Vector Physics and Relative Velocity

Vector Fundamentals and Graphical Representation

• Definition of a Vector:

  • A vector quantity is completely defined by two fundamental properties: its magnitude ($\text{مقداره}$) and its direction ($\text{اتجاهه}$).

  • Vectors are represented graphically using a directed line segment (an arrow).

• Graphical Rules for Vector Representation ($\text{التمثيل البياني للمتجهات}$):

  • The magnitude of the vector is directly proportional to the length of the arrow ($\text{مقدار المتجه يعبر عنه طول السهم}$).

  • The orientation and arrowhead define the vector's direction in space.

  • Scale Mapping Examples:

  • A speed vector $V = 3\,\text{m/s}$ pointing East can be represented graphically by a line of length $3\,\text{cm}$ using a scale of $1\,\text{cm} = 1\,\text{m/s}$.

  • A force vector $F = 5\,\text{N}$ pointing North ($5\,\text{نيوتن شمالاً}$) is represented by an arrow of proportional length pointing vertically upwards.

  • A velocity vector $V = 3\,\text{m/s}$ pointing East ($3\,\text{م/ث شرقاً}$) is drawn horizontally to the right.

• Vector Shift/Translation Property ($\text{ملاحظة نقل المتجه}$):

  • Any vector can be freely shifted or translated in space provided that its magnitude (length of the arrow) and its direction are strictly preserved.

Resultant Vectors and Vector Addition Methods

• Concept of the Resultant Vector ($\text{المحصلة}$):

  • The resultant vector ($\vec{M}_{res}$ or $\vec{V}_{res}$) represents the single combined vector that produces the same effect as two or more individual vectors acting together.

  • Vector addition formula: $\vec{M}_{res} = \vec{A} + \vec{B}$ or $\vec{V}_{res} = \vec{v}_A + \vec{v}_B$.

  • Head-to-Tail Method ($\text{طريقة الرأس والذيل}$):

  • To add vectors graphically, place the tail ($\text{الذيل}$) of the second vector at the head ($\text{الرأس}$) of the first vector.

  • The angle $\theta$ of the resultant with the horizontal axis is determined by $\tan(\theta) = \frac{v_B}{v_A}$.

• Rules for Calculating Resultant Force Magnitude ($\text{إيجاد مقدار محصلة القوى}$):

  • Vectors in the Same Direction (Parallel):

  • Perform simple algebraic addition: $f = f_1 + f_2$.

  • Vectors in Opposite Directions (Anti-parallel):

  • Perform simple algebraic subtraction: $f = f_1 - f_2$ (or $f = f_2 - f_1$).

  • Mutually Perpendicular Vectors (Orthogonal, $90^\circ$ angle):

  • Apply the Pythagorean theorem: $f = \sqrt{f_1^2 + f_2^2}$.

Vector Resolution and Component Analysis

• Concept of Vector Resolution ($\text{تحليل المتجهات}$):

  • Vector resolution is the inverse operation of finding the resultant vector. It decomposes a single vector into two perpendicular components along the horizontal ($x$) and vertical ($y$) axes.

• General Equations for Vector Components:

  • Horizontal Component ($\text{المركبة الأفقية}$ / $\text{المسقط الأفقي}$):

  • For force: $F_x = F \cos(\alpha)$

  • For velocity: $V_x = V \cos(\alpha)$

  • Vertical Component ($\text{المركبة الرأسية}$ / $\text{المسقط الرأسي}$):

  • For force: $F_y = F \sin(\alpha)$

  • For velocity: $V_y = V \sin(\alpha)$

• Concrete Examples and Case Studies:

  • Case Study 1: Calculating Displacement Magnitude and Direction

  • Given horizontal displacement $x = 3\,\text{m}$ East ($\text{شرقاً}$) and vertical displacement $y = 4\,\text{m}$ North ($\text{شمالاً}$).

  • Displacement magnitude ($d$):       $d = \sqrt{x^2 + y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\,\text{m}$

  • Direction angle $\theta$ relative to horizontal:       $\tan(\theta) = \frac{y}{x} = \frac{4}{3} \implies \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 36.89^\circ$

  • Complete displacement statement: $5\,\text{m}$ North of East making an angle of $36.89^\circ$ with the horizontal ($5\,\text{m \text{ شمال شرقه وتصنع زاوبة } 36.89 \text{ مع الأفقي}}$).

  • Case Study 2: Component Reconstruction from Hypotenuse

  • Given a hypotenuse $h = 5\,\text{m}$ at an angle $\theta = 53.13^\circ$ with the horizontal.

  • Horizontal side ($X$):       $\cos(53.13^\circ) = \frac{X}{5} \implies X = 5 \cos(53.13^\circ) = 3\,\text{m}$

  • Vertical side ($Y$):       $\sin(53.13^\circ) = \frac{Y}{5} \implies Y = 5 \sin(53.13^\circ) = 4\,\text{m}$

  • Angle verification:       $\tan(\theta) = \frac{4}{3} \implies \theta = 53.13010235^\circ$

  • Case Study 3: Velocity Components at $30^\circ$

  • Given velocity magnitude $V = 5\,\text{m/s}$ at an angle $\alpha = 30^\circ$ to the horizontal.

  • Horizontal velocity component ($V_x$):       $V_x = V \cos(30^\circ) = 5 \cos(30^\circ) = 2.5\sqrt{3}\,\text{m/s} \approx 4.33\,\text{m/s}$

  • Vertical velocity component ($V_y$):       $V_y = V \sin(30^\circ) = 5 \sin(30^\circ) = 2.5\,\text{m/s}$

  • Case Study 4: Vertical Component Determination via Tangent

  • Given opposite horizontal leg $5\,\text{m/s}$ and angle $30^\circ$:       $\tan(30^\circ) = \frac{5}{V_y} \implies 5 = V_y \tan(30^\circ) \implies V_y = \frac{5}{\tan(30^\circ)} = 5\sqrt{3}\,\text{m/s}$

Vector Subtraction and Directional Conventions

• Principles of Vector Subtraction:

  • True mathematical vector subtraction does not exist as a separate operation in vector physics; subtraction is defined as the addition of a vector pointing in the opposite direction.

  • The negative sign ($-$) strictly indicates an opposite directional sense relative to a chosen positive reference axis.

• Directional Sign Conventions:

  • East direction ($\text{الشرق}$) is defined as positive ($+$).

  • West direction ($\text{الغرب}$) is defined as negative ($---$).

• Illustrative Examples:

  • An object moving with speed $3\,\text{m/s}$ East is denoted as $V = +3\,\text{m/s}$.

  • An object moving with speed $5\,\text{m/s}$ West is denoted as $V = -5\,\text{m/s}$.

  • Combining a positive force/velocity of $+10$ with an opposing vector of $-2$:     $V_{res} = (+10) + (-2) = +8$     The positive result ($+8$) signifies a net vector directed East ($8\,\text{شرقاً}$).

Relative Velocity Calculations

• Definition of Relative Velocity ($\text{السرعة النسبية}$):

  • Relative velocity is the velocity of a body (e.g., $B$) as observed from the reference frame of another moving body (e.g., $A$).

• General Relative Velocity Formulas:

  • Velocity of $B$ relative to $A$:     $v_{AB} = v_B - v_A$

  • Velocity of $A$ relative to $B$:     $v_{BA} = v_A - v_B$

  • Oppositional Identity:     $v_{AB} = -v_{BA}$

• Quantitative Scenarios:

  • Scenario A: Objects Moving in the Same Direction (East)

  • Object $A$ moves at $v_A = +3\,\text{m/s}$ East ($3\,\text{شرقاً}$).

  • Object $B$ moves at $v_B = +4\,\text{m/s}$ East ($4\,\text{شرقاً}$).

  • Relative position difference from coordinates $A(1,1)$ and $B(4,5)$ gives vector $(4-1, 5-1) = (3,4)$.

  • Relative velocity of $B$ relative to $A$ ($v_{AB}$):       $v_{AB} = v_B - v_A = +4 - (+3) = +1\,\text{m/s}$

  • Relative velocity of $A$ relative to $B$ ($v_{BA}$):       $v_{BA} = v_A - v_B = +3 - (+4) = -1\,\text{m/s}$

  • Scenario B: Objects Moving in Opposite Directions

  • Object $A$ moves at $v_A = +4\,\text{m/s}$ East ($4\,\text{شرقاً}$).

  • Object $B$ moves at $v_B = -5\,\text{m/s}$ West ($5\,\text{غرباً}$).

  • Relative velocity of $B$ relative to $A$ ($v_{AB}$):       $v_{AB} = v_B - v_A = -5 - (+4) = -9\,\text{m/s}$

  • Relative velocity of $A$ relative to $B$ ($v_{BA}$):       $v_{BA} = v_A - v_B = +4 - (-5) = +9\,\text{m/s}$

River Crossing Dynamics and Boat Kinetics

• Kinematic System Setup ($\text{حركة القارب وضفة النهر}$):

  • A boat crosses a river of width $d = 60\,\text{m}$ at an angle of $60^\circ$ with the riverbank ($\text{ضفة النهر}$).

  • Parallel component / adjacent speed along bank = $3\,\text{m/s}$.

  • Total boat velocity along the angled trajectory = $6\,\text{m/s}$.

• Trigonometric Verification and Kinetic Solutions:

  • Verification of bank angle cosine:     $\cos(60^\circ) = \frac{3}{6} = \frac{1}{2}$

  • Perpendicular velocity component across the river ($V$):     $\sin(60^\circ) = \frac{V}{6} \implies V = 6 \sin(60^\circ) = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\,\text{m/s}$

  • Time required to cross the river ($t$):     Using the distance-speed-time relationship $d = v \times t$:     $t = \frac{d}{V} = \frac{60}{3\sqrt{3}} = \frac{20}{\sqrt{3}} \approx 11.55\,\text{sec}$