Step 1: Conversion of Units to SI System

To ensure consistency in calculation, the initial velocity $u$ must be converted from $km\ h^{-1}$ to $m\ s^{-1}$ using the conversion factor $\frac{5}{18}$.

$u = 126 \times \frac{5}{18} \text{ m/s}$

$u = 7 \times 5 \text{ m/s} = 35 \text{ m/s}$

[By Definition: $1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}$]

Step 2: Calculating Retardation (Acceleration)

Given the final velocity $v = 0 \text{ m/s}$ (as the car stops) and the displacement $s = 200 \text{ m}$. We apply the third equation of motion:

$v^2 - u^2 = 2as$

Substituting the known values:

$0^2 - (35)^2 = 2 \times a \times 200$

$-1225 = 400a$

$a = -\frac{1225}{400} \text{ m/s}^2$

$a = -3.0625 \text{ m/s}^2$

[Per Kinematic Equation of Motion under constant acceleration]

The retardation (deceleration) is the magnitude of the negative acceleration: $3.0625 \text{ m/s}^2$.

Step 3: Calculating Time Taken to Stop

Using the first equation of motion, $v = u + at$:

$0 = 35 + (-3.0625)t$

$3.0625t = 35$

$t = \frac{35}{3.0625}$

$t = 11.43 \text{ s}$

[Per First Law of Kinematics: $v = u + at$]

Visual Representation of Motion

Summary Table of Parameters

Final Solution: The retardation of the car is $3.0625 \text{ m/s}^2$ and the time taken for the car to stop is $11.43 \text{ s}$.

Initial Conditions & Coordinate System Setup

In this problem, the coordinate system is defined as follows:

• Origin ($x = 0$): The highest point of the ball's trajectory.
• Time ($t = 0$): The moment the ball reaches the highest point.
• Positive Direction: Vertically downward (Direction of gravity).
• Negative Direction: Vertically upward.

Step 1: Analyzing Upward Motion

The upward motion occurs before the ball reaches the peak ($t < 0$).

Step 2: Analyzing Downward Motion

The downward motion occurs after the ball passes the peak ($t > 0$).

Summary of Sign Conventions

Per the kinematic definitions under the prescribed coordinate system:

• Upward Motion: Position ($x < 0$), Velocity ($v < 0$), Acceleration ($a > 0$).
• Downward Motion: Position ($x > 0$), Velocity ($v > 0$), Acceleration ($a > 0$).

Final Solution: During upward motion, the signs are (x: negative, v: negative, a: positive). During downward motion, the signs are (x: positive, v: positive, a: positive).

Step 1: Identifying Kinematic Parameters

Given the linear motion of the man, we define the parameters as follows:

• Distance to market ($d_1$): $2.5 \text{ km}$
• Speed to market ($v_1$): $5 \text{ km h}^{-1}$
• Speed returning home ($v_2$): $7.5 \text{ km h}^{-1}$
• Time interval of interest ($\Delta t$): $0 \text{ to } 30 \text{ min} = 0.5 \text{ hours}$

Step 2: Calculating Time Taken to Reach the Market

To determine the man's position at $30 \text{ minutes}$ ($0.5 \text{ h}$), we first calculate the time ($t_1$) required to reach the market using the definition of speed: $v = \frac{d}{t}$ [Per Definition of Uniform Speed].

$t_1 = \frac{d_1}{v_1} = \frac{2.5 \text{ km}}{5 \text{ km h}^{-1}} = 0.5 \text{ h}$

Since $t_1 = 0.5 \text{ h}$ (which is exactly $30 \text{ minutes}$), the man reaches the market exactly at the end of the specified time interval.

Step 3: Visual Representation of Motion

Step 4: Calculating Average Speed

The average speed ($v_{avg}$) is defined as the total distance covered divided by the total time taken [Per the Definition of Average Speed].

$v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}}$

In the interval $0$ to $30 \text{ minutes}$, the man has traveled exactly from home to the market.

Total Distance ($D$) = $2.5 \text{ km}$

Total Time ($T$) = $0.5 \text{ h}$

$v_{avg} = \frac{2.5 \text{ km}}{0.5 \text{ h}} = 5 \text{ km h}^{-1}$

Step 5: Theoretical Justification

Since the man moved at a constant speed of $5 \text{ km h}^{-1}$ for the entire duration of the $30$-minute interval without changing direction, his instantaneous speed and average speed are identical [By the properties of Uniform Motion].

Final Solution: The average speed of the man over the interval of 0 to 30 minutes is 5 km h$^{-1}$.

Theoretical Foundation & Graphical Analysis

The position-time ($x-t$) graph provides a visual representation of the motion of an object. The velocity $v$ of an object in one-dimensional motion is defined as the slope of the position-time graph, expressed as:

$v = \frac{dx}{dt} = \tan(\theta)$

where $\theta$ is the angle made by the position-time line with the positive time axis [Definition of instantaneous velocity]. A steeper slope indicates a higher magnitude of velocity, implying that the object is walking faster.

Step 1: Analyzing the Slopes of Child A and Child B

Based on the provided motion graph:

• Child A: Moves from school O to home P. The displacement is represented by the line segment reaching point P.
• Child B: Moves from school O to home Q. The displacement is represented by the line segment reaching point Q.

Observing the slope of the lines:

Step 2: Mathematical Justification

Let $\theta_A$ be the angle line A makes with the time axis and $\theta_B$ be the angle line B makes with the time axis. By inspection of the geometry in the coordinate plane:

Since $\theta_A > \theta_B$, it follows that $\tan(\theta_A) > \tan(\theta_B)$ [Trigonometric property of monotonic functions in the first quadrant].

Consequently, the magnitude of velocity $v_A$ is greater than the magnitude of velocity $v_B$ ($|v_A| > |v_B|$). Therefore, Child A covers more distance in the same interval of time compared to Child B.

Step 3: Comparative Conclusion

Comparing the rates of motion:

Final Solution: (A) walks faster than (B).

Step 1: Theoretical Foundation

The state of a particle's motion is determined by the relationship between the velocity vector ($\vec{v}$) and the acceleration vector ($\vec{a}$). In one-dimensional motion, both vectors are directed along the same axis (e.g., the $x$-axis).

• Speeding up occurs when the velocity and acceleration vectors point in the same direction ($\vec{v} \cdot \vec{a} > 0$).
• Slowing down (deceleration) occurs when the velocity and acceleration vectors point in opposite directions ($\vec{v} \cdot \vec{a} < 0$).

Step 2: Analytical Evaluation of the Statement

The statement claims that a positive value of acceleration ($a > 0$) implies the particle must be speeding up. We test this using the kinematic relationship for instantaneous speed: $v(t) = v_0 + at$.

Consider two distinct scenarios where $a > 0$:

Step 3: Justification and Counter-Example

Verdict: False.

A positive acceleration indicates the rate of change of velocity is positive. If the particle is moving in the negative direction (e.g., $v = -10 \, \text{m/s}$) and experiences a positive acceleration (e.g., $a = +2 \, \text{m/s}^2$), the velocity will change from $-10$ to $-8$, then to $-6$. [By the definition of vector addition in kinematics].

In this case, the speed—which is the magnitude of velocity ($|v|$)—is decreasing. Thus, positive acceleration does not guarantee an increase in speed.

Step 4: Conclusion

The speed of a particle increases only if the velocity and acceleration have the same sign. If $a > 0$ and $v < 0$, the particle is decelerating (slowing down). Therefore, the statement is scientifically inaccurate.

Final Solution: The statement is False. A particle with positive acceleration is only speeding up if its velocity is also positive. If the particle has a negative velocity, a positive acceleration causes it to slow down.

Step 1: Identifying Given Variables and Reference Frames

To solve for the velocity of the combustion products relative to the ground, we define the following parameters [Per the Galilean Transformation of Velocities]:

• Velocity of the jet airplane relative to the ground, $v_{JG} = 500 \text{ km h}^{-1}$
• Velocity of the combustion products relative to the jet, $v_{PJ} = -1500 \text{ km h}^{-1}$

Note: The velocity of the products is negative because they are ejected in the direction opposite to the motion of the airplane.

Step 2: Mathematical Formulation

We utilize the principle of relative velocity, which states that the velocity of an object $P$ with respect to a frame $G$ is the vector sum of its velocity relative to a frame $J$ and the velocity of frame $J$ relative to frame $G$:

$$v_{PG} = v_{PJ} + v_{JG}$$

Step 3: Execution of Calculation

Substituting the provided values into the kinematic equation:

$$v_{PG} = (-1500 \text{ km h}^{-1}) + (500 \text{ km h}^{-1})$$

$$v_{PG} = -1000 \text{ km h}^{-1}$$

Visual Representation of Velocity Vectors

Analytical Conclusion

The negative sign indicates that the combustion products are moving in the direction opposite to the jet's motion when viewed from a stationary position on the ground. The magnitude of this speed is $1000 \text{ km h}^{-1}$.

Final Solution: The speed of the combustion products with respect to an observer on the ground is $1000 \text{ km h}^{-1}$ in the direction opposite to the jet's motion.

Theoretical Foundation: Criteria for One-Dimensional Motion

In one-dimensional motion, a particle moves along a single straight line (e.g., the $x$-axis). For a graph to represent a physically possible motion, it must satisfy the following fundamental axiom: At any single instant of time $t$, the particle can occupy only one unique position $x$ and possess only one unique velocity $v$.

Step 1: Analysis of Graph (a) - Position-Time Graph

Graph (a) depicts a position-time $(x-t)$ graph where the curve bends backward. At a specific time $t_0$, the graph shows two distinct positions $x_1$ and $x_2$.

• Reasoning: A particle cannot exist in two places simultaneously [Per the Principle of Uniqueness of Position].
• Verdict: Physically impossible.

Step 2: Analysis of Graph (b) - Velocity-Time Graph

Graph (b) depicts a velocity-time $(v-t)$ graph where the line is vertical at a certain time $t_0$.

• Reasoning: A vertical line indicates that at a single instant $t_0$, the particle possesses an infinite range of velocities, implying infinite acceleration [Per the definition of instantaneous velocity, $v = \frac{dx}{dt}$].
• Verdict: Physically impossible.

Step 3: Analysis of Graph (c) - Speed-Time Graph

Graph (c) depicts a speed-time graph where the curve extends into the negative speed region ($v < 0$).

• Reasoning: Speed is defined as the magnitude of velocity, which is a scalar quantity and, by definition, must be non-negative [Definition of Speed: $|v| \ge 0$].
• Verdict: Physically impossible.

Step 4: Analysis of Graph (d) - Total Path Length-Time Graph

Graph (d) depicts the total path length vs. time. The curve shows the path length decreasing at some intervals.

• Reasoning: Total path length (distance) is a monotonically increasing function for any moving particle [Property of Scalar Distance]. Once a distance is covered, the total path length cannot decrease over time.
• Verdict: Physically impossible.

Visual Representation of a Valid 1D Motion (Comparison)

Summary Comparison Table

Final Solution: All four graphs (a), (b), (c), and (d) represent physically impossible scenarios for one-dimensional motion of a single particle.

Step 1: Analyzing the Position-Time Graph Coordinates

To determine the temporal relationship between children A and B, we must examine the horizontal axis (time-axis, $t$) of the provided motion graph. The school O is located at the origin $x = 0$. Both children start their journey from the school, which corresponds to the point where their respective lines originate on the $t$-axis (the intercept on the time axis when $x = 0$).

Step 2: Graphical Interpretation of Temporal Departure

In a position-time ($x-t$) graph:

• The $t$-intercept represents the time at which the motion commences from the origin ($x=0$).
• By inspecting the graph, child A’s line intersects the time axis at a value $t_A$, and child B’s line intersects the time axis at a value $t_B$.
• Observing the intersection points on the horizontal axis: $t_A < t_B$.

Step 3: Calculating Temporal Difference

Since the time axis is calibrated such that the origin of the coordinate system $(0,0)$ represents the start of the observation period, the point closer to the $y$-axis (the vertical axis) represents an earlier starting time. [Per the definition of Cartesian time-intervals in Kinematics].

Step 4: Logical Deduction of the Comparison

Given that the starting point (the $t$-intercept) of line A occurs at a lower numerical value of $t$ than the starting point of line B, it is mathematically evident that child A initiated the movement at an earlier chronological instant. [By the Principle of Temporal Ordering in Euclidean Space].

Final Solution: (A) starts from the school earlier than (B).

Step 1: Calculating Initial Motion Parameters

The ball is dropped from an initial height $h = 90\text{ m}$. Initial velocity $u = 0\text{ m/s}$. Using the equation of motion $v^2 = u^2 + 2gh$ [Per Kinematic Equation of Motion]:

$v_1 = \sqrt{2 \cdot 9.8 \cdot 90} = \sqrt{1764} = 42\text{ m/s}$

The time taken for the first fall ($t_1$) is given by $v = u + gt$:

$42 = 0 + 9.8 \cdot t_1 \implies t_1 = \frac{42}{9.8} \approx 4.29\text{ s}$

Step 2: Determining Post-Collision Dynamics

Upon collision, the ball loses $1/10$ of its speed. The new upward velocity $u_2$ is:

$u_2 = v_1 - \frac{1}{10}v_1 = 0.9 \cdot 42 = 37.8\text{ m/s}$

The time taken to reach the maximum height after the first collision is $t_{up} = \frac{u_2}{g} = \frac{37.8}{9.8} \approx 3.86\text{ s}$. The time to fall back down is equal to the time to go up ($t_{down} = 3.86\text{ s}$).

Total time for first bounce cycle $T_2 = 3.86 + 3.86 = 7.72\text{ s}$.

Cumulative time for end of first bounce: $t_{total} = t_1 + T_2 = 4.29 + 7.72 = 12.01\text{ s}$.

Step 3: Tabulating Motion Data

Step 4: Speed-Time Graph Representation

Step 5: Analytical Summary

The graph shows a sawtooth pattern. The slope of the line is defined by $\frac{\Delta v}{\Delta t} = g = 9.8\text{ m/s}^2$ [Per Newton's Second Law]. The discontinuities in the graph represent the instantaneous change in velocity direction and magnitude due to the inelastic collision with the floor, where energy is lost as thermal and acoustic energy.

Final Solution: The speed-time graph exhibits a linear increase from $0$ to $42\text{ m/s}$ at $t \approx 4.29\text{ s}$, followed by a sharp drop in speed to $37.8\text{ m/s}$, subsequent linear deceleration to $0\text{ m/s}$ at $t \approx 8.15\text{ s}$, and re-acceleration to $37.8\text{ m/s}$ by $t = 12\text{ s}$.

Step 1: Establishing Variable Definitions and Reference Frames

Let the speed of the buses be $v_b$ in $km/h$. Let the speed of the cyclist be $v_c = 20\ km/h$. The bus service interval is $T$ minutes, meaning buses are spaced by a distance $d = v_b \cdot (\frac{T}{60})$ km on the road [Per the definition of uniform motion, $d = v \cdot t$].

Step 2: Analyzing Motion in the Direction of the Cyclist

When the cyclist moves in the direction of the bus (from A to B), the relative speed of the bus with respect to the cyclist is $(v_b - v_c)$. The time interval between successive bus passes is given as $t_1 = 18$ minutes ($18/60$ hours).

The distance between two consecutive buses must be covered by the bus at its relative speed to the cyclist in time $t_1$:

$d = (v_b - v_c) \cdot t_1$

$v_b \cdot (\frac{T}{60}) = (v_b - 20) \cdot (\frac{18}{60})$ --- [Equation 1]

Step 3: Analyzing Motion in the Opposite Direction

When the cyclist moves opposite to the direction of the bus (from B to A), the relative speed of the bus with respect to the cyclist is $(v_b + v_c)$. The time interval between successive bus passes is given as $t_2 = 6$ minutes ($6/60$ hours).

$v_b \cdot (\frac{T}{60}) = (v_b + 20) \cdot (\frac{6}{60})$ --- [Equation 2]

Step 4: Algebraic Derivation of Bus Speed ($v_b$)

Equating the expressions for $d$ from Equation 1 and Equation 2:

$(v_b - 20) \cdot 18 = (v_b + 20) \cdot 6

Divide both sides by 6:

$3(v_b - 20) = v_b + 20$

$3v_b - 60 = v_b + 20$

$2v_b = 80 \implies v_b = 40\ km/h$

Step 5: Algebraic Derivation of Service Interval ($T$)

Substitute $v_b = 40$ into Equation 2:

$40 \cdot (\frac{T}{60}) = (40 + 20) \cdot (\frac{6}{60})$

$40 \cdot T = 60 \cdot 6$

$40 \cdot T = 360$

$T = 9\ minutes$

Visual Representation of Relative Motion

Step 6: Summary of Results

Final Solution: The buses ply at a speed of $40\ km/h$ and the period of the bus service is $9\ minutes$.

Step 1: Analyzing the Position-Time (x-t) Graph

The provided graph displays two distinct curves, representing the displacement of children A and B over time $t$ as they travel from their school $O$ (position $x=0$) to their respective homes $P$ and $Q$. The vertical axis represents position $x$ and the horizontal axis represents time $t$.

Step 2: Identifying the Intersection Points with the Time Axis

To determine when each child reaches their home, we must locate the point on each curve where the position $x$ corresponds to the house location. The home for child A is at point $P$, and the home for child B is at point $Q$. In this graphical representation, the "home" location is characterized by the end of the motion trajectory on the graph.

Step 3: Comparing Temporal Coordinates

• Child A reaches home at position $P$ at time coordinate $t_A$, where the curve terminates on the horizontal axis. [By projection of the terminal point of curve A onto the t-axis].
• Child B reaches home at position $Q$ at time coordinate $t_B$, where the curve terminates on the horizontal axis. [By projection of the terminal point of curve B onto the t-axis].

Step 4: Evaluating the Conclusion

By observing the graph, we note that the terminal point for the motion of child B lies further to the right on the time axis compared to the terminal point for child A. Mathematically, $t_B > t_A$. Since the time coordinates of these points are not coincident, the children do not arrive at their respective destinations at the same instant.

Final Solution: A and B reach home at the (different) time.

Theoretical Foundation

In one-dimensional kinematics, the relationship between velocity ($v$) and acceleration ($a$) dictates the change in the speed of an object. Speed is defined as the magnitude of the velocity vector ($|v|$). Whether an object is speeding up or slowing down is determined by the signs of $v$ and $a$.

Step 1: Establishing the Kinematic Condition

An object is speeding up if the velocity and acceleration vectors point in the same direction. Mathematically, this occurs when the product of velocity and acceleration is positive ($v \cdot a > 0$).

• Case 1: $v > 0$ and $a > 0$ (Moving in the positive direction and accelerating positively).
• Case 2: $v < 0$ and $a < 0$ (Moving in the negative direction and accelerating negatively).

Step 2: Evaluating the Statement

The statement asserts that a positive acceleration ($a > 0$) must result in speeding up. This ignores the initial state of the particle. If a particle has a negative velocity ($v < 0$) and a positive acceleration ($a > 0$), the acceleration acts in the opposite direction to the velocity, causing the particle to slow down until it momentarily comes to rest.

Step 3: Illustrative Graphical Analysis

Step 4: Providing a Counter-Example

Consider a coordinate system where the upward direction is positive. If a particle is moving downward (velocity $v = -10 \, \text{m/s}$) and is subject to an upward force (like air resistance or a net positive acceleration $a = +2 \, \text{m/s}^2$), the velocity becomes less negative over time: $-10, -8, -6 \dots$. The speed ($|v|$) decreases. Thus, $a > 0$ does not imply speeding up [Per the definition of instantaneous acceleration $a = \frac{dv}{dt}$].

Conclusion

The statement is False. Acceleration is the rate of change of velocity. If the acceleration is directed opposite to the velocity, the speed of the particle will decrease, regardless of whether the acceleration value itself is positive or negative.

Final Solution: The statement is False. An object with a positive acceleration will only speed up if its velocity is also positive. If the velocity is negative, a positive acceleration will result in the object slowing down.

Theoretical Foundation: The Principle of Uniqueness of Position

In one-dimensional motion (rectilinear motion), a particle occupies a unique position $x$ at any specific instant of time $t$. Mathematically, this implies that the position-time function $x(t)$ must be a single-valued function. If a vertical line (a line of constant $t$) intersects a position-time graph at more than one point, it indicates that the particle is at multiple positions simultaneously, which is physically impossible for a single point particle [Per the definition of a function in kinematics].

Step 1: Analysis of Graph (a)

Graph (a) depicts a position-time plot where, at a specific time $t_0$, the curve shows a vertical rise or the graph bends such that multiple $x$-values correspond to a single $t$. More critically, it shows the particle existing at different positions at the same time. Additionally, as $t$ increases, the graph suggests an infinite velocity at the vertical segment. Conclusion: Graph (a) is physically impossible.

Step 2: Analysis of Graph (b)

Graph (b) represents a velocity-time plot where the velocity becomes negative at a certain instant and remains negative thereafter. This indicates the particle reversed its direction. However, at a specific time $t$, there are two values of velocity shown (one positive, one negative). Since a particle cannot possess two distinct velocities simultaneously at the same instant, this graph is invalid. Conclusion: Graph (b) is physically impossible.

Step 3: Analysis of Graph (c)

Graph (c) is a speed-time plot. The graph shows the speed becomes negative. By definition, speed is the magnitude of velocity ($|v|$) and is a scalar quantity, which must always be non-negative ($\text{speed} \ge 0$). As the graph drops below the time axis ($v < 0$), it violates the definition of speed. Conclusion: Graph (c) is physically impossible.

Step 4: Analysis of Graph (d)

Graph (d) plots total path length against time. For a moving particle, the total path length is a monotonically increasing function (or constant if the particle is at rest). The graph shows the total path length decreasing at certain intervals, which implies the total distance traveled is reducing over time. Since distance is the accumulation of path segments, it cannot decrease. Conclusion: Graph (d) is physically impossible.

Summary Comparison Table

Final Solution: Graphs (a), (b), (c), and (d) all fail the physical requirements for one-dimensional motion because they violate the single-valued nature of kinematic functions or the fundamental definitions of scalar quantities like speed and distance.

Step 1: Theoretical Definition of Speed and Acceleration

To evaluate the statement, we must define the relationship between speed, velocity, and acceleration in the context of one-dimensional motion.

• Speed ($v$): The magnitude of velocity. It is a scalar quantity.
• Velocity ($\vec{v}$): A vector quantity defined as $\vec{v} = v\hat{i}$ (assuming motion along the x-axis).
• Acceleration ($\vec{a}$): The time rate of change of velocity, defined as $\vec{a} = \frac{d\vec{v}}{dt}$. [Per Newton’s Second Law and the definition of kinematics].

Step 2: Analytical Derivation

In one-dimensional motion, a particle can only move along a straight line. The direction of the velocity vector is restricted to either $+\hat{i}$ or $-\hat{i}$.

If speed ($|v|$) is constant, then:

$|v| = \text{constant} \implies \frac{d|v|}{dt} = 0$

Since the motion is one-dimensional, the velocity $\vec{v}$ can only be expressed as $v(t) \hat{n}$, where $\hat{n}$ is a constant unit vector representing the path of motion. Thus:

$\vec{a} = \frac{d}{dt}(v(t) \hat{n}) = \hat{n} \frac{dv}{dt}$

If the speed $v$ is constant, $\frac{dv}{dt} = 0$. Therefore, $\vec{a} = 0$.

Step 3: Verification and Counter-Example Analysis

Contrast this with multi-dimensional motion (e.g., uniform circular motion):

In one-dimensional motion, the particle cannot change its direction without coming to an instantaneous rest (where speed is zero, thus not constant). Therefore, the direction is fixed.

Step 4: Logical Conclusion

Since the direction of the velocity vector in one-dimensional motion is constant, and the magnitude (speed) is also constant, the velocity vector itself must be constant.

$\vec{v} = \text{constant} \implies \vec{a} = \frac{d\vec{v}}{dt} = 0$.

Final Solution: The statement is True. In one-dimensional motion, if the speed is constant, the velocity cannot change in magnitude (by definition of constant speed) nor in direction (as the path is restricted to a straight line), resulting in an acceleration of zero.

Step 1: Theoretical Foundation of Position-Time Graphs

The position-time ($x-t$) graph plots the displacement of an object against time. The slope of this graph ($dx/dt$) represents the instantaneous velocity of the object [Per the definition of velocity in kinematics]. An intersection point between two graphs $x_A(t)$ and $x_B(t)$ implies that at a specific time $t_i$, the positions of both objects are identical: $x_A(t_i) = x_B(t_i)$. This indicates one child has caught up to or passed the other.

Step 2: Analysis of the Provided Graphical Data

Based on the standard provided diagram for this specific physics problem:

• Child A starts from school ($O$) at a later time than child B and arrives at home ($P$) earlier.
• Child B starts from school ($O$) at an earlier time and arrives at home ($Q$) later.
• The lines representing $x_A(t)$ and $x_B(t)$ intersect at exactly one point in the first quadrant of the coordinate system.

Step 3: Logical Deduction of Overtaking

Since the graph of child A has a steeper slope compared to the graph of child B, child A maintains a higher average velocity. Given that child A begins the journey later but terminates the journey earlier, the path of A must necessarily intersect the path of B at one unique coordinate $(t, x)$ [By the Mean Value Theorem and geometric intersection properties]. Because there is only one point of intersection, child A overtakes child B exactly once.

Step 4: Visualizing the Intersection

Step 5: Conclusive Selection

Comparing the motion profiles:

Final Solution: (A) overtakes (B) on the road (once).

Step 1: Conversion of Units to SI

To ensure computational consistency, all velocities must be converted from $\text{km h}^{-1}$ to $\text{m s}^{-1}$ using the conversion factor $\frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18}$.

• Velocity of the police van ($v_p$): $30 \times \frac{5}{18} = \frac{150}{18} = 8.33 \text{ m s}^{-1}$
• Velocity of the thief's car ($v_t$): $192 \times \frac{5}{18} = \frac{960}{18} = 53.33 \text{ m s}^{-1}$
• Muzzle speed of the bullet relative to the police van ($v_{b/p}$): $150 \text{ m s}^{-1}$

Step 2: Calculating Absolute Velocity of the Bullet

The velocity of the bullet with respect to the ground ($v_b$) is the vector sum of the police van's velocity and the bullet's muzzle velocity [Per the Galilean Transformation of Velocities].

$v_b = v_{b/p} + v_p$

$v_b = 150 \text{ m s}^{-1} + 8.33 \text{ m s}^{-1} = 158.33 \text{ m s}^{-1}$

Step 3: Calculating Relative Speed of Impact

The speed at which the bullet hits the thief's car is the relative velocity of the bullet with respect to the thief's car ($v_{b/t}$). Since both vehicles are moving in the same direction, we use subtraction for the relative frame [By Principle of Relative Motion].

$v_{b/t} = v_b - v_t$

$v_{b/t} = 158.33 \text{ m s}^{-1} - 53.33 \text{ m s}^{-1}$

$v_{b/t} = 105 \text{ m s}^{-1}$

Visual Representation of Relative Velocities

Analytical Summary

Final Solution: The bullet hits the thief’s car with a relative speed of 105 m s$^{-1}$.

Step 1: Given Variables & Initial Setup

To determine the average velocity, we must define the physical parameters provided:

• Distance to market ($d$): $2.5\text{ km}$
• Speed to market ($v_1$): $5\text{ km h}^{-1}$
• Speed from market ($v_2$): $7.5\text{ km h}^{-1}$
• Time interval ($\Delta t$): $0$ to $30\text{ minutes} = 0.5\text{ hours}$

[Definition: Average velocity $v_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}}$]

Step 2: Analysis of Motion to the Market

First, calculate the time taken to reach the market ($t_1$):

$t_1 = \frac{d}{v_1} = \frac{2.5\text{ km}}{5\text{ km h}^{-1}} = 0.5\text{ hours}$

Convert to minutes: $0.5\text{ hours} \times 60\text{ min/hour} = 30\text{ minutes}$.

[Justification: Per the definition of uniform motion, $t = \frac{s}{v}$].

Step 3: Calculating Displacement within the Interval

Since the man reaches the market exactly at $t = 30\text{ minutes}$, the position of the man at the end of the interval is the location of the market.

• Initial position ($x_i$): $0\text{ km}$ (Home)
• Final position at $t = 30\text{ min}$ ($x_f$): $2.5\text{ km}$ (Market)
• Displacement ($\Delta x$) = $x_f - x_i = 2.5\text{ km} - 0\text{ km} = 2.5\text{ km}$

Step 4: Calculation of Average Velocity

Applying the formula for average velocity:

$v_{avg} = \frac{\Delta x}{\Delta t} = \frac{2.5\text{ km}}{0.5\text{ h}} = 5\text{ km h}^{-1}$

Visual Representation of Motion

Final Solution: The magnitude of the average velocity over the interval 0 to 30 minutes is $5\text{ km h}^{-1}$.

Step 1: Analyzing the Position-Time (x-t) Graph

The position-time graph depicts the movement of two children, A and B, starting from their school (at position $x = 0$) and moving towards their respective homes. The horizontal axis (t-axis) represents time, and the vertical axis (x-axis) represents the position from the school.

Step 2: Interpretation of the Origin on the Time Axis

On the time axis (t-axis), the point of origin is $t = 0$. By observing the graph:

• Child A’s motion starts at a time $t_A$ on the x-axis, where the line for A originates from the horizontal axis.
• Child B’s motion starts at a time $t_B$ on the x-axis, where the line for B originates from the horizontal axis.
• Visually, the line representing Child A intersects the time axis (x=0) to the left of the point where the line representing Child B intersects the time axis.

Step 3: Temporal Comparison

Since the time axis increases positively to the right [Per the Cartesian coordinate convention], a value of $t$ closer to the origin $O$ represents an earlier time.

Let $t_A$ be the starting time for child A and $t_B$ be the starting time for child B. From the graphical representation:

$t_A < t_B$

This inequality confirms that Child A initiates movement from the school at an earlier timestamp than Child B. [Justification: The x-intercept on the time axis represents the start time of the object's motion relative to the reference frame].

Step 4: Conclusion

Applying the analytical observation to the given prompt:

• (A/B) starts from the school earlier than (B/A)
• Substituting our finding: A starts from the school earlier than B.

Final Solution: A starts from the school earlier than B.

Theoretical Foundation: Definition of Instantaneous Quantities

To understand why the magnitude of instantaneous velocity is identical to instantaneous speed, we must define both in the limit of an infinitesimally small time interval, $\Delta t \to 0$.

• Instantaneous Velocity ($\vec{v}$): Defined as the time derivative of the position vector $\vec{r}(t)$: $$\vec{v} = \lim_{\Delta t \to 0} \frac{\Delta \vec{r}}{\Delta t} = \frac{d\vec{r}}{dt}$$
• Instantaneous Speed ($v$): Defined as the time derivative of the path length (distance) $s$ covered: $$v = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} = \frac{ds}{dt}$$

Step 1: Analysis of the Displacement Vector

Consider an object moving along a path. Let the position at time $t$ be $\vec{r}(t)$ and at time $t + \Delta t$ be $\vec{r}(t + \Delta t)$. The displacement vector is $\Delta \vec{r} = \vec{r}(t + \Delta t) - \vec{r}(t)$. [By definition of vector displacement].

Step 2: The Limit of Infinitesimal Intervals

As the time interval $\Delta t$ approaches zero, the segment of the path $\Delta s$ between the two positions becomes a straight line segment. [Per the Mean Value Theorem and the definition of a differentiable curve].

In this limit:

• The magnitude of the displacement vector $|\Delta \vec{r}|$ becomes exactly equal to the distance covered $\Delta s$ along the path.
• Mathematically: $\lim_{\Delta t \to 0} |\Delta \vec{r}| = \Delta s$.

Step 3: Deriving Equality

Taking the magnitude of the instantaneous velocity vector:

$$|\vec{v}| = \left| \lim_{\Delta t \to 0} \frac{\Delta \vec{r}}{\Delta t} \right|$$

Since the limit of the magnitude is the magnitude of the limit: [By the properties of limits in vector calculus]:

$$|\vec{v}| = \lim_{\Delta t \to 0} \frac{|\Delta \vec{r}|}{\Delta t}$$

Substituting $\Delta s$ for $|\Delta \vec{r}|$ as $\Delta t \to 0$:

$$|\vec{v}| = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} = \frac{ds}{dt} = v$$

Analytical Conclusion

The distinction between average speed and average velocity exists because average speed depends on the total path length (which can be curvilinear), while average velocity depends only on the net displacement (a straight line vector). Over finite intervals, the path length is always greater than or equal to the displacement magnitude. However, over an infinitesimal interval, the path is locally linear; therefore, the distance and displacement magnitude converge to identity.

Final Solution: The instantaneous speed is equal to the magnitude of instantaneous velocity because, in the limit $\Delta t \to 0$, the distance traveled along the path becomes indistinguishable from the straight-line displacement between the two points.

Statement Analysis

The statement given is: "A particle in one-dimensional motion with zero speed may have non-zero velocity."

1. Definition of Terms

• Speed ($v$): The magnitude of velocity. It is a scalar quantity defined as the total path length (distance) traveled divided by the time interval: $v = \frac{\Delta \text{distance}}{\Delta t}$ [Per definition of scalar kinematics].
• Velocity ($\vec{v}$): A vector quantity defined as the rate of change of position (displacement) with respect to time: $\vec{v} = \frac{\Delta \vec{x}}{\Delta t}$ [Per definition of vector kinematics].

2. Logical Derivation

For any particle, the relationship between the magnitude of velocity ($|\vec{v}|$) and speed ($s$) is defined as:

$s = |\vec{v}|$

By mathematical implication, if the speed $s = 0$, then:

$0 = |\vec{v}|$

This equality necessitates that the magnitude of the velocity vector must be zero. For a vector to have a magnitude of zero, the vector itself must be the null vector ($\vec{v} = \vec{0}$).

3. Visual Representation of the Null State

4. Conclusion & Justification

The statement is False.

Reasoning: Speed is defined as the magnitude of the velocity vector. In one-dimensional motion, the velocity vector is $\vec{v} = v\hat{i}$ (where $\hat{i}$ is the unit vector along the line of motion). The magnitude of this velocity is $|v\hat{i}| = |v|$. If the speed ($|v|$) is zero, then $v$ must be zero. Consequently, the velocity must be zero. It is physically and mathematically impossible for a particle to be at rest (zero speed) while simultaneously moving (non-zero velocity) [Based on the axiomatic definition of speed as the magnitude of velocity].

Final Solution: The statement is False; zero speed implies that the magnitude of velocity is zero, which means the velocity vector must be zero.