Initial Categorization of Physical Quantities

To identify the vector quantity, we must evaluate each given physical quantity based on the requirement of having both a magnitude and a specific direction, following the laws of vector addition [Per the Axioms of Vector Algebra].

Mathematical Analysis of Impulse

Impulse ($J$) is defined by the change in linear momentum ($\Delta p$). Since momentum $p = mv$ (where $v$ is velocity, a vector), it follows that:

$\vec{J} = \vec{F}_{avg} \cdot \Delta t$

[By Newton's Second Law of Motion in terms of momentum: $\vec{F} = \frac{d\vec{p}}{dt} \implies d\vec{p} = \vec{F} dt$].

Because the average force $\vec{F}_{avg}$ is a vector quantity, the resulting impulse must possess the same directional orientation as the force applied during the time interval $\Delta t$.

Deduction

Upon reviewing the list provided against the criteria for vectors, only Impulse satisfies the requirement of possessing magnitude and a specific direction in three-dimensional space.

Final Solution: The only vector quantity in the provided list is Impulse.

Step 1: Theoretical Classification of Physical Quantities

Physical quantities are categorized into two fundamental types based on their mathematical properties:

• Scalar Quantities: Physical quantities that possess only magnitude and satisfy the algebraic laws of addition, subtraction, and multiplication. They are independent of directional orientation.
• Vector Quantities: Physical quantities that possess both magnitude and a specific direction in space, and must satisfy the laws of vector addition (e.g., triangle law or parallelogram law).

Step 2: Analytical Evaluation of the List

Step 3: Verification of Scalar Identity

1. Work ($W$): [Per the scalar product theorem], the dot product of any two vectors $\vec{A} \cdot \vec{B} = AB\cos(\theta)$ produces a magnitude without directional properties. Hence, work is a scalar.

2. Current ($I$): [Per Kirchhoff's Current Law], current entering a node is added algebraically ($I_1 + I_2 = I_3$) regardless of the geometric angle between the wires. Because it fails to adhere to the triangle law of vector addition, it is classified as a scalar [Ref: Fundamentals of Electrodynamics].

Final Solution: The two scalar quantities in the provided list are Work and Current.

Step 1: Theoretical Foundation of Vector Summation

Given the vector equation: $\vec{a} + \vec{b} + \vec{c} + \vec{d} = \vec{0}$

To analyze the relationship between the magnitude of vector $\vec{a}$ and the sum of the magnitudes of the other vectors, we isolate $\vec{a}$: $\vec{a} = -(\vec{b} + \vec{c} + \vec{d})$

[By the properties of vector addition and additive inverse]

Step 2: Applying the Triangle Inequality Theorem (Generalization)

The generalized triangle inequality states that the magnitude of the resultant of any number of vectors is less than or equal to the sum of the magnitudes of the individual vectors: $|\vec{R}| = |\vec{v_1} + \vec{v_2} + ... + \vec{v_n}| \leq |\vec{v_1}| + |\vec{v_2}| + ... + |\vec{v_n}|$

Applying this to our isolated equation: $|\vec{a}| = |-(\vec{b} + \vec{c} + \vec{d})|$ $|\vec{a}| = |\vec{b} + \vec{c} + \vec{d}|$

[By the property $|-\vec{V}| = |\vec{V}|$]

Per the generalized triangle inequality: $|\vec{a}| \leq |\vec{b}| + |\vec{c}| + |\vec{d}|$

Step 3: Logical Deduction

The mathematical derivation shows that the magnitude of $\vec{a}$ is at most equal to the sum of the magnitudes of $\vec{b}$, $\vec{c}$, and $\vec{d}$. It can never exceed this sum because the shortest distance between the starting point and the ending point of the vector chain formed by $\vec{b}, \vec{c},$ and $\vec{d}$ is the straight-line vector $\vec{a}$.

Step 4: Visualizing the Vector Polygon

Step 5: Comparative Analysis

Final Solution: The statement is correct. The magnitude of $\vec{a}$ cannot be greater than the sum of the magnitudes of $\vec{b}$, $\vec{c}$, and $\vec{d}$ because it represents the closing side of a vector polygon, which must satisfy the inequality $|\vec{a}| \leq |\vec{b}| + |\vec{c}| + |\vec{d}|$.

Statement Analysis

The statement is: "The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point."

The status of this statement is: True.

Step 1: Defining the Velocity Vector

In the framework of kinematics in a plane, the position of a particle at any time $t$ is defined by its position vector $\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$. The instantaneous velocity vector $\vec{v}$ is defined as the first time derivative of the position vector [Per the Definition of Instantaneous Velocity]:

$$\vec{v} = \frac{d\vec{r}}{dt} = \lim_{\Delta t \to 0} \frac{\vec{r}(t + \Delta t) - \vec{r}(t)}{\Delta t}$$

Step 2: Geometric Interpretation of the Derivative

Consider a path defined in the $xy$-plane. Let points $P$ and $Q$ be two positions of the particle at times $t$ and $t + \Delta t$, respectively. The displacement vector is $\Delta \vec{r} = \vec{r}(t + \Delta t) - \vec{r}(t)$.

The vector $\frac{\Delta \vec{r}}{\Delta t}$ points in the direction of the displacement (the secant line connecting $P$ and $Q$). As $\Delta t \to 0$, point $Q$ approaches point $P$ along the trajectory. [By the Fundamental Theorem of Calculus and Geometric Limits].

Step 3: Visual Representation

Step 4: Logical Conclusion

• As $\Delta t$ approaches $0$, the secant line segment $PQ$ becomes the tangent to the curve at point $P$.
• Since $\vec{v}$ is the limiting value of $\frac{\Delta \vec{r}}{\Delta t}$, the direction of $\vec{v}$ must coincide with the direction of the tangent line at that point.
• If the velocity vector were not tangent to the path, it would imply a component of motion perpendicular to the current trajectory, which contradicts the definition of a path as the locus of successive positions occupied by the particle.

Final Solution: The statement is True, because the instantaneous velocity is defined as the derivative of the position vector with respect to time, and geometrically, the derivative of a position vector function $\vec{r}(t)$ represents the tangent vector to the curve traced by the particle.

Step 1: Definition and Theoretical Foundation

A scalar quantity is defined as a physical quantity that is completely described by its magnitude (a numerical value along with its unit) and has no associated direction in space. Examples include mass, temperature, energy, and work.

The concept of "conservation" refers to a quantity remaining constant over time within a closed system or during a physical process [Per the Law of Conservation of Energy/Mass].

Step 2: Logical Analysis of the Statement

The statement asserts that "a scalar quantity is one that is conserved in a process." To evaluate the truth value of this statement, we must determine if being a scalar is synonymous with being a conserved quantity.

Step 3: Derivation of the Verdict

1. Counter-example Analysis: Consider the temperature of a substance during a heating process. Temperature is a scalar quantity, yet it changes as heat energy is added to the system. Thus, temperature is not conserved. [Reference: Thermodynamic State Functions].

2. Logical Implication: The property of being "scalar" and the property of being "conserved" are mutually independent attributes in physics. Being a scalar describes the mathematical nature of the quantity (a rank-0 tensor), whereas being conserved describes the temporal behavior of the quantity in a system.

3. Conclusion: Since there exist many scalar quantities that are not conserved in a process, the statement is categorically false.

Step 4: Summary of Reasoning

• Reason: Conservation is a property related to the dynamics and interaction of a system, whereas scalar status is a property of the physical quantity's transformation under coordinate rotation.
• Example: In an inelastic collision, the total kinetic energy ($K$) is a scalar quantity, yet it is not conserved (it is converted into internal energy or heat).

Final Solution: The statement is False. Being a scalar quantity describes the nature of the quantity (magnitude only), while conservation is a distinct dynamical property; many scalars, such as temperature or kinetic energy in inelastic collisions, are not conserved in physical processes.

Statement Analysis

The assertion provided is: "A scalar quantity is one that can never take negative values."

Conclusion

This statement is False.

Theoretical Justification

By definition, a scalar quantity is a physical quantity that is completely described by its magnitude (a numerical value) along with its appropriate unit. While some scalars are restricted to positive values due to their physical nature (e.g., mass, length, speed, time), there is no inherent mathematical rule within the definition of a scalar that precludes the assignment of a negative sign.

In physics, a negative sign associated with a scalar quantity typically indicates one of two things:

• Relative Reference: The value is measured relative to an arbitrary reference point or "zero level" (e.g., temperature).
• Directional Convention: The scalar represents a net value where a negative sign denotes a deficit, loss, or orientation opposite to a chosen sign convention (e.g., work, electric potential).

Illustrative Examples

Visual Representation of Scalar Measurement

The following SVG illustrates a temperature scale (a scalar), demonstrating that values can exist on both sides of a defined zero point.

Final Solution: The statement is False. A scalar quantity may take negative values if the physics of the quantity permits values below a chosen reference point or if the scalar represents a signed magnitude (such as Work or Electric Potential).

Given Variables & Initial Setup

To determine the maximum horizontal range ($R$) of a projectile under a vertical constraint (the ceiling height $H_{max} = 25\text{ m}$), we must first identify the initial conditions:

• Initial velocity, $u = 40\text{ m s}^{-1}$
• Acceleration due to gravity, $g = 9.8\text{ m s}^{-2}$
• Maximum vertical clearance (ceiling height), $H = 25\text{ m}$

Step 1: Determining the Vertical Component of Velocity

For the projectile to graze the ceiling without hitting it, its maximum vertical height must be equal to or less than $25\text{ m}$. The formula for maximum height attained by a projectile is given by [Per the Kinematic Equation for vertical motion $v_y^2 = u_y^2 - 2gH$ where $v_y = 0$]:

$H = \frac{u_y^2}{2g}$

Rearranging to find the required vertical component of initial velocity ($u_y$):

$u_y = \sqrt{2gH}$

$u_y = \sqrt{2 \times 9.8 \times 25} = \sqrt{490} \approx 22.136\text{ m s}^{-1}$

Step 2: Determining the Horizontal Component of Velocity

Since the magnitude of the resultant velocity vector is $u = \sqrt{u_x^2 + u_y^2}$, we solve for the horizontal component $u_x$ [By the Pythagorean Theorem]:

$u_x = \sqrt{u^2 - u_y^2}$

$u_x = \sqrt{40^2 - 490} = \sqrt{1600 - 490} = \sqrt{1110} \approx 33.317\text{ m s}^{-1}$

Step 3: Calculating Total Time of Flight

The time taken to reach the maximum height is $t_a = \frac{u_y}{g}$. The total time of flight ($T$) for a symmetric trajectory is [By the derivation of projectile motion time]:

$T = \frac{2u_y}{g} = \frac{2 \times 22.136}{9.8} \approx 4.517\text{ s}$

Step 4: Calculating Maximum Horizontal Distance (Range)

The horizontal range is given by the product of the horizontal velocity component and the total time of flight [As horizontal acceleration $a_x = 0$]:

$R = u_x \times T$

$R = 33.317 \times 4.517 \approx 150.5\text{ m}$

Final Solution: The maximum horizontal distance the ball can travel without hitting the ceiling is approximately 150.5 meters.

Theoretical Foundation of Average Acceleration

In kinematics, for any arbitrary motion of a particle in space, the state of the particle is defined by its position vector $\vec{r}(t)$ and its instantaneous velocity vector $\vec{v}(t)$. The average acceleration $\vec{a}_{avg}$ over a finite time interval $\Delta t = t_2 - t_1$ is defined as the total change in velocity divided by the total time elapsed during that interval.

Step 1: Defining the Kinematic Variables

• Let $\vec{v}(t_1)$ be the instantaneous velocity at time $t_1$.
• Let $\vec{v}(t_2)$ be the instantaneous velocity at time $t_2$.
• The change in velocity vector is defined as $\Delta \vec{v} = \vec{v}(t_2) - \vec{v}(t_1)$.
• The time interval is defined as $\Delta t = t_2 - t_1$.

Step 2: Mathematical Derivation

By the fundamental definition of average acceleration in vector calculus [Per the definition of average rate of change]:

$\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t}$

Substituting the defined variables into the expression:

$\vec{a}_{avg} = \frac{\vec{v}(t_2) - \vec{v}(t_1)}{t_2 - t_1}$

Step 3: Logical Justification

This relation holds true for any arbitrary motion, whether it be rectilinear, circular, or general curvilinear motion in three-dimensional space. The vector nature of velocity ensures that the difference $\vec{v}(t_2) - \vec{v}(t_1)$ accounts for changes in both the magnitude and the direction of the velocity vector [By the principle of Vector Addition and Subtraction].

Vector Representation of Velocity Change

Summary Analysis

Since the equation $\vec{a}_{avg} = \frac{\vec{v}(t_2) - \vec{v}(t_1)}{t_2 - t_1}$ strictly adheres to the definition of the average rate of change of a vector quantity with respect to time, it is mathematically valid for all forms of motion.

Final Solution: The relation $\vec{a}_{avg} = \frac{\vec{v}(t_2) - \vec{v}(t_1)}{t_2 - t_1}$ is true for any arbitrary motion in space.

Step 1: Definition and Physical Validity

The operation of multiplying a vector $\vec{A}$ by a scalar $s$ is mathematically defined as the scaling of the vector's magnitude while preserving or inverting its direction. The resulting quantity is a new vector $\vec{B}$, defined by the relation:

$$\vec{B} = s\vec{A}$$

[Per the axioms of Vector Algebra in Euclidean Space].

Step 2: Analytical Conditions for Meaningfulness

For this operation to be physically meaningful, two conditions must be satisfied:

• Dimensional Consistency: The scalar $s$ may be a dimensionless constant (e.g., $2\vec{v}$), or it may possess dimensions such that the product $s\vec{A}$ represents a physically recognizable quantity [Per the Principle of Dimensional Homogeneity].
• Logical Interpretation: The product must represent a relationship between physical quantities. For example, multiplying the velocity vector $\vec{v}$ (units: $m/s$) by a scalar time interval $\Delta t$ (units: $s$) results in a displacement vector $\vec{d} = \Delta t \vec{v}$ (units: $m$).

Step 3: Visual Representation of Scalar Multiplication

Step 4: Categorization of Meaningful Operations

Conclusion

The operation is meaningful. Multiplying a vector by a scalar is a fundamental operation in physics that allows for the transformation of vectors into other physical quantities (like multiplying mass by acceleration to obtain force) or simply scaling the magnitude of the vector while maintaining its direction (or reversing it if the scalar is negative).

Final Solution: The operation of multiplying a vector by a scalar is physically meaningful as it satisfies the laws of vector algebra and dimensional analysis, allowing for the derivation of new vector quantities or the scaling of existing ones.

1. Theoretical Foundation and Geometric Interpretation

To establish the inequality $|\vec{a} - \vec{b}| \geq ||\vec{a}| - |\vec{b}||$, we utilize the properties of a triangle in vector space. Let $\vec{a}$ and $\vec{b}$ be two vectors represented as adjacent sides of a triangle. The vector $(\vec{a} - \vec{b})$ represents the third side of the triangle connecting the head of $\vec{b}$ to the head of $\vec{a}$.

2. The Triangle Inequality Proof

According to the triangle inequality theorem in geometry, the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the third side. Consider the triangle formed by vectors $\vec{b}$, $(\vec{a} - \vec{b})$, and $\vec{a}$.

The vector sum relation is: $\vec{a} = \vec{b} + (\vec{a} - \vec{b})$

Taking the magnitude of both sides: $|\vec{a}| = |\vec{b} + (\vec{a} - \vec{b})|$

By the Triangle Inequality, $|\vec{x} + \vec{y}| \leq |\vec{x}| + |\vec{y}|$. Applying this to the right-hand side:

$|\vec{a}| \leq |\vec{b}| + |\vec{a} - \vec{b}|$

Rearranging the terms to isolate $|\vec{a} - \vec{b}|$: $|\vec{a} - \vec{b}| \geq |\vec{a}| - |\vec{b}|$ — (Equation 1)

Similarly, by symmetry, since $|\vec{a} - \vec{b}| = |\vec{b} - \vec{a}|$: $|\vec{b} - \vec{a}| \geq |\vec{b}| - |\vec{a}|$ $|\vec{a} - \vec{b}| \geq - (|\vec{a}| - |\vec{b}|)$ — (Equation 2)

Combining Equation 1 and Equation 2, we obtain the absolute difference inequality: $|\vec{a} - \vec{b}| \geq ||\vec{a}| - |\vec{b}||$

3. Geometric Representation

4. Conditions for Equality

The equality sign holds when the "triangle" collapses into a straight line (degenerate triangle). This occurs when the vectors $\vec{a}$ and $\vec{b}$ are collinear and point in the same direction.

• Condition: The vectors $\vec{a}$ and $\vec{b}$ are parallel (i.e., the angle $\theta$ between them is $0^\circ$).
• Mathematical verification: If $\vec{a} = k\vec{b}$ where $k > 0$: $|\vec{a} - \vec{b}| = |k\vec{b} - \vec{b}| = |k-1||\vec{b}|$ $||\vec{a}| - |\vec{b}|| = ||k\vec{b}| - |\vec{b}|| = |k-1||\vec{b}|$ Since both sides equal $|k-1||\vec{b}|$, the equality holds.

Final Solution: The inequality $|\vec{a} - \vec{b}| \geq ||\vec{a}| - |\vec{b}||$ holds as it is a restatement of the Triangle Inequality. The equality holds if and only if the vectors $\vec{a}$ and $\vec{b}$ are collinear and in the same direction (parallel).

Given Statement Analysis

The statement to be evaluated is: "The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre."

Step 1: Decomposition of Acceleration in Circular Motion

In circular motion, the acceleration vector $\vec{a}$ of a particle is the vector sum of two mutually perpendicular components:

• Centripetal Acceleration ($\vec{a}_c$): Directed towards the center of the circle. Its magnitude is given by $a_c = \frac{v^2}{r}$, where $v$ is the instantaneous speed and $r$ is the radius of the path [Per the definition of uniform circular motion kinematics].
• Tangential Acceleration ($\vec{a}_t$): Directed tangentially along the path of motion. Its magnitude is given by $a_t = \frac{dv}{dt}$. This component represents the rate of change of the magnitude of velocity (speed) [By Newton's Second Law applied to tangential dynamics].

Step 2: Vectorial Synthesis of Total Acceleration

The net acceleration $\vec{a}$ is expressed as:

$\vec{a} = \vec{a}_c + \vec{a}_t$

The magnitude of the net acceleration is determined by the Pythagorean theorem, as the components are orthogonal:

$|\vec{a}| = \sqrt{a_c^2 + a_t^2}$

Step 3: Evaluation of Scenarios

Visual Representation of Acceleration Components

Logical Conclusion

The statement is False. While the centripetal acceleration is always directed towards the center, the net acceleration only points towards the center in the specific case of uniform circular motion (where speed is constant). In non-uniform circular motion, the presence of tangential acceleration ($a_t = \frac{dv}{dt} \neq 0$) shifts the resultant acceleration vector away from the radial direction.

Final Solution: False; the net acceleration is only directed towards the center in uniform circular motion. In all other cases, the tangential component of acceleration causes the net acceleration vector to deviate from the radial direction.

Step 1: Analyzing the Geometric Path

The motorist follows a regular hexagonal path where each side length $s = 500\text{ m}$. Since the motorist turns $60^{\circ}$ to the left after every 500 m, the path describes a closed regular hexagon. The interior angle of a regular hexagon is $120^{\circ}$, and the exterior turning angle is $60^{\circ}$.

Step 2: Calculations for Specific Turns

Step 3: Analytical Derivation

1. Third Turn: The motorist has covered 3 sides of a hexagon. The displacement is the diagonal connecting the starting vertex to the vertex after the 3rd side. In a regular hexagon, this connects opposite vertices, spanning $2s$. Thus, $r = 2 \times 500 = 1000\text{ m}$. [By Geometry of Regular Polygons]

2. Sixth Turn: The motorist has completed one full hexagonal circuit. The final position coincides with the initial position. Thus, $r = 0\text{ m}$. [Per Definition of Displacement in Closed Loops]

3. Eighth Turn: The motorist has completed one full cycle and traveled 2 more sides. The position is the same as the position after the 2nd turn. The distance between the start and the 2nd vertex in a regular hexagon is given by the law of cosines or geometric projection: $r = \sqrt{s^2 + s^2 - 2s^2 \cos(120^{\circ})} = \sqrt{s^2 + s^2 + s^2} = s\sqrt{3}$.
$r = 500\sqrt{3} \approx 500 \times 1.732 = 866\text{ m}$.

Summary Table of Comparison

• 3rd Turn: Path length $1500\text{ m}$, Displacement $1000\text{ m}$. Ratio: $0.67$.
• 6th Turn: Path length $3000\text{ m}$, Displacement $0\text{ m}$. Ratio: $0$.
• 8th Turn: Path length $4000\text{ m}$, Displacement $866\text{ m}$. Ratio: $0.22$.

Final Solution: The displacements at the 3rd, 6th, and 8th turns are 1000 m, 0 m, and 866 m respectively. The ratios of displacement to path length are 0.67, 0, and 0.22 respectively.

Theoretical Foundation

In kinematics, the relationship $\vec{v}(t) = \vec{v}(0) + \vec{a}t$ represents the velocity of an object at time $t$. To determine its validity for arbitrary motion in space, we must analyze the definition of instantaneous acceleration [Per the Fundamental Theorem of Calculus in Kinematics]:

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

Step 1: Derivation for Constant Acceleration

If we assume acceleration $\vec{a}$ is constant over the time interval $[0, t]$, we perform a definite integration of the acceleration vector with respect to time:

$$\int_{\vec{v}(0)}^{\vec{v}(t)} d\vec{v} = \int_{0}^{t} \vec{a} \, dt$$

Since $\vec{a}$ is constant, it is treated as a scalar constant relative to the integral operator:

$$\vec{v}(t) - \vec{v}(0) = \vec{a} [t]_{0}^{t}$$

$$\vec{v}(t) = \vec{v}(0) + \vec{a}t$$

[By the Fundamental Theorem of Calculus: $\int_{a}^{b} f'(x) dx = f(b) - f(a)$].

Step 2: Analysis of Arbitrary Motion

For arbitrary motion, the acceleration vector $\vec{a}(t)$ is generally a function of time, position, or velocity—it is not necessarily constant. Therefore, the relation must be written in its integral form:

$$\vec{v}(t) = \vec{v}(0) + \int_{0}^{t} \vec{a}(t') \, dt'$$

Comparing the given expression $\vec{v}(t) = \vec{v}(0) + \vec{a}t$ to the general integral form, it is evident that the given relation holds only if $\vec{a}(t) = \text{constant}$.

Visual Representation of Velocity Vectors

Conclusion

The relation $\vec{v}(t) = \vec{v}(0) + \vec{a}t$ is False for arbitrary motion. It is a specific solution valid exclusively for the case of Uniformly Accelerated Motion (where acceleration remains constant in both magnitude and direction).

Final Solution: The relation is false for arbitrary motion because it assumes $\vec{a}$ is constant, whereas in arbitrary motion, $\vec{a}$ is a function of time $\vec{a}(t)$.

Theoretical Foundation & Definitions

To evaluate the truth value of the statement, we define the two primary kinematic quantities over a time interval $\Delta t = t_2 - t_1$:

• Average Speed ($v_{avg}$): Defined as the ratio of total path length (distance) $d$ to the time interval $\Delta t$.
  $v_{avg} = \frac{\text{Total Path Length}}{\Delta t}$
• Average Velocity ($\vec{v}_{avg}$): Defined as the ratio of the displacement vector $\Delta \vec{r}$ to the time interval $\Delta t$.
  $\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t}$

Step 1: Mathematical Comparison

The magnitude of the average velocity is given by:

$|\vec{v}_{avg}| = \left| \frac{\Delta \vec{r}}{\Delta t} \right| = \frac{|\Delta \vec{r}|}{\Delta t}$

[Per the property of scalars in vector division: $|k\vec{A}| = |k||\vec{A}|$ where $k = 1/\Delta t > 0$].

Step 2: Relation Between Distance and Displacement

By the definition of path length and displacement in space, the total path length $d$ is the summation of the magnitude of individual infinitesimal displacement vectors along the trajectory. Geometrically, the straight-line distance between the initial position $\vec{r}_1$ and final position $\vec{r}_2$ (the magnitude of displacement, $|\Delta \vec{r}|$) is the shortest possible path between two points.

Therefore, for any motion: Total Path Length ($d$) $\geq$ Displacement Magnitude ($|\Delta \vec{r}|$) [Per the Triangle Inequality and the definition of a straight line].

Step 3: Logical Synthesis

Since $\Delta t > 0$, we divide both sides of the inequality $d \geq |\Delta \vec{r}|$ by $\Delta t$:

$\frac{d}{\Delta t} \geq \frac{|\Delta \vec{r}|}{\Delta t}$

$v_{avg} \geq |\vec{v}_{avg}|$

Analytical Conclusion

The inequality $v_{avg} \geq |\vec{v}_{avg}|$ holds because the path length is always greater than or equal to the straight-line displacement. Equality occurs strictly in the case of unidirectional, linear motion where the particle does not change its direction [Per the definition of collinear vectors].

Final Solution: The statement is True. The average speed is always greater than or equal to the magnitude of average velocity because the total path length traveled is always greater than or equal to the magnitude of the displacement vector.

Step 1: Categorization Framework

To classify the physical quantities as scalars or vectors, we apply the standard definitions:

• Scalar: A quantity defined solely by its magnitude (a real number with appropriate units). It is invariant under coordinate rotation. [Reference: Scalar Property of Magnitude]
• Vector: A quantity defined by both magnitude and a specific direction in space, obeying the laws of vector addition (e.g., triangle law or parallelogram law). [Reference: Law of Vector Addition]

Step 2: Analytical Classification of Quantities

Step 3: Visual Representation of Vector vs Scalar Logic

The distinction in motion is best illustrated by comparing displacement (vector) to distance (scalar).

Step 4: Summary of Results

Based on the rigorous application of spatial laws and definitions:

• Scalars: Volume, Mass, Speed, Density, Number of moles, Angular frequency.
• Vectors: Acceleration, Velocity, Displacement, Angular velocity.

Final Solution: The provided list contains six scalar quantities (Volume, Mass, Speed, Density, Number of moles, Angular frequency) and four vector quantities (Acceleration, Velocity, Displacement, Angular velocity).

Step 1: Theoretical Analysis of the Operation

The operation in question is the multiplication of two scalar quantities, represented mathematically as $s_1 \cdot s_2$, where $s_1$ and $s_2$ are physical scalars. In physics, a scalar is a quantity defined entirely by its magnitude and associated unit [Per definition of Scalar Physical Quantities].

Step 2: Mathematical and Dimensional Validity

When two scalars are multiplied, the resultant $S$ is defined as:

$S = s_1 \times s_2$

From the perspective of dimensional analysis, if $[s_1] = M^a L^b T^c$ and $[s_2] = M^d L^e T^f$, then the product has dimensions of $[S] = M^{a+d} L^{b+e} T^{c+f}$. This is mathematically consistent and well-defined [By the Principle of Dimensional Homogeneity].

Step 3: Evaluation of Physical Meaningfulness

An operation is considered "meaningful" in physics if it describes a relationship between physical phenomena or defines a new derived physical quantity. Consider the following examples:

Step 4: Geometric Representation of Scalar Product

Since scalar multiplication produces a magnitude without directional components, it can be visualized as an area or a scaling factor.

Step 5: Conclusion on Meaningfulness

The operation is meaningful. The multiplication of two scalars is a standard algebraic process in physics used to derive new physical quantities. Unlike the addition of scalars with different dimensions (which is physically undefined), the multiplication of two scalars with different dimensions is perfectly permissible and yields a valid resultant quantity.

Final Solution: The operation of multiplying any two scalars is physically meaningful. It is a fundamental algebraic procedure used to derive new physical quantities from existing ones, provided the multiplication follows the laws of dimensional algebra.

Step 1: Given Variables & Initial Setup

The position vector of the particle is defined as a function of time $t$ by the equation:

$\boldsymbol{r}(t) = 3.0t \hat{i} - 2.0t^2 \hat{j} + 4.0 \hat{k} \text{ m}$

To determine the velocity ($\boldsymbol{v}$) and acceleration ($\boldsymbol{a}$), we must utilize the definitions of instantaneous kinematic quantities in three-dimensional space.

Step 2: Calculating Velocity Vector $\boldsymbol{v}(t)$

By definition, the instantaneous velocity is the first time derivative of the position vector [Per the definition of velocity in kinematics, $\boldsymbol{v} = \frac{d\boldsymbol{r}}{dt}$]:

$\boldsymbol{v} = \frac{d}{dt}(3.0t \hat{i} - 2.0t^2 \hat{j} + 4.0 \hat{k})$

Applying the power rule of differentiation $\frac{d}{dt}(ct^n) = n \cdot ct^{n-1}$ component-wise:

• $x$-component: $\frac{d}{dt}(3.0t) = 3.0 \text{ m/s}$
• $y$-component: $\frac{d}{dt}(-2.0t^2) = -2.0 \cdot 2t = -4.0t \text{ m/s}$
• $z$-component: $\frac{d}{dt}(4.0) = 0 \text{ m/s}$ [Constant rule: derivative of a constant is zero]

Therefore, the velocity vector is:

$\boldsymbol{v}(t) = (3.0 \hat{i} - 4.0t \hat{j}) \text{ m/s}$

Step 3: Calculating Acceleration Vector $\boldsymbol{a}(t)$

By definition, the instantaneous acceleration is the time derivative of the velocity vector [Per the definition of acceleration, $\boldsymbol{a} = \frac{d\boldsymbol{v}}{dt}$]:

$\boldsymbol{a} = \frac{d}{dt}(3.0 \hat{i} - 4.0t \hat{j})$

Applying the derivative operator:

• $x$-component: $\frac{d}{dt}(3.0) = 0 \text{ m/s}^2$
• $y$-component: $\frac{d}{dt}(-4.0t) = -4.0 \text{ m/s}^2$
• $z$-component: $0 \text{ m/s}^2$

Therefore, the acceleration vector is:

$\boldsymbol{a} = -4.0 \hat{j} \text{ m/s}^2$

Step 4: Vector Visualization

Summary of Results

Final Solution: The velocity of the particle is $\boldsymbol{v} = (3.0 \hat{i} - 4.0t \hat{j}) \text{ m/s}$ and the acceleration is $\boldsymbol{a} = -4.0 \hat{j} \text{ m/s}^2$.

Statement Analysis

The statement "A scalar quantity must be dimensionless" is False.

Theoretical Justification

In classical physics and the SI system of units, a scalar quantity is defined as a physical quantity that is completely described by its magnitude (a numerical value) along with appropriate units. It is invariant under coordinate transformation. [Per the definition of Scalar Quantities in Vector Algebra and Classical Mechanics].

Reasoning & Counter-Examples

The requirement for a quantity to be a scalar depends solely on its lack of directional dependence in space. Dimensionality is a separate property governed by the fundamental units (Mass, Length, Time, etc.) of the physical quantity.

Visual Representation of Scalar Properties

Logical Derivation

• Let $Q$ be a physical quantity.
• $Q$ is a scalar if $Q_{coord1} = Q_{coord2}$ under rotation or reflection. [By the Principle of Frame Invariance for Scalars].
• Let $dim(Q)$ be the dimensions of $Q$. If $dim(Q) = [M^a L^b T^c]$, then if $a, b, c \neq 0$, the quantity has dimensions.
• Mass, Energy, Temperature, and Density are all scalars, yet $dim(Mass) = [M]$, $dim(Density) = [ML^{-3}]$, and $dim(Temperature) = [\Theta]$.
• Since scalars can possess physical dimensions (e.g., $J/kg$, $kg$, $s$), the condition "must be dimensionless" is logically insufficient and empirically incorrect.

Final Solution: The statement is False. Scalar quantities are defined by the presence of magnitude and the absence of directional properties; they are not required to be dimensionless, as evidenced by mass, energy, and work, which are scalars possessing non-zero dimensional formulas.

Step 1: Identifying Given Parameters

To determine the average speed of the taxi, we identify the physical quantities provided in the problem statement:

• Total Distance Covered ($d$): $23\text{ km}$ [This is the actual path length traversed by the taxi].
• Total Time Taken ($\Delta t$): $28\text{ min}$.

Step 2: Unit Conversion for Consistency

To express the average speed in standard SI units (or commonly used $\text{km/h}$ for transport metrics), we convert the time duration from minutes to hours [By Dimensional Analysis].

$\Delta t = 28\text{ min} \times \frac{1\text{ hour}}{60\text{ min}} = \frac{28}{60}\text{ h} = \frac{7}{15}\text{ h}$

Step 3: Theoretical Definition of Average Speed

The average speed ($v_{avg}$) is defined as the ratio of the total path length traveled to the total time interval taken to cover that distance [Per the definition of Scalar Kinematics].

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

Step 4: Mathematical Calculation

Substituting the values into the formula:

$v_{avg} = \frac{23\text{ km}}{\frac{7}{15}\text{ h}}$

$v_{avg} = 23 \times \frac{15}{7}\text{ km/h}$

$v_{avg} = \frac{345}{7}\text{ km/h}$

$v_{avg} \approx 49.2857\text{ km/h}$

Visual Representation of Path Geometry

Final Solution: The average speed of the taxi is approximately $49.29\text{ km/h}$.

1. Theoretical Foundation: The Triangle Inequality

In vector algebra, any two vectors $\vec{a}$ and $\vec{b}$ form the sides of a triangle when placed tip-to-tail. The vector difference $\vec{d} = \vec{a} - \vec{b}$ represents the third side of the triangle. [Per the Triangle Inequality Theorem for vectors, the sum of two sides of a triangle is always greater than the third side].

2. Geometric Representation

3. Analytical Derivation

To prove $|\vec{a} - \vec{b}| \leq |\vec{a}| + |-\vec{b}|$, we use the properties of the magnitude of vectors:

1. Consider the magnitude of the difference vector: $|\vec{a} - \vec{b}|$.
2. By the definition of vector subtraction, this is equivalent to the magnitude of the resultant of $\vec{a}$ and $(-\vec{b})$.
3. [Per the Triangle Inequality]: The magnitude of the resultant of two vectors is always less than or equal to the sum of their individual magnitudes.
4. Since $|-\vec{b}| = |\vec{b}|$, we obtain:
$|\vec{a} + (-\vec{b})| \leq |\vec{a}| + |-\vec{b}|$ $|\vec{a} - \vec{b}| \leq |\vec{a}| + |\vec{b}|$

4. Conditions for Equality

The equality $|\vec{a} - \vec{b}| = |\vec{a}| + |\vec{b}|$ holds if and only if the vectors $\vec{a}$ and $-\vec{b}$ are collinear and point in the same direction.

• Since $-\vec{b}$ is in the opposite direction of $\vec{b}$, $\vec{a}$ and $-\vec{b}$ being in the same direction implies that $\vec{a}$ and $\vec{b}$ are in opposite directions.
• Mathematically, this occurs when the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is $180^\circ$.
• When $\vec{a}$ and $\vec{b}$ are anti-parallel, $|\vec{a} - \vec{b}| = |\vec{a}| - (-|\vec{b}|) = |\vec{a}| + |\vec{b}|$.

5. Summary Table of Vector Magnitudes

Final Solution: The inequality $|\vec{a} - \vec{b}| \leq |\vec{a}| + |\vec{b}|$ is a direct consequence of the triangle inequality. The equality sign applies if and only if the vectors $\vec{a}$ and $\vec{b}$ are anti-parallel, i.e., the angle between them is $180^\circ$.