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 $6 \, \text{m/s}$.

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: 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: 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 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: Establishing the Coordinate System and Velocity Definitions

To analyze the velocity of the child relative to a stationary observer on the platform, we define the following vectors based on the principle of Galilean transformation of velocities:

• Let $v_b$ be the velocity of the belt relative to the stationary platform: $v_b = 4 \text{ km h}^{-1}$.
• Let $v_{cb}$ be the velocity of the child relative to the belt: $v_{cb} = 9 \text{ km h}^{-1}$.
• Let $v_c$ be the velocity of the child relative to the stationary platform.

[Per the Galilean Transformation of Velocities: $\vec{v}_c = \vec{v}_{cb} + \vec{v}_b$]

Step 2: Defining Directional Vectors

We assign the direction of the belt's motion as positive ($+$). The problem asks for the speed of the child running opposite to the direction of the belt.

• Velocity of belt: $v_b = +4 \text{ km h}^{-1}$
• Velocity of child relative to belt (opposite to belt): $v_{cb} = -9 \text{ km h}^{-1}$

Step 3: Calculating Velocity relative to the Stationary Platform

Applying the vector addition formula for the child moving against the belt:

$v_c = v_{cb} + v_b$

$v_c = (-9 \text{ km h}^{-1}) + (4 \text{ km h}^{-1})$

$v_c = -5 \text{ km h}^{-1}$

Visual Representation of Vector Addition

Step 4: Determining Scalar Speed

Speed is defined as the magnitude of the velocity vector [By definition of Speed]:

$\text{Speed} = |v_c| = |-5 \text{ km h}^{-1}| = 5 \text{ km h}^{-1}$

Final Solution: The speed of the child running opposite to the direction of motion of the belt, as observed from the stationary platform, is 5 km h⁻¹.

Step 1: Establishing the Velocity Reference Frames

To analyze the motion of the child, we must define the reference frames clearly using vector notation. Let the direction of the motion of the belt be the positive $x$-direction ($+$). Given:

• Velocity of the belt relative to the ground: $v_b = +4 \text{ km h}^{-1}$
• Velocity of the child relative to the belt: $v_{cb} = \pm 9 \text{ km h}^{-1}$

[By the Principle of Galilean Relativity, the velocity of an object in a frame of reference is the vector sum of its velocity relative to another frame and the velocity of that frame relative to the ground: $\vec{v}_{cg} = \vec{v}_{cb} + \vec{v}_{bg}$]

Step 2: Calculating Speed in the Direction of Motion (a)

When the child runs in the direction of the belt's motion, the relative velocity of the child with respect to the belt is $+9 \text{ km h}^{-1}$.

$v_{g(a)} = v_{cb} + v_b$

$v_{g(a)} = 9 \text{ km h}^{-1} + 4 \text{ km h}^{-1} = 13 \text{ km h}^{-1}$

[As per the Law of Addition of Velocities for collinear motion]

Step 3: Calculating Speed Opposite to the Direction of Motion (b)

When the child runs opposite to the belt's motion, the relative velocity of the child with respect to the belt is $-9 \text{ km h}^{-1}$.

$v_{g(b)} = v_{cb} + v_b$

$v_{g(b)} = -9 \text{ km h}^{-1} + 4 \text{ km h}^{-1} = -5 \text{ km h}^{-1}$

The speed (magnitude) is $|-5| = 5 \text{ km h}^{-1}$.

Step 4: Analyzing Motion Viewed by the Parents

The parents are located on the moving belt. Therefore, the parents and the child share the same frame of reference (the belt). In this frame:

• The velocity of the belt relative to itself is $0$.
• The child's speed is constant at $9 \text{ km h}^{-1}$ regardless of direction, as the belt's velocity $v_b$ cancels out relative to the parents.

[By the Principle of Inertial Frames: Physics laws are invariant in all inertial frames; a change in the observer's frame from ground to belt shifts the velocity measurement by a constant vector $v_b$]

Visual Representation of Relative Velocities

Final Solution: (a) The speed of the child with respect to the ground while running in the direction of the belt is $13 \text{ km h}^{-1}$. (b) The speed of the child with respect to the ground while running opposite to the belt is $5 \text{ km h}^{-1}$. When viewed by the parents, both speeds are $9 \text{ km h}^{-1}$ because the parents are in the same frame of reference as the belt.

Step 1: Analyzing the Kinematics of the Motion

The motion is divided into two phases:

• Phase 1 (Accelerated Motion): From $t = 0$ to $t = 10 \text{ s}$, with $u = 0$ and $a = 1 \text{ m/s}^2$.
• Phase 2 (Uniform Velocity): For $t > 10 \text{ s}$, the vehicle maintains the velocity attained at $t = 10 \text{ s}$.

Step 2: Deriving the Expression for Distance in the $n^{th}$ Second

The distance covered in the $n^{th}$ second, denoted as $s_n$, is given by the kinematic formula:

Substituting $u = 0$ and $a = 1 \text{ m/s}^2$:

This linear relationship holds for $n = 1, 2, ..., 10$.

Step 3: Calculating Values for the Accelerated Phase

Step 4: Analyzing Motion for $n > 10$

First, calculate velocity at $t = 10 \text{ s}$:

For $n > 10$, the distance covered in the $n^{th}$ second is simply the distance covered at uniform velocity:

Step 5: Graphical Representation

Step 6: Analytical Conclusion

The plot of $s_n$ versus $n$ for the accelerated motion ($n \le 10$) follows the relation $s_n = n - 0.5$. Since this is a first-degree polynomial in $n$ ($y = mx + c$), the plot is a straight line with a slope of $1 \text{ m/s}^2$ and a $y$-intercept of $-0.5 \text{ m}$.

Final Solution: The plot of the distance covered during the $n^{th}$ second versus $n$ is a straight line during the accelerated motion, not a parabola.

Part 1: Motion in a Stationary Lift

Given Variables:

• Initial velocity of the ball ($u$) = $49 \, \text{m s}^{-1}$
• Acceleration due to gravity ($g$) = $-9.8 \, \text{m s}^{-2}$ [Taking upward direction as positive]
• Final displacement ($\Delta y$) = $0 \, \text{m}$ (The ball returns to the boy's hand)

Step 1: Applying Kinematic Equation for Displacement

The displacement of an object moving under constant acceleration is given by the kinematic equation: $\Delta y = ut + \frac{1}{2}at^2$ [Per the Equation of Motion under Constant Acceleration]

Substituting the known values:

Step 2: Solving for Time ($t$)

This yields two solutions: $t = 0$ (time of projection) and $t = \frac{49}{4.9}$.

Result: The ball takes $10 \, \text{seconds}$ to return to the boy's hand when the lift is stationary.

Part 2: Motion in a Uniformly Moving Lift

Given Variables:

• Velocity of the lift ($v_L$) = $5 \, \text{m s}^{-1}$
• Initial velocity of the ball relative to the lift ($u_{rel}$) = $49 \, \text{m s}^{-1}$
• Initial velocity of the ball relative to the ground ($u_{ground}$) = $u_{rel} + v_L = 49 + 5 = 54 \, \text{m s}^{-1}$

Step 3: Calculating Relative Acceleration and Velocity

Since the lift moves at a uniform speed, its acceleration is zero. The relative acceleration of the ball with respect to the lift remains $g = -9.8 \, \text{m s}^{-2}$ [Per the Galilean transformation of acceleration].

The boy and the ball are both moving in the frame of reference of the lift. Because the lift is moving with a constant velocity, the physics within the lift remains identical to the stationary case due to the Principle of Relativity.

Step 4: Calculation in the Lift Frame

In the frame of the lift, the displacement of the ball to return to the hand is $\Delta y' = 0$. The initial velocity relative to the lift is $u' = 49 \, \text{m s}^{-1}$.

Note: The time taken is independent of the uniform velocity of the lift because the lift's constant motion does not introduce inertial forces (fictitious forces) that would alter the relative acceleration of the ball.

Final Solution: The ball takes 10 seconds to return to the boy's hand in both the stationary lift and the lift moving at a uniform speed of 5 m s⁻¹.

Theoretical Foundation & Definition of Average Velocity

The average velocity ($v_{avg}$) of a particle moving in one dimension over a specific time interval $[t_1, t_2]$ is defined as the ratio of the total displacement to the total time taken. Displacement is defined as the change in the position coordinate, represented as $\Delta x = x(t_2) - x(t_1)$.

Step 1: Analyzing the Kinematic Definition

According to the fundamental definition of velocity in kinematics, the instantaneous velocity $v(t)$ is the derivative of position with respect to time: $v(t) = \frac{dx}{dt}$ [Per the Calculus of Kinematics].

To find the displacement over an interval, we integrate the velocity function:

$\int_{x(t_1)}^{x(t_2)} dx = \int_{t_1}^{t_2} v(t) dt$

$\Delta x = x(t_2) - x(t_1) = \int_{t_1}^{t_2} v(t) dt$

Step 2: Evaluating the Average Velocity Formula

By the Mean Value Theorem for Integrals, the average value of a function $v(t)$ over an interval $[t_1, t_2]$ is given by:

$v_{avg} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} v(t) dt$ [By Mathematical Definition of Average Function]

Substituting the result from Step 1 into this equation:

$v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$

Step 3: Verification Against Graph Mechanics

In a velocity-time graph, the integral $\int_{t_1}^{t_2} v(t) dt$ corresponds to the area under the curve between $t_1$ and $t_2$. Since the displacement is equal to the area under the $v-t$ graph, the ratio of this area to the time interval $(t_2 - t_1)$ yields the average velocity.

Conclusion on the Statement

The provided formula $v_{average} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$ is a standard definition of average velocity. It is independent of the shape of the velocity-time graph, provided the position function $x(t)$ is continuous and differentiable over the interval.

Final Solution: The formula $v_{average} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$ is correct for describing the motion of the particle over the time-interval $t_1$ to $t_2$, as it is the fundamental definition of average velocity.

Theoretical Foundation of Kinematic Equations

The equation $v(t_2) = v(t_1) + a (t_2 - t_1)$ is derived from the fundamental definition of instantaneous acceleration, $a = \frac{dv}{dt}$ [Per the definition of derivative in calculus]. For this specific equation to hold true, the acceleration $a$ must be constant over the interval $[t_1, t_2]$.

Analysis of the Velocity-Time ($v-t$) Graph

In a $v-t$ graph, the instantaneous acceleration is represented by the slope of the tangent to the curve at any given time $t$. Specifically, the average acceleration over an interval $[t_1, t_2]$ is defined as:

$\bar{a} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$

The equation $v(t_2) = v(t_1) + a(t_2 - t_1)$ can be rearranged as $a = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$. This implies that the acceleration $a$ must be equivalent to the slope of the secant line connecting points $(t_1, v(t_1))$ and $(t_2, v(t_2))$.

Geometrical Interpretation

Evaluation of Validity

• Scenario 1 (Non-linear motion): If the graph is a curve (as shown in the provided visual concept), the slope changes continuously. Therefore, the instantaneous acceleration $a(t)$ is not equal to the constant average acceleration $\bar{a}$ over the interval. Thus, the formula is incorrect for non-uniform motion.
• Scenario 2 (Uniform motion): If the graph is a straight line, the acceleration is constant throughout the interval. In this specific case, the formula is correct as it satisfies the kinematic equation for constant acceleration.

Logical Conclusion

The equation $v(t_2) = v(t_1) + a(t_2 - t_1)$ is only valid for motion with constant acceleration. Looking at the provided velocity-time graph (Fig 3.29), which exhibits non-linear behavior (a curve), the acceleration is not constant. Therefore, the formula is not applicable to the motion represented in the figure for the arbitrary interval $[t_1, t_2]$.

Final Solution: The formula is incorrect for the motion shown in the graph because the acceleration is not constant; the slope of the velocity-time graph varies with time.

Analytical Premise & Theoretical Foundation

The equation provided is $x(t_2) = x(t_1) + v_{average}(t_2 - t_1) + \frac{1}{2} a_{average}(t_2 - t_1)^2$. To determine the validity of this expression for a non-uniformly accelerating particle (as depicted in a general velocity-time graph), we must analyze the kinematic definitions of average velocity and average acceleration over a finite time interval $\Delta t = (t_2 - t_1)$.

Step 1: Analyzing the Definitions

The displacement $\Delta x$ over an interval is given by the integral of the velocity function [By the Fundamental Theorem of Calculus]:

$\Delta x = x(t_2) - x(t_1) = \int_{t_1}^{t_2} v(t) \, dt$

The average velocity is defined as:

$v_{avg} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} v(t) \, dt = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$

Rearranging this yields: $x(t_2) = x(t_1) + v_{avg}(t_2 - t_1)$. This identity is universally correct for any motion.

Step 2: Evaluating the Acceleration Term

The proposed equation adds a second-order term: $\frac{1}{2} a_{avg} (t_2 - t_1)^2$. Here, $a_{avg} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$. Substituting this into the proposed expression:

Proposed RHS = $x(t_1) + v_{avg}(t_2 - t_1) + \frac{1}{2} \left[ \frac{v(t_2) - v(t_1)}{t_2 - t_1} \right] (t_2 - t_1)^2$

Proposed RHS = $x(t_1) + v_{avg} \Delta t + \frac{1}{2} [v(t_2) - v(t_1)] \Delta t$

Step 3: Verification via Kinematic Equivalence

For the equation to be correct, the term $\frac{1}{2} [v(t_2) - v(t_1)] \Delta t$ must equal zero for all arbitrary motions, because $x(t_2) = x(t_1) + v_{avg} \Delta t$ is already the complete description of displacement. The additional term $\frac{1}{2} a_{avg} (\Delta t)^2$ is only valid under the specific condition of Uniformly Accelerated Motion (where $a$ is constant, implying $v_{avg} = \frac{v(t_1) + v(t_2)}{2}$).

Step 4: Graphical Representation

In the graph above, the curve represents non-uniform acceleration. Since the area under the curve (displacement) is not represented by the parabolic formula for general motion, the equation is only a definition-based identity if $a$ were constant. In the context of a general $v-t$ graph where acceleration varies, the inclusion of the $\frac{1}{2}a_{avg}(\Delta t)^2$ term introduces an error unless the motion is linear acceleration.

Conclusion on Validity

The equation is incorrect for general motion as it assumes a kinematic structure that only applies to constant acceleration scenarios. Specifically, for general curvilinear $v-t$ graphs, the displacement is determined solely by the first two terms relative to the average velocity; the third term is mathematically inconsistent with the definition of the average acceleration over a non-linear interval.

Final Solution: The formula is incorrect for general motion, as it is only valid under the condition of constant acceleration.

Theoretical Foundation of Average Acceleration

The concept of average acceleration ($a_{avg}$) in kinematics is defined as the total change in velocity divided by the total time interval over which that change occurs. For any one-dimensional motion, the average acceleration represents the slope of the secant line connecting two specific points $(t_1, v(t_1))$ and $(t_2, v(t_2))$ on a velocity-time ($v$-$t$) graph.

Step 1: Formal Definition of Average Acceleration

By the definition of kinematics in classical mechanics, the average acceleration over the interval $\Delta t = t_2 - t_1$ is given by the ratio of the change in velocity $\Delta v$ to the change in time $\Delta t$:

$a_{average} = \frac{\Delta v}{\Delta t}$ [Per the definition of average acceleration]

Step 2: Substitution of Temporal Coordinates

Given the specific points at time $t_1$ and $t_2$, the velocities are represented as $v(t_1)$ and $v(t_2)$ respectively. The change in velocity is expressed as:

$\Delta v = v(t_2) - v(t_1)$

The duration of the interval is expressed as:

$\Delta t = t_2 - t_1$

Step 3: Verification of the Formula

Substituting the expressions from Step 2 into the definition established in Step 1:

$a_{average} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$

This formula represents the arithmetic mean rate of change of velocity. [Mathematical Axiom: Difference quotient of a function $f(t)$ over the interval $[t_1, t_2]$].

Visual Representation of the Slope

Conclusion on Formula Validity

The provided formula $a_{average} = \frac{v(t_2) - v(t_1)}{t_2 - t_1}$ is correct for describing the average acceleration of a particle over the interval $[t_1, t_2]$. It is derived directly from the fundamental definition of average acceleration in one-dimensional linear motion.

Final Solution: The formula is correct, as it accurately represents the slope of the secant line between $t_1$ and $t_2$ on a $v$-$t$ graph, which is the standard definition of average acceleration.

Step 1: Establishing Coordinate Systems and Variables

To determine the velocity of the child with respect to a stationary observer, we apply the principles of relative velocity in one dimension [Per the Galilean Transformation of velocities].

• Velocity of the belt relative to the stationary ground ($v_{b}$): $+4 \text{ km h}^{-1}$
• Velocity of the child relative to the belt ($v_{c/b}$): $+9 \text{ km h}^{-1}$ (running in the direction of the belt's motion)
• Velocity of the child relative to the stationary ground ($v_{c}$): To be determined.

Step 2: Vector Addition of Relative Velocities

The velocity of an object $c$ with respect to a stationary frame $g$ is the vector sum of its velocity with respect to a moving frame $b$ and the velocity of that moving frame with respect to the stationary frame $g$. [By the Law of Addition of Velocities]:

$v_{c} = v_{c/b} + v_{b}$

Step 3: Algebraic Calculation

Substitute the given values into the equation:

$v_{c} = 9 \text{ km h}^{-1} + 4 \text{ km h}^{-1}$

$v_{c} = 13 \text{ km h}^{-1}$

Visual Representation of Vector Addition

Step 4: Verification and Unit Consistency

The calculation assumes the motion is strictly linear and collinear. Since both the belt and the child move in the positive direction relative to the ground, the magnitudes are added arithmetically.

Final Solution: The speed of the child running in the direction of motion of the belt, as observed from the stationary platform, is $13 \text{ km h}^{-1}$.

Given Variables & Initial Setup

The provided speed-time graph illustrates a particle moving in a fixed direction. To determine the distance traversed between $t = 2\text{ s}$ and $t = 6\text{ s}$, we utilize the fundamental relationship between speed and time: the area under a speed-time graph represents the total distance traveled [Per the definition of displacement/distance in kinematics].

Based on the provided graph:

• At $t = 2\text{ s}$, the speed $v = 0\text{ m/s}$.
• At $t = 5\text{ s}$, the speed reaches a peak of $v = 12\text{ m/s}$.
• At $t = 6\text{ s}$, the speed is $v = 12\text{ m/s}$ (constant velocity from $t=5$ to $t=6$).

Step 1: Segmenting the Area Calculation

The distance traversed between $t = 2\text{ s}$ and $t = 6\text{ s}$ corresponds to the area under the curve in two distinct intervals:

1. Interval 1 ($t = 2\text{ s}$ to $t = 5\text{ s}$): A right-angled triangle.
2. Interval 2 ($t = 5\text{ s}$ to $t = 6\text{ s}$): A rectangle.

Step 2: Calculating Area of Interval 1 ($t = 2$ to $5$)

The shape is a triangle with base $b = (5 - 2) = 3\text{ s}$ and height $h = 12\text{ m/s}$.

$\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height}$ [Geometric formula for triangle area]

$\text{Area}_1 = \frac{1}{2} \times 3 \times 12 = 18\text{ m}$

Step 3: Calculating Area of Interval 2 ($t = 5$ to $6$)

The shape is a rectangle with width $w = (6 - 5) = 1\text{ s}$ and height $h = 12\text{ m/s}$.

$\text{Area}_2 = \text{width} \times \text{height}$ [Geometric formula for rectangular area]

$\text{Area}_2 = 1 \times 12 = 12\text{ m}$

Step 4: Total Distance Traversed

Total Distance = $\text{Area}_1 + \text{Area}_2$

Total Distance = $18\text{ m} + 12\text{ m} = 30\text{ m}$

Final Solution: The total distance traversed by the particle between $t = 2\text{ s}$ and $t = 6\text{ s}$ is $30\text{ meters}$.

Step 1: Analyzing the Velocity-Time Graph

In the provided graph (Fig 3.29), the velocity $v(t)$ is represented as a function of time $t$. The graph is non-linear, indicating that the acceleration $a = \frac{dv}{dt}$ is not constant. In one-dimensional kinematics, the position $x(t)$ at any time $t$ is defined by the integral of the velocity function:

$x(t_2) = x(t_1) + \int_{t_1}^{t_2} v(t) dt$ [Per the Fundamental Theorem of Calculus]

Step 2: Evaluating the Validity of the Proposed Formula

The formula provided in the question is:

$x(t_2) = x(t_1) + v(t_1)(t_2 - t_1) + \frac{1}{2}a(t_2 - t_1)^2$

This specific kinematic equation is derived from the definition of average acceleration under the strict condition that acceleration is constant ($a = \text{constant}$).

• Case A (Constant Acceleration): If the graph were a straight line, the slope would be constant, and the formula would be valid.
• Case B (Variable Acceleration): In the provided graph, the velocity-time relationship is curved. The slope $\frac{dv}{dt}$ changes continuously over the interval $[t_1, t_2]$.

Step 3: Geometrical Interpretation

The displacement $\Delta x = x(t_2) - x(t_1)$ is equal to the area under the $v-t$ curve between $t_1$ and $t_2$.

Because the curve is non-linear, the area under the curve is not equivalent to the area of a trapezoid defined by the initial velocity and a single acceleration value ($v_1\Delta t + \frac{1}{2}a\Delta t^2$).

Step 4: Conclusion

Since the motion depicted in the graph involves a changing slope (variable acceleration), the kinematic equation $x(t_2) = x(t_1) + v(t_1)(t_2 - t_1) + \frac{1}{2}a(t_2 - t_1)^2$ is incorrect for this specific particle. This formula is restricted to the specific condition where $a = \text{constant}$ [Per the Equations of Uniformly Accelerated Motion].

Final Solution: The formula is incorrect because it is only valid for motion with constant acceleration, whereas the provided velocity-time graph shows non-linear motion, implying variable acceleration.