Kinematics-1D

Taking East along the x-axis and North along the y-axis, the displacement vector can be written as $\vec{r} = 20\hat{i} + 10\hat{j}$.
The magnitude of the displacement is calculated as:
$\vert{}\vec{r}\vert{} = \sqrt{20^2 + 10^2}$
$\vert{}\vec{r}\vert{} = \sqrt{400 + 100}$
$\vert{}\vec{r}\vert{} = \sqrt{500}$
$\vert{}\vec{r}\vert{} = 10\sqrt{5}$
$\vert{}\vec{r}\vert{} \approx 22.36 \text{ m} \approx 22.5 \text{ m}$
Therefore, the magnitude of the displacement is approximately 22.5 m.

Let the particle start from point $A$ on the x-axis with position vector $\vec{r}_{OA} = r\hat{i}$. After covering one fourth of a circular arc, it reaches point $B$ on the y-axis with position vector $\vec{r}_{OB} = r\hat{j}$.
The distance travelled is one fourth of the total circumference of the circle:
$\text{Distance} = \frac{2\pi r}{4} = \frac{\pi r}{2}$
The displacement vector is given by:
$\vec{\Delta r} = \vec{r}_{OB} - \vec{r}_{OA} = r\hat{j} - r\hat{i}$
The magnitude of displacement is:
$\vert{}\vec{\Delta r}\vert{} = \sqrt{(-r)^2 + r^2} = \sqrt{2r^2} = r\sqrt{2}$
Therefore, the distance travelled and displacement are $\frac{\pi r}{2}$ and $r\sqrt{2}$ respectively.

Let the initial position of the contact point on the ground be at the origin $(0, 0)$.
When the wheel completes half a revolution:
Horizontal displacement along the ground = $\pi R$
Vertical height above the ground = $2R$
The final coordinates of the point become $(\pi R, 2R)$.
The magnitude of the displacement is given by the distance formula:
$\text{Displacement} = \sqrt{(\pi R - 0)^2 + (2R - 0)^2}$
$\text{Displacement} = \sqrt{\pi^2 R^2 + 4R^2}$
$\text{Displacement} = R\sqrt{\pi^2 + 4}$

Let the total distance covered be $x$.
Time taken to cover the first part of the distance $\left( \frac{2}{5}x \right)$ at speed $v_1$ is:
$t_1 = \frac{\frac{2}{5}x}{v_1} = \frac{2x}{5v_1}$
Time taken to cover the second part of the distance $\left( \frac{3}{5}x \right)$ at speed $v_2$ is:
$t_2 = \frac{\frac{3}{5}x}{v_2} = \frac{3x}{5v_2}$
Total time taken is:
$t = t_1 + t_2 = \frac{2x}{5v_1} + \frac{3x}{5v_2} = \frac{x}{5} \left( \frac{2}{v_1} + \frac{3}{v_2} \right) = \frac{x(2v_2 + 3v_1)}{5v_1 v_2}$
Average speed is:
$v_{\text{av}} = \frac{x}{t} = \frac{x}{\frac{x(2v_2 + 3v_1)}{5v_1 v_2}} = \frac{5v_1 v_2}{3v_1 + 2v_2}$

Since the car returns to its starting point, the initial position and final position are identical.
The net displacement of the car is:
$\text{Displacement} = 0$
Therefore, the velocity is given by:
$\text{Velocity} = \frac{\text{Displacement}}{\text{Time}} = 0$
However, as the car moves, the total path length covered increases continuously with time, which means the speed increases.
Hence, velocity is zero but speed increases.

Time taken to reach the market from home:
$t_1 = \frac{2.5 \text{ km}}{5 \text{ km/h}} = 0.5 \text{ h} = 30 \text{ min}$
Since we need the average speed over 40 min, the remaining time for the return trip is:
$t_2 = 40 \text{ min} - 30 \text{ min} = 10 \text{ min} = \frac{10}{60} \text{ h} = \frac{1}{6} \text{ h}$
Distance travelled during the return journey in 10 min:
$d_2 = 7.5 \text{ km/h} \times \frac{1}{6} \text{ h} = 1.25 \text{ km}$
Total distance covered in 40 min:
$d_{\text{total}} = 2.5 \text{ km} + 1.25 \text{ km} = 3.75 \text{ km}$
Total time in hours:
$t_{\text{total}} = \frac{40}{60} \text{ h} = \frac{2}{3} \text{ h}$
Average speed:
$v_{\text{av}} = \frac{3.75}{\frac{2}{3}} = \frac{11.25}{2} = \frac{45}{8} \text{ km/h}$

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.