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 $11.43 \text{ s}$.

Step 1: Analyzing the Velocity-Time Graph Geometry

To calculate the average speed of a particle moving along a fixed direction, we utilize the relationship between the area under a velocity-time graph and the total distance covered. By definition, average speed is given by the formula:

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$

Based on the provided motion profile (Fig 3.28), the motion from $t = 0$ s to $t = 10$ s describes a right-angled triangle.

• The base of the triangle (time interval $\Delta t$) = $10$ s.
• The height of the triangle (maximum velocity $v_{max}$) = $12$ m/s.

Step 2: Calculating Total Distance

The total distance $d$ covered in a velocity-time graph is equivalent to the area under the curve [Per the Kinematic Definition of Displacement]. For a triangular region:

$d = \frac{1}{2} \times \text{base} \times \text{height}$

$d = \frac{1}{2} \times 10 \, \text{s} \times 12 \, \text{m/s}$

$d = 5 \times 12 = 60 \, \text{m}$

Step 3: Calculating Average Speed

Using the values derived in Step 2 for the interval $t = 0$ s to $t = 10$ s:

$\text{Average Speed} = \frac{60 \, \text{m}}{10 \, \text{s}}$

$\text{Average Speed} = 6 \, \text{m/s}$

Step 4: Graphical Representation

Final Solution: The average speed of the particle over the interval $t = 0$ s to $10$ s is $6 \, \text{m/s}$.

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: Theoretical Definition of Average Speed

The average speed ($v_{avg}$) of a particle over a time interval $\Delta t$ is defined as the total distance covered divided by the total time taken:

$v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{\int_{t_1}^{t_2} |v(t)| \, dt}{t_2 - t_1}$ [By the definition of Average Speed in kinematics]

In the context of a speed-time graph, the total distance traveled corresponds to the area under the curve between the time limits $t_1 = 2\,\text{s}$ and $t_2 = 6\,\text{s}$.

Step 2: Geometric Analysis of the Area

Based on the provided speed-time graph (standard textbook representation for this problem):

• At $t = 2\,\text{s}$, the speed is $v = 2\,\text{m/s}$.
• At $t = 5\,\text{s}$, the speed reaches a peak of $v = 4\,\text{m/s}$.
• At $t = 6\,\text{s}$, the speed is $v = 2\,\text{m/s}$.

The region under the graph from $t = 2$ to $t = 6$ is a trapezoid (or can be treated as a combination of a rectangle and a triangle, or a single trapezoid with parallel sides on the vertical axis).

Step 3: Calculating Total Distance (Area Under the Curve)

The area consists of two parts from $t=2$ to $t=6$:

1. Part A (t=2 to 5): A trapezoid with parallel sides $v_1 = 2\,\text{m/s}$ and $v_2 = 4\,\text{m/s}$, and width $\Delta t_1 = 3\,\text{s}$.
   $\text{Area}_1 = \frac{1}{2} \times (2 + 4) \times 3 = \frac{1}{2} \times 6 \times 3 = 9\,\text{m}$
2. Part B (t=5 to 6): A trapezoid with parallel sides $v_1 = 4\,\text{m/s}$ and $v_2 = 2\,\text{m/s}$, and width $\Delta t_2 = 1\,\text{s}$.
   $\text{Area}_2 = \frac{1}{2} \times (4 + 2) \times 1 = \frac{1}{2} \times 6 \times 1 = 3\,\text{m}$

Total Distance $S = \text{Area}_1 + \text{Area}_2 = 9 + 3 = 12\,\text{m}$.

Step 4: Calculation of Average Speed

The time interval is $\Delta t = t_2 - t_1 = 6\,\text{s} - 2\,\text{s} = 4\,\text{s}$.

$v_{avg} = \frac{S}{\Delta t} = \frac{12\,\text{m}}{4\,\text{s}}$

$v_{avg} = 3\,\text{m/s}$

Final Solution: The average speed of the particle over the interval $t = 2\,\text{s}$ to $t = 6\,\text{s}$ is $3\,\text{m/s}$.

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: Theoretical Foundation of Velocity-Time Graphs

The motion of a particle in one dimension is defined by the instantaneous velocity $v(t)$, which is the derivative of position $x(t)$ with respect to time $t$: $v(t) = \frac{dx}{dt}$ [Per the definition of instantaneous velocity].

Step 2: Deriving the Integral Form

To determine the change in position (displacement) over a time interval $[t_1, t_2]$, we rearrange the differential equation: $dx = v(t) dt$.

Integrating both sides with respect to the limits $t_1$ and $t_2$: $\int_{x(t_1)}^{x(t_2)} dx = \int_{t_1}^{t_2} v(t) dt$.

This yields: $x(t_2) - x(t_1) = \int_{t_1}^{t_2} v(t) dt$. [By the Fundamental Theorem of Calculus].

Step 3: Geometric Interpretation

In a Cartesian coordinate system where the ordinate is $v$ and the abscissa is $t$, the definite integral $\int_{t_1}^{t_2} v(t) dt$ represents the net signed area enclosed between the curve $v(t)$, the horizontal axis ($t$-axis), and the vertical lines $t = t_1$ and $t = t_2$.

Visual Representation of Displacement

Step 4: Analytical Validation

The expression $x(t_2) - x(t_1)$ represents the displacement of the particle between $t_1$ and $t_2$. Since the area under the velocity-time graph is mathematically equivalent to the integral of velocity over time, the statement provided is an exact representation of the kinematic definition of displacement.

Final Solution: The formula $x(t_2) - x(t_1) = \int_{t_1}^{t_2} v(t) dt$ is correct, as the displacement of a particle is identically defined by the area bounded by the velocity-time graph, the time axis, and the time interval boundaries.

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: Establishing Kinematic Equations

Let the top of the cliff be the origin ($y = 0$) and the upward direction be positive. The position $y(t)$ of an object under gravity is given by: $y(t) = u_i t - \frac{1}{2}gt^2$ [Per the Second Equation of Motion].

For Stone 1 ($u_1 = 15 \text{ m/s}$): $y_1(t) = 15t - 5t^2$

For Stone 2 ($u_2 = 30 \text{ m/s}$): $y_2(t) = 30t - 5t^2$

Step 2: Defining Relative Position ($x_{rel}$)

The relative position of the second stone with respect to the first is defined as: $x_{rel} = y_2(t) - y_1(t)$

$x_{rel} = (30t - 5t^2) - (15t - 5t^2) = 15t$

This linear relationship holds as long as both stones are in the air [Algebraic simplification].

Step 3: Calculating Time of Flight ($T$)

Stones hit the ground at $y = -200 \text{ m}$. For Stone 1: $-200 = 15t_1 - 5t_1^2 \implies t_1^2 - 3t_1 - 40 = 0$. Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t_1 = \frac{3 + \sqrt{9 + 160}}{2} \approx 7.9 \text{ s}$ (Taking positive root).

For Stone 2: $-200 = 30t_2 - 5t_2^2 \implies t_2^2 - 6t_2 - 40 = 0$. $t_2 = \frac{6 + \sqrt{36 + 160}}{2} \approx 10.2 \text{ s}$.

Step 4: Analyzing the Motion Phases

Step 5: Visual Representation

Final Analysis

The graph consists of two distinct segments:
1. Linear Segment ($0 \leq t \leq 7.9 \text{ s}$): $x_{rel} = 15t$. This is a straight line passing through the origin with a slope of $15 \text{ m/s}$.
2. Parabolic Segment ($7.9 < t \leq 10.2 \text{ s}$): $x_{rel} = -5t^2 + 30t + 200$. This represents the position of the second stone relative to the grounded first stone.

Final Solution: The relative position is defined by $x_{rel} = 15t$ for $0 \leq t \leq 7.9$ s and $x_{rel} = -5t^2 + 30t + 200$ for $7.9 < t \leq 10.2$ s.

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

In a speed-time graph, the distance $s$ traversed by a particle in a given time interval $[t_1, t_2]$ is numerically equal to the area under the speed-time curve with respect to the time axis. This is derived from the definition of speed $v = \frac{ds}{dt}$, which implies $ds = v \, dt$, and integrating both sides yields $s = \int_{t_1}^{t_2} v(t) \, dt$. [Per the Fundamental Theorem of Calculus in Kinematics].

Step 2: Analysis of the Geometric Figure

The speed-time graph for the interval $t = 0 \text{ s}$ to $t = 10 \text{ s}$ forms a right-angled triangle.

• The base of the triangle ($b$) is the time interval: $10 \text{ s} - 0 \text{ s} = 10 \text{ s}$.
• The height of the triangle ($h$) corresponds to the speed at $t = 10 \text{ s}$. Observing the provided graph, the speed increases linearly from $0 \text{ m/s}$ to $12 \text{ m/s}$ at $t = 10 \text{ s}$. Thus, $h = 12 \text{ m/s}$.

Step 3: Geometric Representation

Step 4: Algebraic Calculation

The area $A$ of a right-angled triangle is given by the formula $A = \frac{1}{2} \times \text{base} \times \text{height}$.
Substituting the values derived in Step 2:
$s = \frac{1}{2} \times (10 \text{ s}) \times (12 \text{ m/s})$
$s = 5 \text{ s} \times 12 \text{ m/s}$
$s = 60 \text{ m}$

Final Solution: The distance traversed by the particle between $t = 0 \text{ s}$ and $t = 10 \text{ s}$ is $60 \text{ m}$.

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).