Comprehensive University Kinematics, Vector Mechanics, and Newtonian Dynamics Study Guide

Foundations of Kinematics and One-Dimensional Motion

• Definition and Scope of Kinematics:

  • Kinematics is the branch of classical mechanics dedicated to describing the motion of points, bodies, and systems of bodies without consideration of the forces that cause the motion.

  • Fully characterizing one-dimensional and multi-dimensional motion requires tracking primary physical quantities: distance, displacement, speed, velocity, acceleration, time duration, average speed, and average velocity.

• Distance versus Displacement:

  • Distance ($x$): A scalar quantity representing the total length of the continuous path traversed by an object between its initial and final positions. It depends entirely on the trajectory taken and is strictly non-negative ($x \ge 0$).

  • Displacement ($\vec{\Delta x}$): A vector quantity defined as the straight-line directed segment extending from an object's initial position ($\vec{x}_0$) to its final position ($\vec{x}$):     $\vec{\Delta x} = \vec{x} - \vec{x}_0$

  • The magnitude of displacement ($|\vec{\Delta x}|$) represents the shortest spatial separation between endpoints. Unlike distance, displacement incorporates directional orientation ($+$, $-$ in one dimension; angles in multiple dimensions).

  • Comparative Path Scenario: An object navigating an irregular step trajectory from an initial origin to a terminal point may cover an approximate path length of $x \approx 140\,\text{m}$, while its direct vector displacement magnitude evaluates to $|\vec{\Delta x}| \approx 90\,\text{m}$.

• Zero Displacement with Non-Zero Distance Traversed:

  • Displacement evaluates strictly to zero ($\vec{\Delta x} = 0$) whenever an object returns exactly to its initial starting coordinates, regardless of the cumulative distance traveled:

  • Rebound Motion: A ball propelled toward a wall located $40\,\text{cm}$ away that rebounds to the hand covers a total distance of $x = 80\,\text{cm}$, yet possesses a net displacement of $\vec{\Delta x} = 0\,\text{m}$.

  • Closed Circular Orbit: An object executing a closed loop of circumference $x = 1.0\,\text{m}$ finishes at its starting point, resulting in $\vec{\Delta x} = 0\,\text{m}$.

  • Oscillatory Path: A skateboarder rolling down and up a symmetrical half-pipe track returning to the release point yields $\vec{\Delta x} = 0\,\text{m}$.

• Scale Diagrams:

  • A scale diagram reduces or enlarges all physical lengths uniformly by an invariant geometric ratio or conversion factor (e.g., $1.9\,\text{cm} = 20.0\,\text{m}$ or $3.2\,\text{cm} = 20.0\,\text{m}$).

  • Standard applications include geographic maps, structural blueprints, architectural building drafts, virtual video game engines, and industrial assembly schematics.

• International System of Units (SI) Conventions:

  • Base Quantities and Base Units:

  • Length: meter ($\text{m}$

  • Mass: kilogram ($\text{kg}$

  • Time: second ($\text{s}$

  • Electric current: ampere ($\text{A}$

  • Thermodynamic temperature: kelvin ($\text{K}$

  • Amount of substance: mole ($\text{mol}$

  • Luminous intensity: candela ($\text{cd}$

  • Selected SI Derived Units:

  • Frequency: hertz ($\text{Hz} = \text{s}^{-1}$

  • Force: newton ($\text{N} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}$

  • Pressure: pascal ($\text{Pa} = \text{N/m}^2$

  • Energy: joule ($\text{J} = \text{N}\cdot\text{m}$

  • Power: watt ($\text{W} = \text{J/s}$

  • Electric Charge: coulomb ($\text{C} = \text{A}\cdot\text{s}$

  • Electric Potential: volt ($\text{V} = \text{W/A}$

  • Electric Resistance: ohm ($\Omega = \text{V/A}$

  • Celsius Temperature: degree Celsius ($^{\circ}\text{C}$, equivalent in interval magnitude to $\text{K}$

  • SI Metric Prefixes:

  • Tera ($\text{T}$): $10^{12}$

  • Giga ($\text{G}$): $10^9$

  • Mega ($\text{M}$): $10^6$

  • Kilo ($\text{k}$): $10^3$

  • Hecto ($\text{h}$): $10^2$

  • Deka ($\text{da}$): $10^1$

  • Deci ($\text{d}$): $10^{-1}$

  • Centi ($\text{c}$): $10^{-2}$

  • Milli ($\text{m}$): $10^{-3}$

  • Micro ($\mu$): $10^{-6}$

  • Nano ($\text{n}$): $10^{-9}$

  • Pico ($\text{p}$): $10^{-12}$

• Unit Conversion Mechanics ($\text{km/h}$ to $\text{m/s}$):

  • Standard road speeds expressed in kilometers per hour ($\text{km/h}$) require conversion to SI base units ($\text{m/s}$) using chain dimensional analysis:     $60\,\frac{\text{km}}{\text{h}} \times \left(\frac{1 \times 10^3\,\text{m}}{1\,\text{km}}\right) \times \left(\frac{1\,\text{h}}{3600\,\text{s}}\right) = \frac{60 \times 1000\,\text{m}}{3600\,\text{s}} = 16.7\,\text{m/s}$

  • Operating rule: Converting $\text{km/h}$ to $\text{m/s}$ requires dividing by the numerical factor $3.6$. The equivalent magnitude in $\text{m/s}$ is always smaller than in $\text{km/h}$.

• Speed, Velocity, and Average Rates:

  • Instantaneous Speed ($v$): Scalar rate of distance covered per unit time interval ($v = \frac{x}{t}$).

  • Instantaneous Velocity ($\vec{v}$): Vector rate of change of displacement evaluated at a specific temporal instant ($t \to 0$), possessing both magnitude and spatial direction (e.g., $16.7\,\text{m/s}$, North).

  • Average Speed ($v_{\text{ave}}$) and Average Velocity ($\vec{v}_{\text{ave}}$):     v_{\text{ave}} = \frac{x_{\text{total}}}{t_{\text{total}}}}     \vec{v}_{\text{ave}} = \frac{\vec{\Delta x}}{t_{\text{total}}}}

  • Orbital Case Study (Lunar Orbit):

  • Orbital distance traveled: $x = 2.42 \times 10^9\,\text{m}$.

  • Orbital time period: $t = 28\,\text{days}$.

  • Time unit conversion: $28\,\text{days} \times \left(\frac{24\,\text{h}}{1\,\text{day}}\right) \times \left(\frac{3600\,\text{s}}{1\,\text{h}}\right) = 2.42 \times 10^6\,\text{s}$.

  • Average orbital speed:       $v_{\text{ave}} = \frac{2.42 \times 10^9\,\text{m}}{2.42 \times 10^6\,\text{s}} = 1000\,\text{m/s}$

  • Average orbital velocity (for a complete closed orbital revolution where $\vec{\Delta x} = 0\,\text{m}$):       $\vec{v}_{\text{ave}} = \frac{0\,\text{m}}{2.42 \times 10^6\,\text{s}} = 0\,\text{m/s}$

  • Mean orbital radius parameter: $r = 0.384\,\text{Gm} = 3.84 \times 10^8\,\text{m}$.

• Acceleration:

  • Vector measure of the rate of change of velocity over time:     $\vec{a} = \frac{\vec{v} - \vec{v}_0}{t}$

  • Acceleration is zero ($\vec{a} = 0\,\text{m/s}^2$) during constant velocity conditions, which represents a special case of constant acceleration.

• 
  

  • $t = 0$ to $10\,\text{s}$: $\vec{a} = +2.0\,\text{m/s}^2$

  • $t = 10$ to $20\,\text{s}$: $\vec{a} = 0\,\text{m/s}^2$

  • $t = 20$ to $24\,\text{s}$: $\vec{a} = -5.0\,\text{m/s}^2$

  • $t = 24$ to $30\,\text{s}$: $\vec{a} = -0.5\,\text{m/s}^2$

• Tracking boundary parameters yields displacement increments of approximately $+100\,\text{m}$ over the first interval, an additional $+200\,\text{m}$ linear rise over the second interval reaching $300\,\text{m}$, a concave flattening peak adding $+40\,\text{m}$ to $340\,\text{m}$, and a slight rollback of $-9\,\text{m}$, terminating at an exact net coordinate of $331\,\text{m}$, North at $t = 30\,\text{s}$.

Kinematic Formulations and Problem-Solving Methodology

• The Four Fundamental Constant Acceleration Equations:

  1. $\mathbf{v} = \mathbf{v}_0 + \mathbf{a}t$

  2. $v^2 = v_0^2 + 2a\Delta x$

  3. $\Delta \mathbf{x} = \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2$

  4. $\Delta \mathbf{x} = \left(\frac{\mathbf{v} + \mathbf{v}_0}{2}\right)t$

• Kinematic Variables Dictionary:

  • $\Delta \mathbf{x}$: Displacement ($\text{m}$)

  • $\mathbf{v}_0$: Initial velocity ($\text{m/s}$)

  • $\mathbf{v}$: Final velocity ($\text{m/s}$)

  • $t$: Time interval duration ($\text{s}$)

  • $\mathbf{a}$: Constant acceleration ($\text{m/s}^2$)

• Rigorous Six-Step Problem-Solving Protocol:

  1. Explicitly enumerate all five kinematic variables ($\Delta \mathbf{x}$, $\mathbf{v}_0$, $\mathbf{v}$, $t$, $\mathbf{a}$).

  2. Establish coordinate orientation and directionality; define which direction corresponds to positive (+$) and negative (-).\n 3. Extract all given numerical quantities and assign appropriate signs consistent with established coordinates.\n 4. Identify the unstated and unrequested variable ("don't know, don't care") and eliminate it from consideration.\n 5. Choose the single constant acceleration formula containing the remaining four variables.\n 6. Substitute numerical values and solve algebraically for the targeted unknown.\n\n- **Comprehensive Bungee Jumping Worked Analysis:**\n - **Phase 1: Free Fall Prior to Cord Tension:**\n - An individual steps off a high platform (\mathbf{v}_0 = 0\,\text{m/s}$) with an unstretched cord length of$l = 15\,\text{m}.\n - Coordinate definition: Upward (+$), Downward ($-$).

  • Parameter setup: $\mathbf{v}_0 = 0\,\text{m/s}$, $\mathbf{a} = -9.8\,\text{m/s}^2$, $\Delta \mathbf{x} = -15\,\text{m}$, $t = \text{"don't know, don't care"}$, $\mathbf{v} = ?$

  • Mathematical solution:       $v^2 = v_0^2 + 2a\Delta x$$v^2 = (0\,\text{m/s})^2 + 2(-9.8\,\text{m/s}^2)(-15\,\text{m}) = +294\,\text{m}^2/\text{s}^2$$v = \pm \sqrt{294\,\text{m}^2/\text{s}^2} = -17.1\,\text{m/s}$

  • Final velocity statement (retaining two significant figures): $\mathbf{v} = 17\,\text{m/s}$, downward.

  • Phase 2: Elastic Deceleration to Complete Rest:

  • The elastic cord stretches until maximum displacement is attained at an overall vertical span of $l_{\text{total}} = 48\,\text{m}$.

  • Deceleration stretch interval: $\Delta \mathbf{x} = -(48\,\text{m} - 15\,\text{m}) = -33\,\text{m}$.

  • Boundary velocities: Initial downward entry velocity $\mathbf{v}_0 = -17.15\,\text{m/s}$; final arrest velocity $\mathbf{v} = 0\,\text{m/s}$.

  • Equation manipulation:       $v^2 = v_0^2 + 2a\Delta x \implies a = \frac{v^2 - v_0^2}{2\Delta x}$$a = \frac{(0\,\text{m/s})^2 - (-17.15\,\text{m/s})^2}{2(-33\,\text{m})} = \frac{-294.1\,\text{m}^2/\text{s}^2}{-66\,\text{m}} = +4.456\,\text{m/s}^2$

  • Expressed to significant digits: $\mathbf{a} = 4.5\,\text{m/s}^2$, upward.

• Inclined Plane Cart Dynamics (Round-Trip Kinematics):

  • A cart is launched up an inclined runway with initial velocity $\mathbf{v}_0 = +1.0\,\text{m/s}$, taking a total round-trip time of $t = 6.0\,\text{s}$ to return to the release point.

  • Boundary Parameters: Since the ramp is symmetric, return velocity is $\mathbf{v} = -1.0\,\text{m/s}$ and net displacement over the full interval is $\Delta \mathbf{x} = 0\,\text{m}$.

  • Acceleration Evaluation:$\mathbf{v} = \mathbf{v}_0 + \mathbf{a}t$$-1.0\,\text{m/s} = 1.0\,\text{m/s} + \mathbf{a}(6.0\,\text{s}) \implies -2.0\,\text{m/s} = \mathbf{a}(6.0\,\text{s})$$\mathbf{a} = -0.33\,\text{m/s}^2 = 0.33\,\text{m/s}^2\text{, down the ramp}$

  • Physical sanity check: Acceleration along an incline without external thrust must satisfy $0 < \|\mathbf{a}\| < 9.8\,\text{m/s}^2$.

  • One-Way Upslope Displacement (Half-Trip Analysis):

  • At maximum upward progression ($t = 3.0\,\text{s}$), instantaneous velocity drops to $\mathbf{v} = 0\,\text{m/s}$.       $\Delta \mathbf{x} = \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2 = (1.0\,\text{m/s})(3.0\,\text{s}) + \frac{1}{2}(-0.33\,\text{m/s}^2)(3.0\,\text{s})^2$$\Delta \mathbf{x} = 3.0\,\text{m} - 1.5\,\text{m} = +1.5\,\text{m} = 1.5\,\text{m}\text{, up the track}$

  • Cumulative Distance vs. Displacement at $t = 6.0\,\text{s}$:

  • Total distance traversed: $x = 2 \times (1.5\,\text{m}) = 3.0\,\text{m}$

  • Net displacement: $\Delta \mathbf{x} = 0\,\text{m}$

Kinematic Graphical Analysis and Calculus Relationships

• Interdependence of Kinematic Curves:

  • Slope Relationships:

  • The instantaneous slope of a position-time ($x-t$) graph defines the instantaneous velocity:       $\mathbf{v} = \frac{d\mathbf{x}}{dt}$

  • The instantaneous slope of a velocity-time ($v-t$) graph defines the instantaneous acceleration:       $\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{\Delta \mathbf{v}}{\Delta t}$

  • Curvature and Polynomial Evolution:

  • An object moving at constant velocity displays a linear $x-t$ plot, a horizontal $v-t$ line ($\text{slope} = 0$), and a zero-magnitude acceleration axis line ($\mathbf{a} = 0\,\text{m/s}^2$).

  • An object undergoing constant non-zero acceleration displays a parabolic $x-t$ curve governed by $\Delta \mathbf{x} = \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2$ (quadratic form $y = ax^2$), a linearly sloped $v-t$ profile, and a horizontal non-zero $a-t$ profile.

• Area Geometric Integration on Velocity-Time Graphs:

  • The definite area bounded between a velocity-time curve and the horizontal time axis over any specified time interval ($[t_1, t_2]$) equals the net displacement ($\Delta \mathbf{x}$) during that window:     $\Delta \mathbf{x} = \int_{t_1}^{t_2} \mathbf{v}(t)\,dt$

• Three-Segment Profile Quantitative Breakdown:

  • Segment A ($t = 0\,\text{s}$ to $t = 5.0\,\text{s}$): Object undergoes uniform positive acceleration starting from rest ($\mathbf{v}_0 = 0\,\text{m/s}$) up to $\mathbf{v} = 10.0\,\text{m/s}$. Evaluated as triangular area:     $\Delta \mathbf{x}_A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(5.0\,\text{s})(10.0\,\text{m/s}) = +25.0\,\text{m}$

  • Acceleration during segment A:       $\mathbf{a}_A = \frac{10.0\,\text{m/s} - 0\,\text{m/s}}{5.0\,\text{s}} = +2.0\,\text{m/s}^2$

  • Segment B ($t = 5.0\,\text{s}$ to $t = 15.0\,\text{s}$): Object coasts at a constant velocity of $\mathbf{v} = 10.0\,\text{m/s}$. Evaluated as rectangular area:     $\Delta \mathbf{x}_B = \text{base} \times \text{height} = (15.0\,\text{s} - 5.0\,\text{s})(10.0\,\text{m/s}) = (10.0\,\text{s})(10.0\,\text{m/s}) = +100\,\text{m}$

  • Acceleration during segment B:       $\mathbf{a}_B = 0\,\text{m/s}^2$

  • Segment C ($t = 15.0\,\text{s}$ to $t = 25.0\,\text{s}$): Object uniformly decelerates from $\mathbf{v} = 10.0\,\text{m/s}$ down to $\mathbf{v} = 5.0\,\text{m/s}$. Evaluated as trapezoidal area:     $\Delta \mathbf{x}_C = \left(\frac{10.0\,\text{m/s} + 5.0\,\text{m/s}}{2}\right)(25.0\,\text{s} - 15.0\,\text{s}) = (7.5\,\text{m/s})(10.0\,\text{s}) = +75.0\,\text{m}$

  • Acceleration during segment C:       $\mathbf{a}_C = \frac{5.0\,\text{m/s} - 10.0\,\text{m/s}}{10.0\,\text{s}} = -0.50\,\text{m/s}^2$

  • Cumulative Displacement: Summing all constituent regions across the full $25.0\,\text{s}$ domain:     $\Delta \mathbf{x}_{\text{total}} = \Delta \mathbf{x}_A + \Delta \mathbf{x}_B + \Delta \mathbf{x}_C = 25.0\,\text{m} + 100\,\text{m} + 75.0\,\text{m} = +200\,\text{m}$

• Multi-Stage Acceleration Profile Integration:

  • Consider an acceleration piecewise profile:

  • $t = 0\,\text{s}$ to $10\,\text{s}$: $\mathbf{a} = +2.0\,\text{m/s}^2$

  • $t = 10\,\text{s}$ to $20\,\text{s}$: $\mathbf{a} = 0\,\text{m/s}^2$

  • $t = 20\,\text{s}$ to $24\,\text{s}$: $\mathbf{a} = -5.0\,\text{m/s}^2$

  • $t = 24\,\text{s}$ to $30\,\text{s}$: $\mathbf{a} = -0.50\,\text{m/s}^2$

  • Tracking boundary parameters yields displacement increments of approximately $+100\,\text{m}$ over the first interval, an additional $+200\,\text{m}$ linear rise over the second interval reaching $300\,\text{m}$, a concave flattening peak adding $+40\,\text{m}$ to $340\,\text{m}$, and a slight rollback of $-9\,\text{m}$, terminating at an exact net coordinate of $331\,\text{m}$, North at $t = 30\,\text{s}$.

Free Fall and Vertical Gravitational Kinematics

• Universal Principles of Free Fall:

  • Free fall describes the idealized unconstrained vertical motion of an object subjected exclusively to gravitational attraction, setting aerodynamic drag and air resistance to zero.

  • Near Earth's surface, gravitational acceleration acts strictly downward toward Earth's center with a constant magnitude:     $g_{\text{Earth}} = 9.8\,\text{m/s}^2$

  • Lunar gravitational comparison:     $g_{\text{Moon}} = 1.6\,\text{m/s}^2$

  • While velocity changes continuously by $-9.8\,\text{m/s}$ each second, the acceleration vector remains completely invariant throughout both the ascent and descent phases.

• Quadratic Distance Accumulation ($\Delta x \propto t^2$):

  • For an object dropped from rest ($\mathbf{v}_0 = 0\,\text{m/s}$), vertical fall distance is directly governed by:     $\Delta \mathbf{x} = \frac{1}{2}\mathbf{a}t^2 = \frac{1}{2}(-9.8\,\text{m/s}^2)t^2 = -(4.9\,\text{m/s}^2)t^2$

  • When fall duration doubles ($\times 2$), accumulated vertical displacement quadruples ($\times 4$); when duration triples ($\times 3$), displacement increases ninefold ($\times 9$).

  • Discrete Second-by-Second Fall Table:

  • At $t = 0\,\text{s}$: $\mathbf{x} = 0\,\text{m}$

  • At $t = 1.0\,\text{s}$: $\mathbf{x} = -4.9\,\text{m}$

  • At $t = 2.0\,\text{s}$: $\mathbf{x} = -19.6\,\text{m}$

  • At $t = 3.0\,\text{s}$: $\mathbf{x} = -44.1\,\text{m}$

  • At $t = 4.0\,\text{s}$: $\mathbf{x} = -78.5\,\text{m}$

  • At $t = 5.0\,\text{s}$: $\mathbf{x} = -122.6\,\text{m}$

  • At $t = 6.0\,\text{s}$: $\mathbf{x} = -176\,\text{m}$

  • At extended duration $t = 10.0\,\text{s}$:       $\Delta \mathbf{x} = \frac{1}{2}(-9.8\,\text{m/s}^2)(10.0\,\text{s})^2 = -490\,\text{m}$

• Comparative Directional Launch Dynamics:

  • Upward Projection Case:

  • A projectile is launched vertically upward from an elevation with $\mathbf{v}_0 = +4.95\,\text{m/s}$. The velocity upon falling to $\Delta \mathbf{x} = -78.5\,\text{m}$ below the starting height is determined by:       $v^2 = v_0^2 + 2a\Delta x = (4.95\,\text{m/s})^2 + 2(-9.8\,\text{m/s}^2)(-78.5\,\text{m})$$v^2 = 24.5\,\text{m}^2/\text{s}^2 + 1538.6\,\text{m}^2/\text{s}^2 = 1563\,\text{m}^2/\text{s}^2$$v = -\sqrt{1563\,\text{m}^2/\text{s}^2} = -39.5\,\text{m/s} = 39.5\,\text{m/s}\text{, downward}$

  • Downward Projection Case:

  • A projectile is propelled directly downward with initial velocity $\mathbf{v}_0 = -4.95\,\text{m/s}$ through the same displacement $\Delta \mathbf{x} = -78.5\,\text{m}$:       $v^2 = (-4.95\,\text{m/s})^2 + 2(-9.8\,\text{m/s}^2)(-78.5\,\text{m}) = 1563\,\text{m}^2/\text{s}^2 \implies v = 39.5\,\text{m/s}\text{, downward}$

  • Physical Distinction: While final velocities at displacement $\Delta \mathbf{x} = -78.5\,\text{m}$ are strictly identical in magnitude and direction due to path symmetry, the downward-thrown projectile reaches that position in significantly less elapsed time.

Vector Operations in Two Dimensions

• Properties of Vectors in Two Dimensions:

  • A vector is fully specified by its magnitude ($R$) and its direction. In polar notation, direction is defined by the polar angle ($\theta$), measured counterclockwise (CCW) from the positive horizontal axis ($+x$-axis).

  • Polar Angle Conversions:

  • $45.0^{\circ}$ North of East corresponds to polar angle $\theta = 45.0^{\circ}$.

  • $60.0^{\circ}$ South of East corresponds to polar angle $\theta = 360.0^{\circ} - 60.0^{\circ} = 300.0^{\circ}$.

  • An angle measured $22^{\circ}$ below the positive $x$-axis corresponds to polar angle $\theta = 360^{\circ} - 22^{\circ} = 338^{\circ}$.

• Graphical Vector Addition (Tip-to-Tail Rule):

  • Steps for graphical construction:

  1. Establish an unambiguous scaling ratio (e.g., $3.2\,\text{cm} = 20.0\,\text{m}$).

  2. Fix an origin and draw the first vector $\mathbf{d}_1$ accurately using a protractor and straightedge.

  3. Position the tail of the second vector $\mathbf{d}_2$ directly at the terminal arrow tip of $\mathbf{d}_1$.

  4. Repeat sequentially for all constituent vectors.

  5. Construct the resultant vector $\mathbf{d} = \sum \mathbf{d}_i$ by connecting a directed vector originating at the tail of the first vector and terminating at the tip of the final vector.

  • Measurement Demonstration: Adding $\mathbf{d}_1 = 20.0\,\text{m}$ at $45.0^{\circ}$ to $\mathbf{d}_2 = 30.0\,\text{m}$ at $300.0^{\circ}$ yields a physical drawing length of $5.1\,\text{cm}$. Converting gives:     $d = 5.1\,\text{cm} \times \left(\frac{20.0\,\text{m}}{3.2\,\text{cm}}\right) = 31.9\,\text{m}$

  • Protractor measurement reveals an orientation $22^{\circ}$ below the positive horizontal axis, giving $\mathbf{d} = 31.9\,\text{m}$ at $338^{\circ}$.

  • Note: The scalar sum ($20.0\,\text{m} + 30.0\,\text{m} = 50.0\,\text{m}$) does not equal the vector resultant magnitude.

• Analytical Vector Resolution into Orthogonal Components:

  • Any two-dimensional vector can be resolved into independent perpendicular projections along the Cartesian axes:     $d_x = d\cos(\theta)$$d_y = d\sin(\theta)$

  • Decomposition Examples:

  • Force vector $\mathbf{F} = 800\,\text{N}$ directed at $10.0^{\circ}$ below the horizontal ($\theta = 350.0^{\circ}$):       $F_x = (800\,\text{N})\cos(350.0^{\circ}) = +788\,\text{N}$$F_y = (800\,\text{N})\sin(350.0^{\circ}) = -139\,\text{N}$

  • Velocity vector $\mathbf{v}_0 = 20.0\,\text{m/s}$ directed at $70.0^{\circ}$ above the horizontal ($\theta = 70.0^{\circ}$):       $v_{0x} = (20.0\,\text{m/s})\cos(70.0^{\circ}) = +6.84\,\text{m/s}$$v_{0y} = (20.0\,\text{m/s})\sin(70.0^{\circ}) = +18.8\,\text{m/s}$

  • Displacement vector $\mathbf{d}_3 = 15.0\,\text{m}$ at polar angle $\theta = 210.0^{\circ}$ (Quadrant III):       $d_{3x} = (15.0\,\text{m})\cos(210.0^{\circ}) = (15.0\,\text{m})(-0.8660) = -12.99\,\text{m} \approx -13.0\,\text{m}$$d_{3y} = (15.0\,\text{m})\sin(210.0^{\circ}) = (15.0\,\text{m})(-0.5000) = -7.50\,\text{m}$

• Analytical Addition via Component Summation Method:

  • Summing $\mathbf{d}_1 = 20.0\,\text{m}$ at $45.0^{\circ}$ and $\mathbf{d}_2 = 30.0\,\text{m}$ at $300.0^{\circ}$:

  • Vector 1 components:       $d_{1x} = (20.0\,\text{m})\cos(45.0^{\circ}) = +14.14\,\text{m}$$d_{1y} = (20.0\,\text{m})\sin(45.0^{\circ}) = +14.14\,\text{m}$

  • Vector 2 components:       $d_{2x} = (30.0\,\text{m})\cos(300.0^{\circ}) = +15.00\,\text{m}$$d_{2y} = (30.0\,\text{m})\sin(300.0^{\circ}) = -25.98\,\text{m}$

  • Resultant orthogonal sums:       $d_x = d_{1x} + d_{2x} = 14.14\,\text{m} + 15.00\,\text{m} = +29.14\,\text{m}$$d_y = d_{1y} + d_{2y} = 14.14\,\text{m} + (-25.98\,\text{m}) = -11.84\,\text{m} \approx -11.9\,\text{m}$

  • Resultant magnitude calculation (Pythagorean Theorem):       $d = \sqrt{d_x^2 + d_y^2} = \sqrt{(29.14\,\text{m})^2 + (-11.84\,\text{m})^2} = \sqrt{849.1 + 140.2} = \sqrt{989.3} = 31.4\,\text{m}$

  • Reference angle calculation:       $\tan(\alpha) = \frac{|d_y|}{|d_x|} = \frac{11.84\,\text{m}}{29.14\,\text{m}} = 0.4063 \implies \alpha = \arctan(0.4063) = 22.1^{\circ}$

  • Since $d_x > 0$ and $d_y < 0$, the vector resides in Quadrant IV:       $\theta = 360^{\circ} - 22.1^{\circ} = 338^{\circ}$$\mathbf{d} = 31.4\,\text{m} \text{ at } 338^{\circ}$

• Analytical Addition via Trigonometric Laws:

  • When adding two vectors tip-to-tail, they form an oblique triangle where the interior angle between $\mathbf{d}_1$ and $\mathbf{d}_2$ is $C = 75.0^{\circ}$.

  • Law of Cosines evaluation:     $c^2 = a^2 + b^2 - 2ab\cos(C)$$d^2 = (20.0)^2 + (30.0)^2 - 2(20.0)(30.0)\cos(75.0^{\circ}) = 400.0 + 900.0 - 1200(0.2588) = 1300.0 - 310.58 = 989.42$$d = \sqrt{989.42} = 31.4\,\text{m}$

  • Law of Sines angle determination:     $\frac{\sin(B)}{b} = \frac{\sin(C)}{c} \implies \frac{\sin(B)}{30.0} = \frac{\sin(75.0^{\circ})}{31.4}$$\sin(B) = \frac{30.0 \times \sin(75.0^{\circ})}{31.4} = \frac{30.0(0.9659)}{31.4} = 0.9228 \implies B = 67.3^{\circ}$

  • Angle with respect to horizontal axis: $\alpha = 67.3^{\circ} - 45.0^{\circ} = 22.3^{\circ}$, confirming $\theta = 338^{\circ}$.

• Addition of Three Non-Collinear Vectors:

  • System vectors: $\mathbf{d}_1 = 20.0\,\text{m}$ at $45.0^{\circ}$, $\mathbf{d}_2 = 30.0\,\text{m}$ at $300.0^{\circ}$, $\mathbf{d}_3 = 15.0\,\text{m}$ at $210.0^{\circ}$.

  • Horizontal sum:     $d_x = d_{1x} + d_{2x} + d_{3x} = 14.14\,\text{m} + 15.00\,\text{m} + (-13.00\,\text{m}) = +16.14\,\text{m}$

  • Vertical sum:     $d_y = d_{1y} + d_{2y} + d_{3y} = 14.14\,\text{m} + (-25.98\,\text{m}) + (-7.50\,\text{m}) = -19.34\,\text{m}$

  • Magnitude:     $\|\mathbf{d}\| = \sqrt{(16.14\,\text{m})^2 + (-19.34\,\text{m})^2} = \sqrt{260.5 + 374.0} = \sqrt{634.5} = 25.2\,\text{m}$

  • Direction:     $\alpha = \arctan\left(\frac{19.34}{16.14}\right) = \arctan(1.198) = 50.2^{\circ} \implies \theta = 360^{\circ} - 50.2^{\circ} = 310^{\circ}$$\mathbf{d} = 25.2\,\text{m} \text{ at } 310^{\circ}$

• Zero Resultant Closed Polygons:

  • When a sequence of vectors forms a closed geometric loop returning to the origin, their net vector sum is identically zero:     $\mathbf{d}_2 + \mathbf{d}_3 + \mathbf{d}_4 + \mathbf{d}_5 = 0$

  • Consequently, the closing vector $\mathbf{d}_5$ must be the exact negative equilibrant of the sum of the preceding vectors:     $\mathbf{d}_5 = -(\mathbf{d}_2 + \mathbf{d}_3 + \mathbf{d}_4)$

Two-Dimensional Projectile Kinematics

• Orthogonal Independence of Projectile Motion:

  • Two-dimensional projectile motion occurs when an object is launched with an initial velocity vector oriented at any angle other than $90^{\circ}$ relative to the horizontal.

  • Under idealized conditions (zero atmospheric drag), horizontal and vertical kinematic components are entirely decoupled and independent:

  • Horizontal Axis ($x$): Experiences zero net acceleration ($a_x = 0\,\text{m/s}^2$). Horizontal velocity remains strictly invariant throughout flight ($v_x = v_{0x} = v_0 \cos(\theta)$). Horizontal position is given by $\Delta x = v_{0x}t$.

  • Vertical Axis ($y$): Subject to constant gravitational acceleration ($a_y = -9.8\,\text{m/s}^2$). Vertical velocity varies continuously ($v_y = v_{0y} + a_y t = v_0 \sin(\theta) - gt$). Vertical displacement is governed by $\Delta y = v_{0y}t - \frac{1}{2}gt^2$.

  • Time of flight ($t$) is the singular scalar variable linking both axes.

• Level-Ground Range Derivation and Angular Optimization:

  • For launch and landing on level ground ($\Delta y = 0\,\text{m}$), trajectory symmetry establishes that the landing vertical velocity equals the negative launch vertical velocity: $v_y = -v_{0y}$.

  • Flight duration derivation:     $v_y = v_{0y} + a_y t \implies -v_0 \sin(\theta) = v_0 \sin(\theta) - gt$$-2v_0 \sin(\theta) = -gt \implies t = \frac{2v_0 \sin(\theta)}{g}$

  • Substitution into horizontal displacement:     $\Delta x = (v_0 \cos(\theta))t = (v_0 \cos(\theta))\left(\frac{2v_0 \sin(\theta)}{g}\right) = \frac{v_0^2 (2\sin(\theta)\cos(\theta))}{g}$

  • Applying the double-angle identity $2\sin(\theta)\cos(\theta) = \sin(2\theta)$ yields the level-ground range formula:     $\Delta x = \frac{v_0^2 \sin(2\theta)}{g}$

  • Range optimization analysis:

  • Maximum range occurs where $\sin(2\theta)$ reaches its mathematical ceiling of $1.0$, corresponding to $2\theta = 90^{\circ} \implies \theta = 45^{\circ}$.

  • Complementary launch angles ($\theta$ and $90^{\circ} - \theta$, such as $15^{\circ}$ and $75^{\circ}$) produce identical horizontal ranges because $\sin(2(90^{\circ} - \theta)) = \sin(180^{\circ} - 2\theta) = \sin(2\theta)$.

• Long Jumper Trajectory and World Record Velocity:

  • An athlete leaps with an initial velocity of $v_0 = 9.50\,\text{m/s}$ at an inclination of $\theta = 40.0^{\circ}$ over level ground.

  • Velocity components:     $v_{0x} = (9.50\,\text{m/s})\cos(40.0^{\circ}) = 7.277\,\text{m/s}$$v_{0y} = (9.50\,\text{m/s})\sin(40.0^{\circ}) = 6.106\,\text{m/s}$

  • Total time aloft:     $t = \frac{2v_{0y}}{g} = \frac{2(6.106\,\text{m/s})}{9.8\,\text{m/s}^2} = 1.246\,\text{s} \approx 1.25\,\text{s}$

  • Peak vertical elevation (at apex, $v_y = 0\,\text{m/s}$):     $v_y^2 = v_{0y}^2 + 2a_y \Delta y_{\text{max}} \implies 0 = (6.106\,\text{m/s})^2 + 2(-9.8\,\text{m/s}^2)\Delta y_{\text{max}}$$\Delta y_{\text{max}} = \frac{37.28\,\text{m}^2/\text{s}^2}{19.6\,\text{m/s}^2} = 1.90\,\text{m}$

  • Horizontal leap distance (Range):     $\Delta x = v_{0x}t = (7.277\,\text{m/s})(1.246\,\text{s}) = 9.07\,\text{m}$

• Women's World Record Parameter Inversion:

  • Target world record distance: $\Delta x = 7.52\,\text{m}$ achieved at $\theta = 40.0^{\circ}$.

  • From horizontal axis: $7.52 = (v_0 \cos(40.0^{\circ}))t \implies t = \frac{7.52}{v_0 \cos(40.0^{\circ})}$.

  • From vertical axis: $-v_0 \sin(40.0^{\circ}) = v_0 \sin(40.0^{\circ}) - 9.8t \implies 2v_0 \sin(40.0^{\circ}) = 9.8t$.

  • Eliminating $t$ by algebraic substitution:     $2v_0 \sin(40.0^{\circ}) = 9.8\left(\frac{7.52}{v_0 \cos(40.0^{\circ})}\right)$$2v_0^2 \sin(40.0^{\circ})\cos(40.0^{\circ}) = (9.8)(7.52) = 73.696$$v_0^2 \sin(80.0^{\circ}) = 73.696 \implies v_0^2 = \frac{73.696}{\sin(80.0^{\circ})} = \frac{73.696}{0.9848} = 74.83\,\text{m}^2/\text{s}^2$$v_0 = \sqrt{74.83\,\text{m}^2/\text{s}^2} = 8.65\,\text{m/s}$

• Non-Level Elevation Trajectory: Diver Leaping from Pier:

  • A diver leaps off a pier of height $h = 3.0\,\text{m}$ with launch speed $v_0 = 3.2\,\text{m/s}$ at $\theta = 25^{\circ}$ above the horizontal. Landing occurs on water below ($\Delta y = -3.0\,\text{m}$).

  • Velocity components:     $v_{0x} = (3.2\,\text{m/s})\cos(25^{\circ}) = 2.90\,\text{m/s}$$v_{0y} = (3.2\,\text{m/s})\sin(25^{\circ}) = 1.352\,\text{m/s}$

  • Method 1: Quadratic Formula Execution:$\Delta y = v_{0y}t + \frac{1}{2}a_y t^2 \implies -3.0 = 1.352t - 4.9t^2$$-4.9t^2 + 1.352t + 3.0 = 0$$t = \frac{-1.352 \pm \sqrt{(1.352)^2 - 4(-4.9)(3.0)}}{2(-4.9)} = \frac{-1.352 \pm \sqrt{1.828 + 58.8}}{-9.8} = \frac{-1.352 \pm \sqrt{60.628}}{-9.8}$$t = \frac{-1.352 \pm 7.786}{-9.8}$

  • Discarding the unphysical negative root leaves $t = \frac{-9.138}{-9.8} = +0.933\,\text{s}$.

  • Method 2: Two-Step Kinematic Evaluation (Avoiding Quadratic):

  • Solve for landing vertical velocity:       $v_y^2 = v_{0y}^2 + 2a_y \Delta y = (1.352\,\text{m/s})^2 + 2(-9.8\,\text{m/s}^2)(-3.0\,\text{m}) = 1.828 + 58.8 = 60.63\,\text{m}^2/\text{s}^2$$v_y = -\sqrt{60.63\,\text{m}^2/\text{s}^2} = -7.79\,\text{m/s}$

  • Compute elapsed time:       $t = \frac{v_y - v_{0y}}{a_y} = \frac{-7.79\,\text{m/s} - 1.352\,\text{m/s}}{-9.8\,\text{m/s}^2} = \frac{-9.142}{-9.8} = 0.933\,\text{s}$

  • Horizontal range from pier base:     $\Delta x = v_{0x}t = (2.90\,\text{m/s})(0.933\,\text{s}) = 2.7\,\text{m}$

• Ascending State Verification (Golf Ball Trajectory):

  • A golf ball is struck at $v_0 = 25.0\,\text{m/s}$ at $\theta = 30.0^{\circ}$ from initial coordinates $(0, 1.0\,\text{m})$. Determining vertical velocity at height $y = 5.0\,\text{m}$ on its way upward:

  • Height displacement: $\Delta y = 5.0\,\text{m} - 1.0\,\text{m} = +4.0\,\text{m}$

  • Component initialization: $v_{0y} = (25.0\,\text{m/s})\sin(30.0^{\circ}) = 12.5\,\text{m/s}$$v_y^2 = v_{0y}^2 + 2a_y \Delta y = (12.5\,\text{m/s})^2 + 2(-9.81\,\text{m/s}^2)(4.0\,\text{m}) = 156.25 - 78.48 = 77.77\,\text{m}^2/\text{s}^2$$v_y = \pm \sqrt{77.77\,\text{m}^2/\text{s}^2} = \pm 8.82\,\text{m/s}$

  • Since the ball is ascending, select the positive root: $v_y = +8.8\,\text{m/s}$.

Dynamics: Fundamental Forces and Newton's Laws

• Foundations of Dynamics:

  • Dynamics investigates the physical mechanisms and interactions (forces) that govern changes in the state of motion of massive bodies.

  • A force is fundamentally defined as a push or a pull resulting from an interaction between bodies, quantified in newtons ($1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$).

• The Four Fundamental Forces of Nature:

  • All natural interactions reduce fundamentally to four action-at-a-distance forces:

  1. Gravitational Force: Universal mutual attraction between masses governed by Newton's Universal Law:        $F_g = \frac{G M m}{r^2}$

     • Near planetary surfaces, this approximates to $F_g = mg$.

     • Governs planetary orbits, stellar structures, and galactic clustering. Hypothetical mediating gauge boson: graviton.

  2. Electromagnetic Force: Governs interactions between electrically charged particles, mediated by the exchange of photons.

     • Dictates atomic structure, chemical bonding, electron behavior, friction, tension, and material elasticity.

  3. Strong Nuclear Force: Mediated by gluons ($8$ color varieties) acting among quarks ($u$, $d$) to assemble protons ($uud$) and neutrons ($udd$), binding atomic nuclei against electromagnetic repulsion.

  4. Weak Nuclear Force: Mediated by massive vector bosons ($W^+$, $W^-$, $Z^0$).

     • Operates over subatomic ranges, governing neutrino interactions, beta decay processes, and thermonuclear burning in stars.

• Everyday Macroscopic Mechanical Forces:

  • Normal Force ($\mathbf{F}_N$): Electromagnetic contact repulsion exerted perpendicular to the interface plane between two surfaces.

  • Tension Force ($\mathbf{F}_T$): Axial pull exerted through an extended flexible cord, rope, or cable, directed strictly along its length.

  • Spring Force ($\mathbf{F}_{\text{spring}}$): Restorative elastic force acting opposite to mechanical displacement from equilibrium (Hooke's Law):     $\mathbf{F}_{\text{spring}} = -k\mathbf{x}$

  • Frictional Force ($\mathbf{F}_f$): Tangential contact force acting parallel to the surface interface, resisting relative sliding motion.

• Newton's Laws of Motion:

  • First Law (Law of Inertia): An object continues in a state of rest or uniform motion with constant velocity in a straight line unless acted upon by a non-zero net external force:     $\sum \mathbf{F} = 0 \iff \mathbf{a} = 0$

  • Inertia is an intrinsic property directly proportional to mass ($m$).

  • Second Law: The acceleration of a body is directly proportional to and in the same direction as the net force ($\mathbf{F}_{\text{net}}$) acting on it, and inversely proportional to its mass:     $\mathbf{F}_{\text{net}} = \sum \mathbf{F} = m\mathbf{a}$

  • Third Law (Action-Reaction Pairs): Whenever one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first:     $\mathbf{F}_{A \text{ on } B} = -\mathbf{F}_{B \text{ on } A}$

  • Action and reaction forces act on separate bodies and never cancel each other within a single free-body diagram.

• Action-Reaction Case Studies:

  • Human Pyramid Equilibrium:

  • Three individuals stacked vertically: Person 3 (top, $F_{g3} = 650\,\text{N}$), Person 2 (middle, $F_{g2} = 700\,\text{N}$), Person 1 (bottom, $F_{g1} = 750\,\text{N}$).

  • Person 3 FBD: Downward gravity $F_{g3} = 650\,\text{N}$, upward support $F_{23} = 650\,\text{N}$.

  • Action-reaction pair: Person 3 pushes down on Person 2 with $F_{32} = 650\,\text{N}$.

  • Person 2 FBD: Downward load $F_{32} = 650\,\text{N}$, downward gravity $F_{g2} = 700\,\text{N}$. Upward support from Person 1:       $F_{12} = F_{32} + F_{g2} = 650\,\text{N} + 700\,\text{N} = 1350\,\text{N}$

  • Action-reaction pair: Person 2 pushes down on Person 1 with $F_{21} = 1350\,\text{N}$.

  • Person 1 FBD: Downward load $F_{21} = 1350\,\text{N}$, downward gravity $F_{g1} = 750\,\text{N}$. Upward normal force from ground:       $F_{N1} = F_{21} + F_{g1} = 1350\,\text{N} + 750\,\text{N} = 2100\,\text{N}$

  • Orbital Gravitational Interaction (Sun-Mercury System):

  • Astronomical constants: $M_{\text{Sun}} = 1.99 \times 10^{30}\,\text{kg}$, $M_{\text{Mercury}} = 3.3 \times 10^{23}\,\text{kg}$, orbital radius $r = 6.8 \times 10^{10}\,\text{m}$, gravitational constant $G = 6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2$.

  • Mutual gravitational force magnitude:       $F_g = \frac{G M_{\text{Sun}} M_{\text{Mercury}}}{r^2} = \frac{(6.67 \times 10^{-11})(1.99 \times 10^{30})(3.3 \times 10^{23})}{(6.8 \times 10^{10})^2} = 9.5 \times 10^{21}\,\text{N}$

  • While the force exerted by the Sun on Mercury equals the force exerted by Mercury on the Sun ($9.5 \times 10^{21}\,\text{N}$), the resulting acceleration of the Sun is negligible due to its massive inertia ($a_{\text{Sun}} \approx 4.8 \times 10^{-9}\,\text{m/s}^2$), whereas Mercury undergoes significant orbital centripetal acceleration ($a_{\text{Mercury}} \approx 0.029\,\text{m/s}^2$).