5 t_1 t_2 = 500$.
Step 3: Solve for $t_1$ using substitution.
Substitute $t_2 = 25 - 2t_1$:
$5 t_1^2 + 5 t_1 (25 - 2t_1) = 500$
$5 t_1^2 + 125 t_1 - 10 t_1^2 = 500$
$-5 t_1^2 + 125 t_1 - 500 = 0$
$t_1^2 - 25 t_1 + 100 = 0$
$(t_1 - 5)(t_1 - 20) = 0$
Selecting $t_1 = 5\text{ s}$ (as total time is $25\text{ s}$, $t_1$ cannot be $20\text{ s}$).
Step 4: Calculate duration of uniform motion $t_2$.
$t_2 = 25 - 2(5) = 15\text{ s}$.

Suppose the pickpocket is caught at a time $t$ after motorcycle starts. The distance travelled by the motorcycle during this interval is
$s = \frac{1}{2}at^2$
During this interval the car travels a distance
$s + d = vt$
$\frac{1}{2}at^2 + d = vt$
$at^2 - 2vt + 2d = 0$
$t = \frac{2v \pm \sqrt{4v^2 - 8ad}}{2a}$
$\implies t = \frac{v \pm \sqrt{v^2 - 2ad}}{a}$
For $t$ to be real:
$v^2 - 2ad \ge 0$
$\implies v \ge \sqrt{2ad}$
Step Solution:
Step 1: Formulate the distance equations for both the motorcycle and the police car.
Distance traveled by the motorcycle starting from rest with constant acceleration $a$ in time $t$:
$s = \frac{1}{2}at^2$
Distance traveled by the police car moving at uniform speed $v$ in time $t$:
$s_{\text{car}} = vt$
Step 2: Relate the two distances based on the initial separation $d$.
$s + d = vt$
$\frac{1}{2}at^2 + d = vt$
Step 3: Rearrange into a quadratic equation in terms of time $t$.
$\frac{1}{2}at^2 - vt + d = 0$
$at^2 - 2vt + 2d = 0$
Step 4: Determine the condition for a real solution of time $t$.
Using the quadratic formula, the discriminant must be greater than or equal to zero:
$D = (-2v)^2 - 4(a)(2d) \ge 0$
$4v^2 - 8ad \ge 0$
$v^2 \ge 2ad$
$v \ge \sqrt{2ad}$

During the reaction time of $0.20\text{ s}$, the car continues to move with a speed of $54\text{ kmh}^{-1}$ i.e. $15\text{ ms}^{-1}$.
So, distance travelled by car during this time is
$s_1 = (15)(0.2) = 3\text{ m}$
Now, when brakes are applied, the distance $s_2$ travelled by the car is
$0^2 - (15)^2 = 2(-6)s_2$
$\implies s_2 = \frac{225}{12}\text{ m} = 18.75\text{ m}$
Total distance travelled by the car is $s = s_1 + s_2 = 3 + 18.75 = 21.75\text{ m}$.
Step Solution:
Step 1: Convert speed from $\text{kmh}^{-1}$ to $\text{ms}^{-1}$.
$v = 54 \times \frac{5}{18} = 15\text{ ms}^{-1}$
Step 2: Calculate distance covered during the reaction time.
$s_1 = v \times t_{\text{reaction}} = 15 \times 0.20 = 3\text{ m}$
Step 3: Calculate braking distance.
Using $v_f^2 - v^2 = 2 a s_2$:
$0^2 - 15^2 = 2(-6)s_2$
$-225 = -12 s_2$
$s_2 = \frac{225}{12} = 18.75\text{ m}$
Step 4: Compute total stopping distance.
$s_{\text{total}} = s_1 + s_2 = 3 + 18.75 = 21.75\text{ m}$

For a body moving with constant acceleration $a$, we use the relation $v^2 - u^2 = 2as$.
Rearranging the equation for displacement gives $s = \frac{v^2 - u^2}{2a}$.
This equation is quadratic in $v$ and linear in $s$.
The graph between displacement $s$ on the y-axis and velocity $v$ on the x-axis is a parabola concave towards the $s$-axis.
Therefore, graph (c) correctly represents the $s-v$ relationship for uniform acceleration.

The total distance travelled is the area under the velocity-time graph from $t = 0$ to $t = 4$ seconds.
From $t = 0$ to $t = 1$ s, the velocity increases from $5 \text{ m/s}$ to $20 \text{ m/s}$. Distance $s_1 = \frac{5 + 20}{2} \times (1 - 0) = 12.5 \text{ m}$.
From $t = 1$ to $t = 2$ s, the velocity remains constant at $20 \text{ m/s}$. Distance $s_2 = 20 \times (2 - 1) = 20 \text{ m}$.
From $t = 2$ to $t = 3$ s, the velocity decreases from $20 \text{ m/s}$ to $10 \text{ m/s}$. Distance $s_3 = \frac{20 + 10}{2} \times (3 - 2) = 15 \text{ m}$.
From $t = 3$ to $t = 4$ s, the velocity decreases from $10 \text{ m/s}$ to $5 \text{ m/s}$. Distance $s_4 = \frac{10 + 5}{2} \times (4 - 3) = 7.5 \text{ m}$.
Total distance travelled = $s_1 + s_2 + s_3 + s_4 = 12.5 + 20 + 15 + 7.5 = 55 \text{ m}$.

Let $t_1$ be the time for which the car accelerates with acceleration $\alpha$.
Let $t_2$ be the time for which the car decelerates with retardation $\beta$.
The maximum velocity acquired during acceleration is $v_{\text{max}} = 0 + \alpha t_1$, which gives $t_1 = \frac{v_{\text{max}}}{\alpha}$.
The car then comes to rest from $v_{\text{max}}$ with deceleration $\beta$, so $0 = v_{\text{max}} - \beta t_2$, which gives $t_2 = \frac{v_{\text{max}}}{\beta}$.
The total time elapsed is $t = t_1 + t_2 = \frac{v_{\text{max}}}{\alpha} + \frac{v_{\text{max}}}{\beta} = v_{\text{max}} \left(\frac{\alpha + \beta}{\alpha \beta}\right)$.
Solving for $v_{\text{max}}$ gives $v_{\text{max}} = \frac{\alpha \beta t}{\alpha + \beta}$.

From the velocity-time graph, the velocity increases to a maximum value of $1000 \text{ m/s}$ at $t = 20 \text{ s}$ and then decreases to $0 \text{ m/s}$ at $t = 120 \text{ s}$.
The maximum height is reached when the velocity becomes zero at $t = 120 \text{ s}$.
This distance corresponds to the area of the triangle $OAB$ under the velocity-time curve.
Base of triangle $OAB = 120 - 0 = 120 \text{ s}$.
Height of triangle $OAB = 1000 \text{ m/s}$.
Maximum height $= \frac{1}{2} \times 120 \times 1000 = 60000 \text{ m} = 60 \text{ km}$.

From the previous problem, the maximum height reached (total displacement) is $60000 \text{ m}$.
The time taken to reach the maximum height (when velocity becomes zero) is $t = 120 \text{ s}$.
Using the formula for mean velocity, $\text{Mean velocity} = \frac{\text{Total displacement}}{\text{Total time}}$.
$\text{Mean velocity} = \frac{60000 \text{ m}}{120 \text{ s}} = 500 \text{ m/s}$.

From the velocity-time graph in problem 41, the rocket accelerates from initial velocity $v_i = 0 \text{ m/s}$ at time $t_i = 0 \text{ s}$ to a maximum velocity $v_f = 1000 \text{ m/s}$ at time $t_f = 20 \text{ s}$.
The acceleration $a$ is the slope of line $OA$.
$a = \frac{v_f - v_i}{t_f - t_i}$.
$a = \frac{1000 - 0}{20 - 0}$.
$a = 50 \text{ m/s}^2$.

The given graph is a trapezoid under the velocity-time curve representing the motion of the lift.
The base on the time axis extends from $t = 0 \text{ s}$ to $t = 12 \text{ s}$, giving total duration $a = 12 \text{ s}$.
The top parallel side represents constant velocity at $3.6 \text{ m/s}$ from $t = 2 \text{ s}$ to $t = 10 \text{ s}$, so duration $b = 10 - 2 = 8 \text{ s}$.
The maximum velocity (height of the trapezoid) is $h = 3.6 \text{ m/s}$.
The total height reached by the lift is equal to the area of the trapezoid.
$\text{Height} = \frac{1}{2} \times (a + b) \times h$.
$\text{Height} = \frac{1}{2} \times (12 + 8) \times 3.6$.
$\text{Height} = \frac{1}{2} \times 20 \times 3.6 = 10 \times 3.6 = 36.0 \text{ m}$.