Kinematics-1D

If we take East as x-axis and North as y-axis, then displacement = $20\hat{i} + 10\hat{j}$
So, magnitude of displacement = $\sqrt{20^2 + 10^2} = 10\sqrt{5} = 22.5$ m.
Step Solution:
Step 1: Define the direction vectors along orthogonal axes, assigning East to the x-axis ($\hat{i}$) and North to the y-axis ($\hat{j}$).
Step 2: Express the total displacement in vector form based on the given movements: $\vec{r} = 20\hat{i} + 10\hat{j}$.
Step 3: Calculate the magnitude of the displacement vector using $\vert{}\vec{r}\vert{} = \sqrt{x^2 + y^2} = \sqrt{20^2 + 10^2}$.
Step 4: Simplify the expression: $\sqrt{400 + 100} = \sqrt{500} = 10\sqrt{5}$ m.
Step 5: Compute the numerical value: $10 \times 2.236 = 22.36$ m $\approx 22.5$ m.

Let particle start from A, its position vector $\vec{r}_{OA} = r\hat{i}$
After one quarter position vector $\vec{r}_{OB} = r\hat{j}$
So displacement $= r\hat{j} - r\hat{i}$
Magnitude of displacement $= r\sqrt{2}$.
and distance = one fourth of circumference $= \frac{2\pi r}{4} = \frac{\pi r}{2}$
Step Solution:
Step 1: Identify the path length (distance) for one-fourth of a circular path of radius $r$. The total circumference of a full circle is $2\pi r$, so one-fourth of the circumference is $\frac{2\pi r}{4} = \frac{\pi r}{2}$.
Step 2: Define the initial position vector $\vec{r}_{OA} = r\hat{i}$ along the x-axis and the final position vector $\vec{r}_{OB} = r\hat{j}$ along the y-axis after traversing one-fourth of the circle.
Step 3: Calculate the displacement vector as the difference between final and initial position vectors: $\vec{\Delta r} = r\hat{j} - r\hat{i}$.
Step 4: Compute the magnitude of the displacement vector using $\vert{}\vec{\Delta r}\vert{} = \sqrt{(-r)^2 + r^2} = \sqrt{2r^2} = r\sqrt{2}$.
Step 5: Combine both results to get distance and displacement respectively as $\frac{\pi r}{2}$ and $r\sqrt{2}$.

Horizontal distance covered by the wheel in half revolution $= \pi R$
So the displacement of the point which was initially in contact with a ground $= \sqrt{(\pi R)^2 + (2R)^2}$
$= R\sqrt{\pi^2 + 4}$
Step Solution:
Step 1: Determine the horizontal distance traveled by the wheel during half a revolution. Since a full revolution covers $2\pi R$, half a revolution covers $x = \pi R$.
Step 2: Determine the vertical distance moved by the point initially at the bottom contact point. After half a revolution, this point reaches the highest position (top) of the wheel, resulting in a vertical displacement $y = 2R$.
Step 3: Calculate the magnitude of the net displacement vector using the two perpendicular components $x$ and $y$: $S = \sqrt{x^2 + y^2}$.
Step 4: Substitute the values of horizontal and vertical components into the formula: $S = \sqrt{(\pi R)^2 + (2R)^2}$.
Step 5: Factor out $R^2$ from under the square root to get $S = \sqrt{R^2(\pi^2 + 4)} = R\sqrt{\pi^2 + 4}$.

Average speed = $\frac{\text{Total distance travelled}}{\text{Total time taken}} = \frac{x}{t_1 + t_2}$
$= \frac{x}{\frac{(2/5)x}{v_1} + \frac{(3/5)x}{v_2}} = \frac{5 v_1 v_2}{2 v_2 + 3 v_1}$
Step Solution:
Step 1: Assume the total distance of the journey is $x$.
Step 2: Calculate the time taken $t_1$ to cover the first distance component $d_1 = \frac{2}{5}x$ at speed $v_1$: $t_1 = \frac{(2/5)x}{v_1}$.
Step 3: Calculate the time taken $t_2$ to cover the remaining distance component $d_2 = \frac{3}{5}x$ at speed $v_2$: $t_2 = \frac{(3/5)x}{v_2}$.
Step 4: Express the average speed equation as $v_{av} = \frac{x}{t_1 + t_2} = \frac{x}{\frac{2x}{5v_1} + \frac{3x}{5v_2}}$.
Step 5: Factor out $x$ from the denominator and simplify the expression: $v_{av} = \frac{1}{\frac{2}{5v_1} + \frac{3}{5v_2}} = \frac{1}{\frac{2v_2 + 3v_1}{5v_1 v_2}} = \frac{5v_1 v_2}{3v_1 + 2v_2}$.

As the net displacement = 0
Hence velocity = 0 ; but speed increases.
Step Solution:
Step 1: Identify the initial and final position of the car. Since the car returns to its starting point, the net displacement $\Delta r$ is $0$.
Step 2: Calculate the velocity of the car using the formula $v = \frac{\Delta r}{t} = \frac{0}{t} = 0$.
Step 3: Analyze the speed of the car. Since the car has travelled a non-zero path length (distance), its speed increases as distance accumulates over time.
Step 4: Conclude that the velocity is zero while the speed increases.

Time taken in going to market $= \frac{2.5}{5} = \frac{1}{2}$ hr $= 30$ min .
As we are told to find average speed for the interval 40 min., so remaining time for consideration of motion is 10 min.
So distance travelled in remaining 10 min $= 7.5 \times \frac{10}{60} = 1.25$ km.
Hence, average speed $= \frac{\text{Total distance}}{\text{Total time}} = \frac{(2.5 + 1.25)\text{km}}{(40 / 60)\text{hr}} = \frac{45}{8}$ km/hr .
Step Solution:
Step 1: Calculate the time taken to reach the market using $t_1 = \frac{d_1}{v_1} = \frac{2.5\text{ km}}{5\text{ km/h}} = 0.5\text{ hr} = 30\text{ min}$.
Step 2: Determine the remaining time interval out of the total 40 minutes: $t_{\text{remaining}} = 40\text{ min} - 30\text{ min} = 10\text{ min} = \frac{10}{60}\text{ hr} = \frac{1}{6}\text{ hr}$.
Step 3: Calculate the distance travelled during these remaining 10 minutes at a speed of 7.5 km/h: $d_2 = v_2 \times t_{\text{remaining}} = 7.5 \times \frac{1}{6} = 1.25\text{ km}$.
Step 4: Find the total distance covered in 40 minutes: $d_{\text{total}} = 2.5\text{ km} + 1.25\text{ km} = 3.75\text{ km}$.
Step 5: Compute the average speed over 40 minutes ($\frac{40}{60}\text{ hr} = \frac{2}{3}\text{ hr}$): $v_{av} = \frac{3.75}{2/3} = 3.75 \times \frac{3}{2} = \frac{11.25}{2} = \frac{45}{8}\text{ km/h}$.

$3t = \sqrt{3x} + 6 \implies \sqrt{3x} = (3t - 6) \implies 3x = (3t - 6)^2 \implies x = 3t^2 - 12t + 12$
$\therefore v = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 12t + 12) = 6t - 12$
If velocity = 0 then, $6t - 12 = 0 \implies t = 2\text{sec}$
Hence at $t = 2$, $x = 3(2)^2 - 12(2) + 12 = 0\text{ metres}$.
Step Solution:
Step 1: Rearrange the given relation $3t = \sqrt{3x} + 6$ to isolate $\sqrt{3x}$: $\sqrt{3x} = 3t - 6$.
Step 2: Square both sides to express $3x$ as a quadratic equation in terms of $t$: $3x = (3t - 6)^2 = 9t^2 - 36t + 36$.
Step 3: Divide by 3 to get the displacement function: $x = 3t^2 - 12t + 12$.
Step 4: Differentiate $x$ with respect to $t$ to obtain the velocity function: $v = \frac{dx}{dt} = 6t - 12$.
Step 5: Set velocity to zero to find the instant of time when the particle comes to rest: $6t - 12 = 0 \implies t = 2\text{ s}$.
Step 6: Substitute $t = 2\text{ s}$ back into the displacement equation: $x = 3(2)^2 - 12(2) + 12 = 12 - 24 + 12 = 0\text{ metres}$.

$x = a + bt^2$
$\therefore v = \frac{dx}{dt} = 0 + 2bt$
At $t = 3\text{ sec}$, $v = 2 \times 3 \times 3 = 18\text{ cm/sec}$ (As $b = 3\text{cm}$)
Step Solution:
Step 1: Write down the position equation of the particle: $x = a + bt^2$.
Step 2: Differentiate the position equation with respect to time $t$ to obtain the instantaneous velocity function: $v = \frac{dx}{dt} = \frac{d}{dt}(a + bt^2) = 2bt$.
Step 3: Substitute the given value of constant $b = 3\text{ cm/s}^2$ into the velocity equation: $v = 2(3)t = 6t$.
Step 4: Calculate the instantaneous velocity at $t = 3\text{ sec}$: $v = 6 \times 3 = 18\text{ cm/sec}$.

Total distance travelled $= 60 \times 1 + 40 \times \frac{1}{2} = 80$ km and Total time taken $= 1\text{ hr} + \frac{1}{2}\text{ hr} = \frac{3}{2}$ hr
$\therefore \text{Average speed} = \frac{80}{3/2} = 53.33$ km/h
Step Solution:
Step 1: Calculate the distance covered in the first part of the journey: $d_1 = 60\text{ km/h} \times 1\text{ h} = 60\text{ km}$.
Step 2: Calculate the distance covered in the second part of the journey: $d_2 = 40\text{ km/h} \times 0.5\text{ h} = 20\text{ km}$.
Step 3: Determine the total distance covered: $d_{\text{total}} = d_1 + d_2 = 60\text{ km} + 20\text{ km} = 80\text{ km}$.
Step 4: Determine the total time elapsed: $t_{\text{total}} = 1\text{ h} + 0.5\text{ h} = 1.5\text{ h} = \frac{3}{2}\text{ h}$.
Step 5: Compute the average speed by dividing total distance by total time: $v_{av} = \frac{80}{3/2} = \frac{160}{3} \approx 53.33\text{ km/h}$.

In this problem total distance is divided into two equal parts. So
$v_{av} = \frac{d_1 + d_2}{\frac{d_1}{v_1} + \frac{d_2}{v_2}} = \frac{\frac{d}{2} + \frac{d}{2}}{\frac{d/2}{v_1} + \frac{d/2}{v_2}} \implies v_{av} = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}} = \frac{2 v_1 v_2}{v_1 + v_2}$
Step Solution:
Step 1: Let the total distance of the journey be $d$. The journey is split into two equal distance segments, $d_1 = \frac{d}{2}$ and $d_2 = \frac{d}{2}$.
Step 2: Calculate the time taken for the first half of the journey at speed $v_1$: $t_1 = \frac{d/2}{v_1} = \frac{d}{2v_1}$.
Step 3: Calculate the time taken for the second half of the journey at speed $v_2$: $t_2 = \frac{d/2}{v_2} = \frac{d}{2v_2}$.
Step 4: Compute the total time taken for the whole journey: $t_{\text{total}} = t_1 + t_2 = \frac{d}{2v_1} + \frac{d}{2v_2} = \frac{d}{2} \left( \frac{1}{v_1} + \frac{1}{v_2} \right) = \frac{d (v_1 + v_2)}{2 v_1 v_2}$.
Step 5: Divide the total distance $d$ by total time $t_{\text{total}}$ to find average speed: $v_{av} = \frac{d}{\frac{d(v_1 + v_2)}{2 v_1 v_2}} = \frac{2 v_1 v_2}{v_1 + v_2}$.

Let total distance travelled $= x$ and total time taken $t_1 + t_2 = \frac{x/3}{20} + \frac{2x/3}{60}$
$\therefore \text{Average speed} = \frac{x}{\frac{(1/3)x}{20} + \frac{(2/3)x}{60}} = \frac{1}{\frac{1}{60} + \frac{2}{180}} = 36$ km / hr
Step Solution:
Step 1: Assume the total distance of the journey to be $x$.
Step 2: Calculate time $t_1$ taken for the first one-third distance ($d_1 = \frac{x}{3}$) at speed $v_1 = 20\text{ km/hr}$: $t_1 = \frac{x/3}{20} = \frac{x}{60}\text{ hr}$.
Step 3: Calculate time $t_2$ taken for the remaining distance ($d_2 = \frac{2x}{3}$) at speed $v_2 = 60\text{ km/hr}$: $t_2 = \frac{2x/3}{60} = \frac{2x}{180} = \frac{x}{90}\text{ hr}$.
Step 4: Express average speed formula as $v_{av} = \frac{x}{t_1 + t_2} = \frac{x}{\frac{x}{60} + \frac{x}{90}}$.
Step 5: Simplify the denominator: $\frac{x}{60} + \frac{x}{90} = x \left( \frac{3 + 2}{180} \right) = \frac{5x}{180} = \frac{x}{36}$.
Step 6: Compute the final average speed: $v_{av} = \frac{x}{x/36} = 36\text{ km/hr}$.

Given $s = 2t^2 + 2t + 4 \implies \text{velocity } (v) = \frac{ds}{dt} = 4t + 2$ and acceleration $(a) = \frac{dv}{dt} = 4(1) + 0 = 4\text{ m/s}^2$
Step Solution:
Step 1: Write down the given displacement equation of the particle: $s = 2t^2 + 2t + 4$.
Step 2: Differentiate $s$ with respect to time $t$ to find the instantaneous velocity function: $v = \frac{ds}{dt} = \frac{d}{dt}(2t^2 + 2t + 4) = 4t + 2$.
Step 3: Differentiate the velocity function $v$ with respect to time $t$ to obtain the acceleration: $a = \frac{dv}{dt} = \frac{d}{dt}(4t + 2) = 4\text{ m/s}^2$.
Step 4: Conclude that the acceleration of the particle is constant and equal to $4\text{ m/s}^2$.

Given $x = at^2 - bt^3 \implies \text{velocity } (v) = \frac{dx}{dt} = 2at - 3bt^2$ and acceleration $(a) = \frac{dv}{dt} = 2a - 6bt$.
When acceleration $= 0 \implies 2a - 6bt = 0 \implies t = \frac{2a}{6b} = \frac{a}{3b}$.
Step Solution:
Step 1: Write down the position equation of the particle: $x = at^2 - bt^3$.
Step 2: Differentiate position $x$ with respect to time $t$ to obtain the velocity function: $v = \frac{dx}{dt} = 2at - 3bt^2$.
Step 3: Differentiate velocity $v$ with respect to time $t$ to derive the acceleration function: $a = \frac{dv}{dt} = 2a - 6bt$.
Step 4: Set the acceleration expression to zero: $2a - 6bt = 0$.
Step 5: Solve for time $t$: $6bt = 2a \implies t = \frac{2a}{6b} = \frac{a}{3b}$.

Given $y = a + bt + ct^2 - dt^4 \implies v = \frac{dy}{dt} = 0 + b + 2ct - 4dt^3$
Putting $t = 0, v_{\text{initial}} = b$
So initial velocity $= b$
Now, acceleration $(a) = \frac{dv}{dt} = 0 + 2c - 12dt^2$
Putting $t = 0, a_{\text{initial}} = 2c$
Step Solution:
Step 1: Write down the given displacement equation of the particle: $y = a + bt + ct^2 - dt^4$.
Step 2: Differentiate displacement $y$ with respect to time $t$ to obtain the velocity function: $v = \frac{dy}{dt} = b + 2ct - 4dt^3$.
Step 3: Evaluate initial velocity by substituting $t = 0$ into the velocity expression: $v_{\text{initial}} = b + 2c(0) - 4d(0)^3 = b$.
Step 4: Differentiate velocity $v$ with respect to time $t$ to obtain the acceleration function: $a = \frac{dv}{dt} = 2c - 12dt^2$.
Step 5: Evaluate initial acceleration by substituting $t = 0$ into the acceleration expression: $a_{\text{initial}} = 2c - 12d(0)^2 = 2c$.
Step 6: Conclude that the initial velocity and acceleration are $b$ and $2c$ respectively.

differentiating time with respect to distance $\frac{dt}{dx} = 2\alpha x + \beta \implies v = \frac{dx}{dt} = \frac{1}{2\alpha x + \beta}$
So, acceleration $(a) = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} = \frac{-v \cdot 2\alpha}{(2\alpha x + \beta)^2} = -2\alpha \cdot v \cdot v^2 = -2\alpha v^3$
Step Solution:
Step 1: Write down the relation between time and distance: $t = \alpha x^2 + \beta x$.
Step 2: Differentiate time $t$ with respect to distance $x$: $\frac{dt}{dx} = 2\alpha x + \beta$.
Step 3: Express velocity $v$ as the reciprocal of $\frac{dt}{dx}$: $v = \frac{dx}{dt} = \frac{1}{2\alpha x + \beta} = (2\alpha x + \beta)^{-1}$.
Step 4: Use the chain rule to find acceleration $a = v \frac{dv}{dx}$. First find $\frac{dv}{dx} = \frac{d}{dx}(2\alpha x + \beta)^{-1} = -(2\alpha x + \beta)^{-2} \cdot 2\alpha = -2\alpha v^2$.
Step 5: Calculate acceleration: $a = v \left(-2\alpha v^2\right) = -2\alpha v^3$.
Step 6: Since retardation is defined as the magnitude of deceleration (retardation $= -a$), the retardation is $2\alpha v^3$.

Given that $x \propto t^2$ or $x = K t^2$ (where $K =$ constant)
$\text{Velocity } v = \frac{dx}{dt} = 2Kt$
$\text{Acceleration } a = \frac{dv}{dt} = 2K = \text{constant}$
Hence the particle is moving with uniform acceleration.
Step Solution:
Step 1: Express the given relationship between displacement $x$ and time $t$ mathematically: $x \propto t^2 \implies x = K t^2$, where $K$ is a constant of proportionality.
Step 2: Differentiate the displacement $x$ with respect to time $t$ to find the velocity: $v = \frac{dx}{dt} = \frac{d}{dt}(K t^2) = 2 K t$.
Step 3: Differentiate the velocity $v$ with respect to time $t$ to find the acceleration: $a = \frac{dv}{dt} = \frac{d}{dt}(2 K t) = 2 K$.
Step 4: Observe that since $K$ is a constant, the acceleration $a = 2 K$ is also constant with respect to time, which defines uniform acceleration.

$\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$
$\Delta v = \sqrt{v_1^2 + v_2^2 - 2v_1v_2\cos 90^\circ} = \sqrt{5^2 + 5^2} = 5\sqrt{2}$
$\Delta v = 5\sqrt{2}$
$\text{Average acceleration} = \frac{\Delta v}{\Delta t} = \frac{5\sqrt{2}}{10} = \frac{1}{\sqrt{2}}\text{ m/s}^2\text{ toward north-west}$
Step Solution:
Step 1: Write down the initial and final velocity vectors: $\vec{v}_1 = 5\hat{i}\text{ m/s}$ (towards East) and $\vec{v}_2 = 5\hat{j}\text{ m/s}$ (towards North).
Step 2: Express the change in velocity vector as $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1 = 5\hat{j} - 5\hat{i}$.
Step 3: Calculate the magnitude of the change in velocity: $\Delta v = \sqrt{v_1^2 + v_2^2 - 2v_1v_2\cos 90^\circ} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}\text{ m/s}$.
Step 4: Determine the direction of $\Delta\vec{v} = 5\hat{j} - 5\hat{i}$, which combines North ($\hat{j}$) and West ($-\hat{i}$), pointing in the North-West direction.
Step 5: Calculate the average acceleration magnitude by dividing $\Delta v$ by the time interval $\Delta t = 10\text{ sec}$: $a_{av} = \frac{\Delta v}{\Delta t} = \frac{5\sqrt{2}}{10} = \frac{1}{\sqrt{2}}\text{ m/s}^2$.
Step 6: Combine magnitude and direction to conclude that the average acceleration is $\frac{1}{\sqrt{2}}\text{ m/s}^2$ toward north-west.

Given that $v = 4t^3 - 2t$
$x = \int v \, dt, \quad x = t^4 - t^2 + C, \text{ at } t = 0, x = 0 \implies C = 0$
When particle is 2m away from the origin
$2 = t^4 - t^2 \implies t^4 - t^2 - 2 = 0 \implies (t^2 - 2)(t^2 + 1) = 0 \implies t = \sqrt{2}\text{ sec}$
$a = \frac{dv}{dt} = \frac{d}{dt}(4t^3 - 2t) = 12t^2 - 2 \implies a = 12t^2 - 2$
for $t = \sqrt{2}\text{ sec} \implies a = 12 \times (\sqrt{2})^2 - 2 \implies a = 22\text{ m/s}^2$
Step Solution:
Step 1: Write down the given velocity equation of the particle: $v = 4t^3 - 2t$.
Step 2: Integrate velocity $v$ with respect to time $t$ to find the position equation $x(t)$: $x = \int (4t^3 - 2t) \, dt = t^4 - t^2 + C$.
Step 3: Apply the initial condition $x(0) = 0$ to find the integration constant $C = 0$, giving position $x = t^4 - t^2$.
Step 4: Set position $x = 2\text{ m}$ to find time $t$: $t^4 - t^2 - 2 = 0 \implies (t^2 - 2)(t^2 + 1) = 0$. Since $t^2 \ge 0$, we have $t^2 = 2 \implies t = \sqrt{2}\text{ sec}$.
Step 5: Differentiate velocity $v$ with respect to time $t$ to find acceleration $a$: $a = \frac{dv}{dt} = \frac{d}{dt}(4t^3 - 2t) = 12t^2 - 2$.
Step 6: Substitute $t = \sqrt{2}\text{ sec}$ into the acceleration equation: $a = 12(\sqrt{2})^2 - 2 = 12(2) - 2 = 22\text{ m/s}^2$.

Let particle moves towards east and by the application of constant force it moves towards west
$\vec{v}_1 = +10\text{ m/s}$ and $\vec{v}_2 = -2\text{ m/s}$. $\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time}} = \frac{\vec{v}_2 - \vec{v}_1}{t}$
$\implies a = \frac{(-2) - (10)}{4} = \frac{-12}{4} = -3\text{ m/s}^2$
Step Solution:
Step 1: Choose a reference direction (e.g., East as positive and West as negative).
Step 2: State the initial velocity vector $\vec{v}_1 = +10\text{ m/s}$ and final velocity vector $\vec{v}_2 = -2\text{ m/s}$.
Step 3: Calculate the change in velocity vector: $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1 = (-2) - (10) = -12\text{ m/s}$.
Step 4: Divide the change in velocity by the time interval $t = 4\text{ sec}$ to calculate acceleration: $a = \frac{\Delta \vec{v}}{t} = \frac{-12}{4} = -3\text{ m/s}^2$.

Instantaneous velocity $v = \frac{\Delta x}{\Delta t}$, By using the data from the table
$v_1 = \frac{0 - (-2)}{1} = 2\text{ m/s}$, $v_2 = \frac{6 - 0}{1} = 6\text{ m/s}$, $v_3 = \frac{16 - 6}{1} = 10\text{ m/s}$
It is clear that velocity increases by $4\text{ m/s}$ in each second, so acceleration is uniform and equal to $4\text{ m/s}^2$.
Step Solution:
Step 1: Calculate the average velocity for the first time interval $t = 0\text{ s}$ to $t = 1\text{ s}$: $v_1 = \frac{x_1 - x_0}{t_1 - t_0} = \frac{0 - (-2)}{1 - 0} = 2\text{ m/s}$.
Step 2: Calculate the average velocity for the second time interval $t = 1\text{ s}$ to $t = 2\text{ s}$: $v_2 = \frac{x_2 - x_1}{t_2 - t_1} = \frac{6 - 0}{2 - 1} = 6\text{ m/s}$.
Step 3: Calculate the average velocity for the third time interval $t = 2\text{ s}$ to $t = 3\text{ s}$: $v_3 = \frac{x_3 - x_2}{t_3 - t_2} = \frac{16 - 6}{3 - 2} = 10\text{ m/s}$.
Step 4: Observe the change in velocity per second: $a_1 = v_2 - v_1 = 6 - 2 = 4\text{ m/s}^2$ and $a_2 = v_3 - v_2 = 10 - 6 = 4\text{ m/s}^2$.
Step 5: Since the acceleration is constant ($4\text{ m/s}^2$) and positive, the motion is uniform and accelerated.

When distance time graph is a straight line with constant slope than motion is uniform.
Step Solution:
Step 1: Identify that uniform motion corresponds to a motion with constant speed/velocity.
Step 2: Note that the slope of a distance-time ($s-t$) graph gives the speed of the object: $v = \frac{ds}{dt}$.
Step 3: Analyze graph (a): The $s-t$ graph is a straight line inclined to the time axis, meaning its slope is constant ($v = \text{constant}$), representing uniform motion.
Step 4: Analyze graph (b): The slope is variable, representing non-uniform motion.
Step 5: Analyze graph (c): The slope is zero, meaning the distance does not change over time and the particle is at rest ($v = 0$).
Step 6: Analyze graph (d): The slope changes direction from positive to negative, representing non-uniform motion.
Step 7: Conclude that graph (a) correctly represents uniform motion.

$v = \tan\theta$ from displacement graph. So $\frac{v_A}{v_B} = \frac{\tan 30^\circ}{\tan 60^\circ} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{\sqrt{3} \times \sqrt{3}} = \frac{1}{3}$
Step Solution:
Step 1: Identify that the slope of a displacement-time graph represents the velocity: $v = \tan\theta$.
Step 2: Calculate the velocity of particle $A$ using its angle $\theta_A = 30^\circ$: $v_A = \tan 30^\circ = \frac{1}{\sqrt{3}}$.
Step 3: Calculate the velocity of particle $B$ using its angle $\theta_B = 60^\circ$: $v_B = \tan 60^\circ = \sqrt{3}$.
Step 4: Take the ratio of the two velocities: $\frac{v_A}{v_B} = \frac{\tan 30^\circ}{\tan 60^\circ} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}$.
Step 5: Conclude that the ratio of velocities $v_A : v_B$ is $1 : 3$.

In first instant you will apply $v = \tan\theta$ and say, $v = \tan 30^\circ = \frac{1}{\sqrt{3}}\text{ m/s}$.
But it is wrong because formula $v = \tan\theta$ is valid when angle is measured with time axis.
Here angle is taken from displacement axis. So angle from time axis $= 90^\circ - 30^\circ = 60^\circ$.
Now $v = \tan 60^\circ = \sqrt{3}$
Step Solution:
Step 1: Observe the graph given in the image. Notice that the vertical axis represents Time (sec) and the horizontal axis represents Displacement (meter).
Step 2: Identify the angle given in the figure: $30^\circ$ is measured relative to the horizontal axis (Displacement axis).
Step 3: Calculate the angle formed by the line with the vertical axis (Time axis): $\theta = 90^\circ - 30^\circ = 60^\circ$.
Step 4: Compute the velocity $v$ using the slope with respect to the time axis: $v = \tan\theta = \tan 60^\circ = \sqrt{3}\text{ m/s}$.

Average velocity $= \frac{\text{Total displacement}}{\text{Total time}} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{-10 - 0}{5 - 0} = -2\text{ m/s}$
Step Solution:
Step 1: Identify the initial position $x_i = 0$ at time $t_i = 0\text{ s}$ from the displacement-time graph.
Step 2: Identify the final position $x_f = -10\text{ m}$ at time $t_f = 5\text{ s}$ from the displacement-time graph.
Step 3: Calculate total displacement $\Delta x = x_f - x_i = -10 - 0 = -10\text{ m}$.
Step 4: Calculate total time elapsed $\Delta t = t_f - t_i = 5 - 0 = 5\text{ s}$.
Step 5: Compute the average velocity $v_{av} = \frac{\Delta x}{\Delta t} = \frac{-10}{5} = -2\text{ ms}^{-1}$.

Speed in first second $= 30$ and Speed in next two seconds $= 15$. So that ratio $2 : 1$
Step Solution:
Step 1: Calculate the speed in the first second (interval $t = 0$ to $t = 1\text{ s}$): The displacement changes from $0\text{ m}$ to $30\text{ m}$. Speed $v_1 = \frac{30 - 0}{1 - 0} = 30\text{ m/s}$.
Step 2: Calculate the speed in the next two seconds (interval $t = 1\text{ s}$ to $t = 3\text{ s}$): The displacement changes from $30\text{ m}$ to $0\text{ m}$. Speed $v_2 = \left\vert{} \frac{0 - 30}{3 - 1} \right\vert{} = \left\vert{} \frac{-30}{2} \right\vert{} = 15\text{ m/s}$.
Step 3: Calculate the ratio of the speed in the first second to the speed in the next two seconds: $\text{Ratio} = \frac{v_1}{v_2} = \frac{30}{15} = \frac{2}{1} = 2 : 1$.

In first half of motion the acceleration is uniform & velocity gradually decreases, so slope will be negative but for next half acceleration is positive. So slope will be positive. Thus graph 'C' is correct.
Not ignoring air resistance means upward motion will have acceleration $(a + g)$ and the downward motion will have $(g - a)$.
Step Solution:
Step 1: During the upward motion, gravity and air resistance act in the same downward direction, leading to a effective retardation magnitude of $a_1 = g + a$.
Step 2: Speed decreases to zero at the peak, represented by a straight line with a steeper negative slope of magnitude $(g + a)$.
Step 3: During the downward motion, gravity acts downwards while air resistance acts upwards, resulting in a net downward acceleration magnitude of $a_2 = g - a$.
Step 4: Speed increases from zero, represented by a straight line with a smaller positive slope of magnitude $(g - a)$.
Step 5: Comparing the slopes of ascent and descent, the slope during ascent is steeper than during descent, which corresponds to graph (c).

Maximum acceleration means maximum slope of $v-t$ graph.
$\text{Max slope} = \frac{60 - 20}{1.25 - 1.00} = \frac{40}{0.25} = 160\text{ km/hr}^2$
Step Solution:
Step 1: Identify the intervals with positive slope (acceleration) from the graph: section $AB$ and section $CD$.
Step 2: Calculate the acceleration for section $AB$ (from $t = 0.25\text{ h}$ to $t = 0.75\text{ h}$): $a_{AB} = \frac{20 - 0}{0.75 - 0.25} = \frac{20}{0.50} = 40\text{ km/h}^2$.
Step 3: Calculate the acceleration for section $CD$ (from $t = 1.00\text{ h}$ to $t = 1.25\text{ h}$): $a_{CD} = \frac{60 - 20}{1.25 - 1.00} = \frac{40}{0.25} = 160\text{ km/h}^2$.
Step 4: Compare the acceleration values: $a_{CD} = 160\text{ km/h}^2 > a_{AB} = 40\text{ km/h}^2$.
Step 5: Conclude that the maximum acceleration during the journey is $160\text{ km/h}^2$.

Displacement-time graph is a parabola opening downwards.
The slope of this curve decreases continuously with time.
Initial slope is positive, becomes zero at the peak, and then turns negative.
Thus, the velocity decreases linearly with time, forming a straight line with a negative slope.
Step Solution:
Step 1: Identify the given graph of displacement $s$ versus time $t$. The graph represents a parabola opening downwards, which can be expressed mathematically as $s(t) = -at^2 + bt + c$ where $a > 0$.
Step 2: Calculate the velocity $v$ by taking the first derivative of displacement with respect to time $t$: $v = \frac{ds}{dt} = -2at + b$.
Step 3: Analyze the equation of velocity $v = -2at + b$. This is a linear equation of the form $y = mx + c$ with a negative slope $-2a$ and a positive y-intercept $b$.
Step 4: At $t = 0$, velocity is positive ($v = b$). As time increases, velocity decreases linearly, becomes zero at the maximum point of displacement, and becomes negative thereafter.
Step 5: Match this behavior with option A, which correctly displays a straight line with a negative slope starting from a positive velocity.

Distance = The area under $v-t$ graph
$S = \frac{1}{2} (30 + 10) \times 10 = 200 \text{ metre}$
Step Solution:
Step 1: Identify the shape of the region under the given velocity-time graph, which is a trapezium with parallel sides along the time axis.
Step 2: Note the length of the bottom base of the trapezium, which corresponds to the total time interval from $t = 0$ to $t = 30\text{ s}$, so base $b_1 = 30\text{ s}$.
Step 3: Note the length of the top base of the trapezium, which corresponds to the time interval where velocity is constant from $t = 10\text{ s}$ to $t = 20\text{ s}$, so base $b_2 = 20 - 10 = 10\text{ s}$.
Step 4: Identify the height of the trapezium, which corresponds to the maximum velocity $v = 10\text{ m/s}$.
Step 5: Apply the formula for the area of a trapezium to calculate the total distance covered: $S = \frac{1}{2} \times (b_1 + b_2) \times h = \frac{1}{2} \times (30 + 10) \times 10 = \frac{1}{2} \times 40 \times 10 = 200\text{ m}$.

Distance covered in total 7 seconds = Area of trapezium $ABCD = \frac{1}{2}(2 + 6) \times 10 = 40\text{ m}$
Distance covered in last 2 second = area of triangle $CDQ = \frac{1}{2} \times 2 \times 10 = 10\text{ m}$
So required fraction = $\frac{10}{40} = \frac{1}{4}$
Step Solution:
Step 1: Calculate the total distance covered in all 7 seconds by finding the area of the trapezium $ABCD$. The parallel sides are of length $(5 - 3) = 2\text{ s}$ and $(7 - 1) = 6\text{ s}$, and the height is $10\text{ m/s}$.
Step 2: Total distance = $\frac{1}{2} \times (2 + 6) \times 10 = \frac{1}{2} \times 8 \times 10 = 40\text{ m}$.
Step 3: Calculate the distance covered in the last two seconds (from $t = 5\text{ s}$ to $t = 7\text{ s}$) by finding the area of triangle $CDQ$. The base is $(7 - 5) = 2\text{ s}$ and the height is $10\text{ m/s}$.
Step 4: Distance in last two seconds = $\frac{1}{2} \times 2 \times 10 = 10\text{ m}$.
Step 5: Determine the required fraction by dividing the distance in the last two seconds by the total distance: $\text{Fraction} = \frac{10}{40} = \frac{1}{4}$.

Area of region $1 = 2 \times 4 = 8\text{ m}$
Area of region $2 = 2 \times (-2) = -4\text{ m}$
Area of region $3 = (6 - 4) \times 2 = 4\text{ m}$
Displacement = $8 - 4 + 4 = 8\text{ m}$
Distance = $8 + 4 + 4 = 16\text{ m}$
Step Solution:
Step 1: Divide the velocity-time graph into three distinct rectangular regions based on the time intervals.
Step 2: Calculate the area of the first region $A_1$ from $t = 0$ to $t = 2\text{ s}$ with velocity $v = 4\text{ m/s}$: $A_1 = 2 \times 4 = 8\text{ m}$.
Step 3: Calculate the area of the second region $A_2$ from $t = 2\text{ s}$ to $t = 4\text{ s}$ with velocity $v = -2\text{ m/s}$: $A_2 = 2 \times (-2) = -4\text{ m}$.
Step 4: Calculate the area of the third region $A_3$ from $t = 4\text{ s}$ to $t = 6\text{ s}$ with velocity $v = 2\text{ m/s}$: $A_3 = (6 - 4) \times 2 = 4\text{ m}$.
Step 5: Calculate total displacement by summing the signed areas: $\text{Displacement} = 8 - 4 + 4 = 8\text{ m}$.
Step 6: Calculate total distance by summing the magnitudes of all areas: $\text{Distance} = 8 + 4 + 4 = 16\text{ m}$.

In the positive region the velocity decreases linearly (during rise) and in negative region velocity increase linearly (during fall) and the direction is opposite to each other during rise and fall, hence fall is shown in negative region.
Step Solution:
Step 1: Consider the initial state where a ball is thrown vertically upwards with an initial positive velocity $u$.
Step 2: Apply the first equation of motion under constant gravitational retardation: $v = u - gt$, where $g$ is the acceleration due to gravity acting downwards.
Step 3: Analyze the motion during the upward journey: as time $t$ increases, velocity $v$ decreases linearly with a constant negative slope $-g$ until it reaches zero at the peak height ($v = 0$).
Step 4: Analyze the motion during the downward journey: the ball reverses its direction, so its velocity becomes negative and continuously increases in magnitude with time ($v = -g(t - t_{\text{peak}})$).
Step 5: Identify the corresponding graph, which is a single straight line with a constant negative slope starting from a positive value on the velocity axis, crossing the time axis at the maximum height, and continuing into the negative velocity region (Option D).

When ball is dropped from height $d$ its velocity will be zero.
As height decreases, magnitude of velocity increases in negative direction.
On hitting the ground ($h = 0$), velocity suddenly reverses direction to positive.
As it ascends to height $\frac{d}{2}$, velocity decreases parabolically to zero.
Step Solution:
Step 1: Consider the downward journey of the ball starting from rest ($v = 0$) at height $h = d$.
Step 2: Using the third equation of motion $v^2 = u^2 + 2gS$, where displacement $S = d - h$ and taking downward direction as negative, we get $v = -\sqrt{2g(d - h)}$. This indicates a parabolic relationship between $v$ and $h$ in the negative velocity region.
Step 3: When the ball reaches the ground ($h = 0$), its velocity reaches maximum magnitude in the negative direction: $v = -\sqrt{2gd}$.
Step 4: After hitting the ground, the ball bounces upwards, reversing its velocity direction to positive. At $h = 0$, its upward velocity becomes positive: $v = \sqrt{2g\left(\frac{d}{2}\right)} = \sqrt{gd}$.
Step 5: During the upward journey, velocity decreases with height according to $v = \sqrt{2g\left(\frac{d}{2} - h\right)}$ until it becomes zero at height $h = \frac{d}{2}$.
Step 6: Comparing these two parabolic segments with the given choices, option A correctly depicts this behavior.

From given $a - t$ graph acceleration is increasing at constant rate
$\frac{dv}{dt} = kt \Rightarrow v = \frac{kt^2}{2}$
i.e. $v - t$ graph should be parabola opening upwards.
After some time $a = 0$ i.e. $v = \text{constant}$.
This interpretation is clearly shown by graph (c).
Step Solution:
Step 1: Analyze the given acceleration-time ($a-t$) graph. In the first region, acceleration increases linearly with time starting from zero, which can be expressed as $a = kt$, where $k$ is a positive constant representing the slope.
Step 2: Relate acceleration to velocity using the differential relationship $a = \frac{dv}{dt} = kt$.
Step 3: Integrate $\frac{dv}{dt} = kt$ with respect to time $t$ to find the expression for velocity: $v = \int kt \, dt = \frac{kt^2}{2} + C$.
Step 4: The quadratic relation $v \propto t^2$ indicates that the velocity-time graph is a parabola opening upwards, meaning the slope of the $v-t$ curve increases as time progresses.
Step 5: Analyze the second region of the acceleration-time graph where acceleration suddenly drops to zero ($a = 0$).
Step 6: When acceleration is zero, the rate of change of velocity is zero ($\frac{dv}{dt} = 0$), which means the velocity becomes constant ($v = \text{constant}$). A constant velocity is represented by a horizontal line parallel to the time axis.
Step 7: Match this combined behavior (an upward-curving parabola followed by a flat horizontal line) with option C.

In graph (d) for one value of time there are two values of velocity which is not possible.
Step Solution:
Step 1: Examine the fundamental physical restriction on motion: time $t$ increases continuously and a particle can possess only one instantaneous velocity $v$ at any given moment.
Step 2: Apply the vertical line test to each of the given velocity-time ($v-t$) graphs.
Step 3: Graphs (a), (b), and (c) represent single-valued functions where each instant of time $t$ corresponds to exactly one velocity $v$.
Step 4: Graph (d) shows a closed loop shape (oval/circular curve). Drawing a vertical line through this graph intersects the curve at two distinct points for a single value of time $t$.
Step 5: Since a body cannot have two different velocities at the same instant of time, graph (d) is physically impossible.

Ratio of applied force = Ratio of acceleration
$= \frac{a_{AB}}{a_{BC}} = \frac{\tan 30^\circ}{\tan(120^\circ)} = \frac{1/\sqrt{3}}{-\sqrt{3}} = -1/3$
Step Solution:
Step 1: Identify that for a body of constant mass $m$, the force acting on it during any interval is proportional to its acceleration in that interval ($F = ma$).
Step 2: Note that the acceleration is given by the slope of the velocity-time graph, $a = \tan\theta$, where $\theta$ is the angle made with the positive time axis.
Step 3: Calculate the acceleration for the interval $AB$: the line segment $AD$ makes an angle of $30^\circ$ with the time axis, so $a_{AB} = \tan 30^\circ = \frac{1}{\sqrt{3}}$.
Step 4: Calculate the acceleration for the interval $BC$: the line segment $DC$ makes an angle of $60^\circ$ with the negative direction of the time axis, which corresponds to an inclination of $180^\circ - 60^\circ = 120^\circ$ with the positive time axis. Thus, $a_{BC} = \tan 120^\circ = -\sqrt{3}$.
Step 5: Compute the ratio of the applied forces: $\frac{F_{AB}}{F_{BC}} = \frac{a_{AB}}{a_{BC}} = \frac{1/\sqrt{3}}{-\sqrt{3}} = -\frac{1}{3}$.

At a certain instant $t$ slope of $A$ is greater than $B$ ($\theta_A > \theta_B$), so acceleration in $A$ is greater than $B$
Step Solution:
Step 1: Observe that both cars start from rest at time $t = 0$, meaning their initial velocities are both zero.
Step 2: Note that acceleration at any time $t$ is represented by the slope of the velocity-time graph, $a = \tan\theta$.
Step 3: At any given instant $t$, draw tangents to both curves $A$ and $B$.
Step 4: Compare the angles made by the tangents with the time axis: the angle for curve $A$ ($\theta_A$) is greater than the angle for curve $B$ ($\theta_B$).
Step 5: Since $\theta_A > \theta_B$, it follows that $\tan\theta_A > \tan\theta_B$, which means the instantaneous acceleration of car $A$ is greater than that of car $B$.

Initially velocity increases in negative direction and after collision velocity becomes positive and then decreases.
Step Solution:
Step 1: Consider the initial motion as the steel ball falls freely under gravity from a height. Taking upward as the positive direction, the downward velocity increases linearly from zero in the negative direction, represented by $v = -gt$.
Step 2: When the ball hits the marble floor, an elastic or semi-elastic collision occurs, instantly reversing the direction of motion. The velocity abruptly jumps from a negative value to a positive value.
Step 3: After rebounding from the floor, the ball moves upward under the influence of gravity. Its velocity starts at a positive maximum and decreases linearly with time according to $v = u - gt$ until it reaches zero at the peak height.
Step 4: Match this pattern—a linearly decreasing negative velocity followed by an instantaneous step-up to a positive velocity that linearly decreases to zero—with option A.

Acceleration along $OA = \frac{v_2 - v_1}{t} = \frac{10 - 0}{10} = 1\text{ m/s}^2$
Acceleration along $AB = \frac{0}{10} = 0$
Acceleration along $BC = \frac{0 - 10}{20} = -0.5\text{ m/s}^2$
Step Solution:
Step 1: Calculate the acceleration along segment $OA$. Initial velocity at $O$ is $v_1 = 0\text{ m/s}$ at $t_1 = 0\text{ s}$, and final velocity at $A$ is $v_2 = 10\text{ m/s}$ at $t_2 = 10\text{ s}$. Acceleration $a_{OA} = \frac{10 - 0}{10 - 0} = 1\text{ m/s}^2$.
Step 2: Calculate the acceleration along segment $AB$. The velocity remains constant at $v = 10\text{ m/s}$ from $t = 10\text{ s}$ to $t = 20\text{ s}$. Acceleration $a_{AB} = \frac{10 - 10}{20 - 10} = 0\text{ m/s}^2$.
Step 3: Calculate the acceleration along segment $BC$. Velocity decreases from $v_1 = 10\text{ m/s}$ at $t_1 = 20\text{ s}$ to $v_2 = 0\text{ m/s}$ at $t_2 = 40\text{ s}$. Acceleration $a_{BC} = \frac{0 - 10}{40 - 20} = \frac{-10}{20} = -0.5\text{ m/s}^2$.
Step 4: Combine the three calculated acceleration values $(1, 0, -0.5)$, which matches option A.

Let they meet after time $t$. Distance covered by body $A = \frac{1}{2}at^2$; Distance covered by body $B = vt$
and $\frac{1}{2}at^2 = vt \therefore t = \frac{2v}{a}$.
Step Solution:
Step 1: Write the equation for the distance $S_A$ travelled by body $A$ moving with initial velocity $u = 0$ and uniform acceleration $a$ in time $t$: $S_A = ut + \frac{1}{2}at^2 = \frac{1}{2}at^2$.
Step 2: Write the equation for the distance $S_B$ travelled by body $B$ moving with a constant velocity $v$ in time $t$: $S_B = vt$.
Step 3: Since both bodies start from the same point and meet after time $t$, set their distances equal to each other: $\frac{1}{2}at^2 = vt$.
Step 4: Divide both sides by $t$ (since $t \neq 0$ at the meeting point): $\frac{1}{2}at = v$.
Step 5: Solve for time $t$: $t = \frac{2v}{a}$, which corresponds to option A.

Let student will catch the bus after $t\text{ sec}$. So it will cover distance $ut$.
Similarly distance travelled by the bus will be $\frac{1}{2}at^2$ for the given condition
$ut = 50 + \frac{1}{2}at^2 = 50 + \frac{t^2}{2} \Rightarrow u = \frac{50}{t} + \frac{t}{2}$ (As $a = 1\text{ m/s}^2$)
To find the minimum value of $u$, $\frac{du}{dt} = 0$, so we get $t = 10\text{ sec}$
then $u = 10\text{ m/s}$.
Step Solution:
Step 1: Set up the equation of motion for both the student and the bus. In time $t$, the student running at constant velocity $u$ covers a distance $s_{\text{student}} = ut$.
Step 2: The bus starts from rest ($u_0 = 0$) with acceleration $a = 1\text{ ms}^{-2}$, so it covers a distance $s_{\text{bus}} = \frac{1}{2}at^2 = \frac{1}{2}(1)t^2 = \frac{t^2}{2}$.
Step 3: Since the student is initially $50\text{ m}$ behind the bus, for the student to catch the bus, $s_{\text{student}} = 50 + s_{\text{bus}}$, which yields $ut = 50 + \frac{t^2}{2}$.
Step 4: Express velocity $u$ as a function of time $t$: $u = \frac{50}{t} + \frac{t}{2}$.
Step 5: Differentiate $u$ with respect to $t$ and set the derivative to zero to find the time for minimum velocity: $\frac{du}{dt} = -\frac{50}{t^2} + \frac{1}{2} = 0 \Rightarrow t^2 = 100 \Rightarrow t = 10\text{ s}$.
Step 6: Substitute $t = 10\text{ s}$ back into the expression for $u$: $u = \frac{50}{10} + \frac{10}{2} = 5 + 5 = 10\text{ ms}^{-1}$.

$v^2 = u^2 - 2as \Rightarrow 0 = u^2 - 2as \Rightarrow s = \frac{u^2}{2a} \Rightarrow s \propto u^2\text{ (As } a = \text{constant)}$
$\frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^2 = \left(\frac{100}{50}\right)^2 = 4 \Rightarrow s_2 = 4 s_1 = 4 \times 6 = 24\text{ m}$
Step Solution:
Step 1: Write down the third equation of motion relating initial speed $u$, final speed $v = 0$, retardation $a$, and stopping distance $s$: $v^2 = u^2 - 2as$.
Step 2: Substitute $v = 0$ to express the stopping distance $s$: $0 = u^2 - 2as \Rightarrow s = \frac{u^2}{2a}$.
Step 3: Since the retarding acceleration $a$ remains constant, the stopping distance is directly proportional to the square of initial speed: $s \propto u^2$.
Step 4: Form the ratio of stopping distances for the two speeds $u_1 = 50\text{ km/hr}$ and $u_2 = 100\text{ km/hr}$: $\frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^2 = \left(\frac{100}{50}\right)^2 = 2^2 = 4$.
Step 5: Calculate the new minimum stopping distance $s_2$: $s_2 = 4 \times s_1 = 4 \times 6\text{ m} = 24\text{ m}$.

$u = 200\text{ m/s}, v = 100\text{ m/s}, s = 0.1\text{ m}$
$a = \frac{u^2 - v^2}{2s} = \frac{(200)^2 - (100)^2}{2 \times 0.1} = 15 \times 10^4\text{ m/s}^2$
Step Solution:
Step 1: Identify the given values from the problem statement: initial velocity $u = 200\text{ m/s}$, final velocity $v = 100\text{ m/s}$, and distance travelled $s = 10\text{ cm} = 0.1\text{ m}$.
Step 2: Use the third equation of motion for uniform retardation: $v^2 = u^2 - 2as$.
Step 3: Rearrange the formula to solve for acceleration/retardation $a$: $a = \frac{u^2 - v^2}{2s}$.
Step 4: Substitute the values into the equation: $a = \frac{(200)^2 - (100)^2}{2 \times 0.1} = \frac{40000 - 10000}{0.2} = \frac{30000}{0.2}$.
Step 5: Simplify the expression to calculate the magnitude of retardation: $a = 150000 = 15 \times 10^4\text{ m/s}^2$, which corresponds to option D.

By using $S_n = u + \frac{a}{2}(2n - 1)$, Distance travelled by body A in $5^{\text{th}}$ second $= 0 + \frac{a_1}{2}(2 \times 5 - 1)$
Distance travelled by body B in $3^{\text{rd}}$ second is $= 0 + \frac{a_2}{2}(2 \times 3 - 1)$
According to problem : $0 + \frac{a_1}{2}(2 \times 5 - 1) = 0 + \frac{a_2}{2}(2 \times 3 - 1) \Rightarrow 9a_1 = 5a_2 \Rightarrow \frac{a_1}{a_2} = \frac{5}{9}$
Step Solution:
Step 1: Write down the formula for the distance travelled by an object in the $n^{\text{th}}$ second of its motion: $S_n = u + \frac{a}{2}(2n - 1)$.
Step 2: Calculate the distance travelled by body A in its $5^{\text{th}}$ second of motion starting from rest ($u = 0$) with acceleration $a_1$: $S_{5, A} = 0 + \frac{a_1}{2}(2 \times 5 - 1) = \frac{9a_1}{2}$.
Step 3: Determine the corresponding time interval for body B. Since body B starts 2 seconds after body A, the $5^{\text{th}}$ second of body A's motion corresponds to the $3^{\text{rd}}$ second of body B's motion ($n = 5 - 2 = 3$).
Step 4: Calculate the distance travelled by body B in its $3^{\text{rd}}$ second of motion starting from rest ($u = 0$) with acceleration $a_2$: $S_{3, B} = 0 + \frac{a_2}{2}(2 \times 3 - 1) = \frac{5a_2}{2}$.
Step 5: Set the two distances equal as given in the problem: $\frac{9a_1}{2} = \frac{5a_2}{2} \Rightarrow 9a_1 = 5a_2$.
Step 6: Rearrange the terms to find the ratio $a_1 : a_2$: $\frac{a_1}{a_2} = \frac{5}{9}$, which corresponds to option A.

$\text{Time} = \frac{\text{Distance}}{\text{Average velocity}} = \frac{3.06}{0.34} = 9\text{ sec}$
and $\text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time}} = \frac{0.18}{9} = 0.02\text{ m/s}^2$.
Step Solution:
Step 1: Identify the given values from the problem statement: distance covered $s = 3.06\text{ m}$, average velocity $v_{\text{av}} = 0.34\text{ ms}^{-1}$, and change in velocity $\Delta v = 0.18\text{ ms}^{-1}$.
Step 2: Calculate the time interval $t$ using the relationship between distance, average velocity, and time: $t = \frac{\text{Distance}}{\text{Average velocity}} = \frac{3.06}{0.34} = 9\text{ s}$.
Step 3: Calculate the uniform acceleration $a$ using the definition of acceleration as the change in velocity per unit time: $a = \frac{\Delta v}{t} = \frac{0.18}{9} = 0.02\text{ ms}^{-2}$.
Step 4: Match the calculated value with option B.

Let initial ($t = 0$) velocity of particle $= u$
for first $5\text{ sec}$ of motion $s_5 = 10\text{ metre}$, so by using $s = ut + \frac{1}{2}at^2$
$10 = 5u + \frac{1}{2}a(5)^2 \Rightarrow 2u + 5a = 4 \quad \text{.... (i)}$
for first $8\text{ sec}$ of motion $s_8 = 20\text{ metre}$
$20 = 8u + \frac{1}{2}a(8)^2 \Rightarrow 2u + 8a = 5 \quad \text{.... (ii)}$
By solving (i) and (ii) $a = \frac{1}{3}\text{ m/s}^2$ and $u = \frac{7}{6}\text{ m/s}$
Total distance travelled in $10\text{ sec}$
$s_{10} = u \times 10 + \frac{1}{2}a(10)^2 = \frac{7}{6} \times 10 + \frac{1}{2} \times \frac{1}{3} \times 100 = \frac{85}{3}\text{ m}$
So distance travelled in next $2\text{ sec}$ ($10^{\text{th}}$ and $9^{\text{th}}\text{ sec}$)
$= s_{10} - s_8 = \frac{85}{3} - 20 = \frac{25}{3} = 8.3\text{ m}$.
Step Solution:
Step 1: Let $u$ be the initial velocity at $t = 0$ and $a$ be the uniform acceleration.
Step 2: Apply $s = ut + \frac{1}{2}at^2$ for the first $5\text{ s}$ where total distance travelled $s_5 = 10\text{ m}$: $10 = 5u + \frac{1}{2}a(5)^2 \Rightarrow 10 = 5u + \frac{25}{2}a \Rightarrow 2u + 5a = 4$ (Equation 1).
Step 3: Consider the motion over the first $8\text{ s}$ ($5\text{ s} + 3\text{ s}$), where total distance travelled $s_8 = 10 + 10 = 20\text{ m}$: $20 = 8u + \frac{1}{2}a(8)^2 \Rightarrow 20 = 8u + 32a \Rightarrow 2u + 8a = 5$ (Equation 2).
Step 4: Subtract Equation 1 from Equation 2 to solve for acceleration $a$: $(2u + 8a) - (2u + 5a) = 5 - 4 \Rightarrow 3a = 1 \Rightarrow a = \frac{1}{3}\text{ m/s}^2$.
Step 5: Substitute $a = \frac{1}{3}\text{ m/s}^2$ back into Equation 1 to find $u$: $2u + 5\left(\frac{1}{3}\right) = 4 \Rightarrow 2u = 4 - \frac{5}{3} = \frac{7}{3} \Rightarrow u = \frac{7}{6}\text{ m/s}$.
Step 6: Calculate total distance covered in $10\text{ s}$ ($8\text{ s} + 2\text{ s}$): $s_{10} = \left(\frac{7}{6}\right)(10) + \frac{1}{2}\left(\frac{1}{3}\right)(10)^2 = \frac{70}{6} + \frac{100}{6} = \frac{170}{6} = \frac{85}{3}\text{ m}$.
Step 7: Find the distance covered specifically in the next $2\text{ s}$ by subtracting $s_8$ from $s_{10}$: $\text{Distance} = s_{10} - s_8 = \frac{85}{3} - 20 = \frac{25}{3} = 8.33\text{ m} \approx 8.3\text{ m}$, matching option A.

Since the body starts from rest. Therefore $u = 0$.
$S_1 = \frac{1}{2}a(5)^2 = \frac{25 a}{2}$
$S_1 + S_2 = \frac{1}{2}a(10)^2 = \frac{100 a}{2} \Rightarrow S_2 = \frac{100 a}{2} - S_1 = 75 \frac{a}{2}$
$S_1 + S_2 + S_3 = \frac{1}{2}a(15)^2 = \frac{225 a}{2} \Rightarrow S_3 = \frac{225 a}{2} - S_2 - S_1 = \frac{125 a}{2}$
Thus Clearly $S_1 = \frac{1}{3}S_2 = \frac{1}{5}S_3$
Step Solution:
Step 1: Write down the equation for displacement covered in time $t$ for a body starting from rest ($u = 0$) under uniform acceleration $a$: $S = \frac{1}{2}at^2$.
Step 2: Calculate the distance covered in the first $5\text{ s}$, which is $S_1 = \frac{1}{2}a(5)^2 = \frac{25a}{2}$.
Step 3: Calculate the total distance covered in the first $10\text{ s}$, which is $S_1 + S_2 = \frac{1}{2}a(10)^2 = \frac{100a}{2}$.
Step 4: Find $S_2$ by subtracting $S_1$ from the distance covered in $10\text{ s}$: $S_2 = \frac{100a}{2} - \frac{25a}{2} = \frac{75a}{2} = 3 S_1$.
Step 5: Calculate the total distance covered in $15\text{ s}$, which is $S_1 + S_2 + S_3 = \frac{1}{2}a(15)^2 = \frac{225a}{2}$.
Step 6: Find $S_3$ by subtracting $(S_1 + S_2)$ from the distance covered in $15\text{ s}$: $S_3 = \frac{225a}{2} - \frac{100a}{2} = \frac{125a}{2} = 5 S_1$.
Step 7: Compare the values of $S_1$, $S_2$, and $S_3$: since $S_1 = \frac{25a}{2}$, $S_2 = 3S_1$, and $S_3 = 5S_1$, we get $S_1 = \frac{1}{3}S_2 = \frac{1}{5}S_3$, which matches option C.

$S_n = u + \frac{1}{2}a(2n - 1) = 0 + \frac{1}{2}(8)[2 \times 5 - 1] = 36\text{ metres}$
Step Solution:
Step 1: Identify the given parameters from the problem: initial velocity $u = 0$, uniform acceleration $a = 8\text{ m/sec}^2$, and second number $n = 5$.
Step 2: State the formula for the distance covered in the $n^{\text{th}}$ second: $S_n = u + \frac{1}{2}a(2n - 1)$.
Step 3: Substitute $u = 0$, $a = 8$, and $n = 5$ into the formula: $S_5 = 0 + \frac{1}{2}(8)(2 \times 5 - 1)$.
Step 4: Calculate the term inside the bracket: $2 \times 5 - 1 = 9$.
Step 5: Multiply to get the final distance: $S_5 = 4 \times 9 = 36\text{ metres}$, which corresponds to option A.

$F = ma \Rightarrow a \propto \frac{1}{m}$ if $F = \text{constant}$. Since the force is same and the effective mass of system becomes double
$\frac{a_2}{a_1} = \frac{m_1}{m_2} = \frac{m}{2m} \Rightarrow a_2 = \frac{a_1}{2} = 2\text{ m/s}^2$
Step Solution:
Step 1: Express Newton's second law of motion relating engine force $F$, mass $m$, and acceleration $a$: $F = ma$.
Step 2: Note that for a constant engine force $F$, acceleration is inversely proportional to mass: $a \propto \frac{1}{m}$.
Step 3: Let $m_1 = m$ be the mass of the single car and $a_1 = 4\text{ m/s}^2$ be its initial acceleration.
Step 4: When the car pulls another car of the same mass, the total effective mass of the system becomes $m_2 = m + m = 2m$.
Step 5: Set up the ratio of accelerations: $\frac{a_2}{a_1} = \frac{m_1}{m_2} = \frac{m}{2m} = \frac{1}{2}$.
Step 6: Solve for the new acceleration $a_2$: $a_2 = \frac{a_1}{2} = \frac{4}{2} = 2\text{ m/s}^2$, which corresponds to option B.

As $S_n \propto (2n - 1)$, $\frac{S_4}{S_3} = \frac{7}{5}$
Step Solution:
Step 1: Write down the formula for the distance travelled by an accelerating body in the $n^{\text{th}}$ second: $S_n = u + \frac{a}{2}(2n - 1)$.
Step 2: Since the body starts from rest ($u = 0$), the expression simplifies to $S_n = \frac{a}{2}(2n - 1)$, which means $S_n \propto (2n - 1)$.
Step 3: Calculate the distance travelled in the $4^{\text{th}}$ second ($n = 4$): $S_4 \propto 2(4) - 1 = 7$.
Step 4: Calculate the distance travelled in the $3^{\text{rd}}$ second ($n = 3$): $S_3 \propto 2(3) - 1 = 5$.
Step 5: Compute the ratio of the distance travelled in the $4^{\text{th}}$ second to that in the $3^{\text{rd}}$ second: $\frac{S_4}{S_3} = \frac{7}{5}$, which matches option A.