Comprehensive Physics Kinematics and 2D Vectors Study Guide

Course Logistics, Homework, and Methodologies

• Deadlines and Schedule:

  • Learning Module 1 is due today at 11:30 pm.

  • CC1/EP1 is due Sunday at 11:30 pm.

  • Current week focus: Velocity Tutorial.

  • Next week focus: Representations of Motion and Acceleration in 1D.

  • Exam 1 Content Cutoff: Mon/Wed content next week covering Projectile Motion completes the content for Exam 1.

• Tutorial Homework Guidelines:

  • Tutorials are completed during the lab period and must be checked off by the Teaching Assistant (TA) prior to leaving ($15/20$ points).

  • Tutorial Homework must be submitted into Canvas as a PDF file by the following Wednesday at 11:30 pm ($5/20$ points).

  • TAs grade tutorial homework based on participation and completion, requiring reasonable work and responses to be present.

  • Official solutions to tutorial homework are posted after the submission due date.

  • For the upcoming 2D Motion Tutorial: Attendance grants $20/20$ points; no written hand-in is required, and solutions will be posted the following Friday.

• Learning Module Exam Pre-Test & Practice Exam:

  • Assess understanding on the first attempt; unlimited attempts are available thereafter.

  • The practice exam releases on Wednesday. Students should review errors and practice using CC1 immediately and CC2 the following week.

• Core Course Methodologies:

  • Focus on the process, not individual problems. Mastering the universal motion process allows any kinematics problem (including 2D projectile motion) to be solved using identical steps.

  • Your calculator provides numerical values, but your brain provides physical directions.

Core Definitions: Position, Velocity, Acceleration, and the Vector Dance

• Position ($\vec{x}$ or $\vec{y}$):

  • Defines the spatial location of an object relative to a chosen origin and coordinate system.

• Velocity ($\vec{v}$):

  • Defined as the rate of change of position per unit time:         $v_{\text{avg}} = \frac{\Delta \vec{x}}{\Delta t}$

  • Velocity indicates how an object changes its position over time; it does not explicitly state the object's current position.

  • Instantaneous Velocity: The velocity of an object at a specific instant in time, geometrically equivalent to the slope of a line tangent to the position-time graph ($x(t)$) at that specific time.

  • Average Velocity: For motion with constant acceleration, average velocity is exactly halfway between the initial velocity ($\vec{v}_i$) and final velocity ($\vec{v}_f$):         $v_{\text{ave}} = \frac{1}{2}(\vec{v}_i + \vec{v}_f)$

• Acceleration ($\vec{a}$):

  • Defined mathematically as the change in velocity divided by the change in time:         $\vec{a} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v}_f - \vec{v}_i}{\Delta t}$

  • Units:         $\frac{\text{m/s}}{\text{s}} = \text{m/s}^2$

  • Conceptual Definition: Acceleration measures how much the velocity vector changes in one second (e.g., an acceleration of $10\,\text{m/s}^2$ downward means the object changes its velocity by $10\,\text{m/s}$ in the downward direction every second).

  • Acceleration does not indicate an object's current velocity, only how that velocity changes over time.

• Interpretation of Acceleration Signs ($+$ and $-$):

  • Positive ($+$) and negative (-$access) signs denote direction relative to the defined coordinate frame.\n * Positive acceleration does **not** inherently mean speeding up.\n * Negative acceleration does **not** inherently mean slowing down.\n\n* **The Vector Dance (Speeding Up vs. Slowing Down)**:\n * **Speeding Up**: Occurs when the velocity vector (\vec{v}$) and acceleration vector ($\vec{a}) point in the **same direction**.\n * **Slowing Down**: Occurs when the velocity vector (\vec{v}$) and acceleration vector ($\vec{a}) point in **opposite directions**.\n * **Turning**: Occurs when the acceleration vector (\vec{a}$) has a component **perpendicular** to the velocity vector ($\vec{v}).\n\n# Graphical Analysis of Motion and Tangent Methods\n\n* **Relationships Between Graphs**:\n * **Slope Transformations**:\n * The slope of a Position vs. Time (x(t)$) graph yields Velocity ($v(t)).\n * The slope of a Velocity vs. Time (v(t)$) graph yields Acceleration ($a(t)).\n * **Area Transformations**:\n * The area under an Acceleration vs. Time (a(t)) curve yields Change in Velocity:\n            \Delta \vec{v}{i,f} = \vec{a} \Delta t\n * The area under a Velocity vs. Time (v(t)) curve yields Change in Position (Displacement):\n            \Delta \vec{x}{i,f} = \vec{v}i \Delta t + \frac{1}{2} \vec{a} (\Delta t)^2\n\n\n\n* **Worked Slope Calculation Example (v(t) Graph)**:\n * Given points on a line: (t_1, v_1) = (0\,\text{s}, 0\,\text{m/s})$and$(t_2, v_2) = (4\,\text{s}, 8\,\text{m/s}).\n * \text{slope} = \frac{\text{rise}}{\text{run}} = \frac{\Delta v}{\Delta t} = \frac{8\,\text{m/s} - 0\,\text{m/s}}{4\,\text{s} - 0\,\text{s}} = 2\,\text{m/s}^2\n\n* **Graphical Tangent Method (Using a Clear Ruler)**:\n * *Step 1 (Mark the point)*: Place a clear dot on the curve at the exact spot where the tangent line is needed.\n * *Step 2 (Place the ruler)*: Lay a clear ruler across the point so it touches the curve.\n * *Step 3 (Balance the angles)*: Adjust the ruler until the open space ("wedge of light") between the ruler edge and the curve looks equal on both sides of the point.\n * *Step 4 (Draw the line)*: Trace a straight line along the ruler edge through the point, touching the curve at only that single point.\n\n# One-Dimensional Kinematic Equations and Problem-Solving Process\n\n* **Universal Problem-Solving Process for Motion**:\n 1. **Draw a physical diagram**: Establish a coordinate system, define origin (x=0$or$y=0$), and mark positive/negative directions ($+\hat{x}$,$-\hat{x}$or$+\hat{y}$,$-\hat{y}).\n 2. **Sketch graphs**: Draw qualitative curves for x(t)$,$v(t)$, and$a(t).\n 3. **Construct a variable list**: List knowns and unknowns among the five key variables (\Delta \vec{x}$,$\vec{a}$,$\vec{v}_f$,$\Delta t$,$\vec{v}_i).\n 4. **Solve**: Select one or two kinematic equations to solve for the target unknown(s).\n\n\n\n* **1D Kinematic Equations Table**:\n * **Equation 1**: \n        \vec{v}_f = \vec{v}_i + \vec{a}\Delta t\n        *(Missing Variable: Displacement \Delta \vec{x})*\n * **Equation 2**: \n        \Delta \vec{x} = \left(\frac{\vec{v}_i + \vec{v}_f}{2}\right)\Delta t\n        *(Missing Variable: Acceleration \vec{a})*\n * **Equation 3**: \n        \Delta \vec{x} = \vec{v}_i \Delta t + \frac{1}{2}\vec{a}(\Delta t)^2\n        *(Missing Variable: Final Velocity \vec{v}_f)*\n * **Equation 4**: \n        (\vec{v}_f)^2 = (\vec{v}_i)^2 + 2\vec{a}\Delta \vec{x}\n        *(Missing Variable: Time Interval \Delta t)*\n * **Displacement Definition**: \n        \Delta \vec{x} = x_f - x_i\n * **Average Velocity Definition**: \n        v{\text{avg}} = \frac{\Delta \vec{x}}{\Delta t}\n\n# Worked Examples in One-Dimensional Kinematics\n\n* **Example 1: Bear Running North**:\n * *Problem*: A bear runs at 4\,\text{m/s}$North with an acceleration of$3\,\text{m/s}^2$North. What is its velocity$2\,\text{s} later?\n * *Variable List*: \n        \vec{v}i = +4\,\text{m/s}\,\hat{x}\n        \vec{a} = +3\,\text{m/s}^2\,\hat{x}\n        \Delta t = 2\,\text{s}\n * *Calculation*: \n        \vec{v}_f = \vec{v}_i + \vec{a}\Delta t = (+4\,\text{m/s}) + (+3\,\text{m/s}^2)(2\,\text{s}) = +10\,\text{m/s}\n * *Result*: 10\,\text{m/s} North.\n\n* **Example 2: Vertical Ball Throw (20\,\text{m/s} Upward)**:\n * *Problem*: A ball is thrown vertically upward with an initial speed of 20\,\text{m/s}$. Assuming gravity acts downward at$10\,\text{m/s}^2:\n * a) Construct a velocity graph.\n * b) Find average velocity on the way up.\n * c) Calculate maximum height reached.\n * *Analysis & Solution*:\n * a) Velocity starts at +20\,\text{m/s}\,\hat{y}$, linearly decreases to$0\,\text{m/s}$at$t = 2\,\text{s}$(peak height), and reaches$-20\,\text{m/s}\,\hat{y}$at$t = 4\,\text{s}.\n * b) Average velocity on upward leg (t = 0\,\text{s}$to$t = 2\,\text{s}):\n            v{\text{ave}} = \frac{+20\,\text{m/s} + 0\,\text{m/s}}{2} = +10\,\text{m/s}\n * c) Maximum height (\Delta y):\n            \Delta y = v_{\text{ave}} \Delta t = (+10\,\text{m/s})(2\,\text{s}) = 20\,\text{m}\n\n* **Example 3: Car Running Out of Gas**:\n * *Problem*: A car travels at 30\,\text{m/s}$and runs out of gas. It slows down at$1\,\text{m/s} every second. How far does it travel before stopping?\n * *Variable List*: \n        x_i = 0\,\text{m}\quad x_f = ?\n        \vec{v}i = +30\,\text{m/s}\,\hat{x}\quad \vec{v}_f = 0\,\text{m/s}\,\hat{x}\n        \vec{a} = -1\,\text{m/s}^2\,\hat{x}\quad \Delta t = ?\n * *Step 1 (Find stopping time \Delta t)*:\n        \vec{v}_f = \vec{v}_i + \vec{a}\Delta t \implies 0\,\text{m/s} = 30\,\text{m/s} + (-1\,\text{m/s}^2)\Delta t \implies \Delta t = 30\,\text{s}\n * *Step 2 (Find stopping distance x_f)*:\n        x_f = x_i + \vec{v}_i \Delta t + \frac{1}{2}\vec{a}(\Delta t)^2 = 0 + (30\,\text{m/s})(30\,\text{s}) + \frac{1}{2}(-1\,\text{m/s}^2)(30\,\text{s})^2 = 900\,\text{m} - 450\,\text{m} = 450\,\text{m}\n * *Alternative Calculation*: \n        v{\text{ave}} = 15\,\text{m/s} \implies x_f = (15\,\text{m/s})(30\,\text{s}) = 450\,\text{m}\n\n* **Example 4: Ball Thrown Upward Off a 30\,\text{m} Cliff**:\n * *Problem*: A ball is thrown vertically upward at 15\,\text{m/s}$from the edge of a$30\text{-meter} cliff. Find impact time and impact velocity.\n * *Variable List*: \n        y_i = +30\,\text{m}\,\hat{y}\quad y_f = 0\,\text{m}\,\hat{y}\n        \vec{v}i = +15\,\text{m/s}\,\hat{y}\quad \vec{v}_f = ?\n        \vec{a} = -10\,\text{m/s}^2\,\hat{y}\quad \Delta t = ?\n * *Step 1 (Find final velocity \vec{v}_f)*:\n        v_f^2 = v_i^2 + 2a\Delta y = (15\,\text{m/s})^2 + 2(-10\,\text{m/s}^2)(0\,\text{m} - 30\,\text{m}) = 225 + 600 = 825\n        v_f = -\sqrt{825} \approx -28.7\,\text{m/s}\,\hat{y}\n * *Step 2 (Find impact time \Delta t)*:\n        \vec{v}_f = \vec{v}_i + \vec{a}\Delta t \implies -28.72\,\text{m/s} = 15\,\text{m/s} + (-10\,\text{m/s}^2)\Delta t\n        -43.72 = -10\Delta t \implies \Delta t \approx 4.37\,\text{s}\n\n* **Example 5: Motorcycle Accelerating**:\n * *Problem*: A motorcycle travels at 12\,\text{m/s}$and accelerates uniformly at$3\,\text{m/s}^2$for$10\,\text{s}. Find distance traveled and final speed.\n * *Variable List*: \n        x_i = 0\,\text{m}\quad x_f = ?\n        \vec{v}_i = +12\,\text{m/s}\,\hat{x}\quad \vec{v}_f = ?\n        \vec{a} = +3\,\text{m/s}^2\,\hat{x}\quad \Delta t = 10\,\text{s}\n * *Part A (Distance Traveled)*:\n        \Delta x = v_i \Delta t + \frac{1}{2}a(\Delta t)^2 = (12\,\text{m/s})(10\,\text{s}) + \frac{1}{2}(3\,\text{m/s}^2)(10\,\text{s})^2 = 120\,\text{m} + 150\,\text{m} = 270\,\text{m}\n * *Part B (Final Speed)*:\n        v_f = v_i + a\Delta t = 12\,\text{m/s} + (3\,\text{m/s}^2)(10\,\text{s}) = 42\,\text{m/s}\n\n* **Example 6: Two Runners Race to Finish Line**:\n * *Problem*: Finish line is at x = +10\,\text{m}\,\hat{x}$. Runner 1 starts at$x = +2\,\text{m}\,\hat{x}$running at a constant$4\,\text{m/s}\,\hat{x}$. Runner 2 starts at$x = 0\,\text{m}\,\hat{x}$running initially at$3\,\text{m/s}\,\hat{x}. What acceleration must Runner 2 have to tie Runner 1 at the finish line?\n * *Step 1 (Find time for Runner 1)*:\n        \Delta x_1 = 10\,\text{m} - 2\,\text{m} = 8\,\text{m}\n        \Delta x_1 = v_1 \Delta t \implies 8\,\text{m} = (4\,\text{m/s})\Delta t \implies \Delta t = 2\,\text{s}\n * *Step 2 (Find acceleration for Runner 2)*:\n        \Delta x_2 = 10\,\text{m} - 0\,\text{m} = 10\,\text{m}\quad \Delta t = 2\,\text{s}\n        \Delta x_2 = v{i,2}\Delta t + \frac{1}{2}a_2(\Delta t)^2 \implies 10\,\text{m} = (3\,\text{m/s})(2\,\text{s}) + \frac{1}{2}a_2(2\,\text{s})^2\n        10 = 6 + 2a_2 \implies 2a_2 = 4 \implies a_2 = 2\,\text{m/s}^2\,\hat{x}\n\n# Two-Dimensional Vector Representation and Mathematical Operations\n\n* **Vector Notation Systems**:\n * Vectors possess both **magnitude** (length/speed) and **direction**.\n * Magnitude of vector \vec{V}$is denoted$|\vec{V}|$or$V:\n        V = \sqrt{V_x^2 + V_y^2}\n * Two standard notation formats:\n * **Component Form**: Written in unit vector notation \vec{V} = V_x\,\hat{x} + V_y\,\hat{y}.\n * **Magnitude-Angle Form**: Specified by length V$and direction angle$\theta measured relative to a reference axis.\n\n\n\n* **Vector Addition Principles**:\n * **Tip-to-Tail Method**: Place the tail of the second vector at the tip of the first vector. The resultant vector points from the tail of the first to the tip of the last.\n * **Commutative Property**: Order of addition does not alter the resultant vector:\n        \vec{a} + \vec{b} + \vec{c} = \vec{c} + \vec{b} + \vec{a}\n\n* **Component Vector Addition Worked Example**:\n * Given vectors:\n        \vec{c} = 1\,\hat{x} + 0\,\hat{y}\n        \vec{b} = 1\,\hat{x} - 1\,\hat{y}\n        \vec{a} = 0\,\hat{x} - 1\,\hat{y}\n * Sum components independently to find Resultant \vec{R} = \vec{a} + \vec{b} + \vec{c}:\n        R_x = 0 + 1 + 1 = +2\,\hat{x}\n        R_y = -1 - 1 + 0 = -2\,\hat{y}\n        \vec{R} = +2\,\hat{x} - 2\,\hat{y}\n * Magnitude:\n        R = \sqrt{(2)^2 + (-2)^2} = \sqrt{8} \approx 2.83\n * Direction:\n        \tan(\theta) = \frac{\text{opp}}{\text{adj}} = \frac{2}{2} = 1 \implies \theta = \tan^{-1}(1) = 45^\circ\quad\text{(below the } +\hat{x} \text{ axis)}\n\n* **Vector Algebra iClicker Questions**:\n * *Question 1*: Given \vec{A} = 0\,\hat{x} + 1\,\hat{y}$and$\vec{A} + \vec{B} = 0\,\hat{x} - 2\,\hat{y}$, determine$\vec{B}.\n        \vec{B} = (0\,\hat{x} - 2\,\hat{y}) - (0\,\hat{x} + 1\,\hat{y}) = 0\,\hat{x} - 3\,\hat{y}\n * *Question 2*: Given \vec{A} = -1\,\hat{x} + 0\,\hat{y}$and$\vec{B} = 0\,\hat{x} - 2\,\hat{y}$, if$\vec{A} + \vec{B} + \vec{C} = 0\,\hat{x} + 0\,\hat{y}$, determine$\vec{C}.\n        \vec{C} = -(\vec{A} + \vec{B}) = -(-1\,\hat{x} - 2\,\hat{y}) = +1\,\hat{x} + 2\,\hat{y}\n * *Question 3*: Given \vec{A} = -1\,\hat{x} + 0\,\hat{y}$,$\vec{B} = 1\,\hat{x} - 1\,\hat{y}$,$\vec{C} = 0\,\hat{x} - 2\,\hat{y}$, if$\vec{A} + \vec{B} + \vec{C} + \vec{D} = 0\,\hat{x} + 0\,\hat{y}$, determine$\vec{D}.\n        (-1 + 1 + 0 + D_x)\,\hat{x} + (0 - 1 - 2 + D_y)\,\hat{y} = 0\,\hat{x} + 0\,\hat{y} \implies D_x = 0, D_y = +3 \implies \vec{D} = 0\,\hat{x} + 3\,\hat{y}\n\n* **Triangle Estimation Rules**:\n * If \theta > 45^\circ$, then$y > x.\n * If \theta = 45^\circ$, then$y = x.\n * If \theta < 45^\circ$, then$y < x.\n * A 30\text{--}40\text{--}50$component right triangle corresponds to angles of$37^\circ$and$53^\circ.\n\n# Two-Dimensional Kinematics and Multi-Leg Motion Examples\n\n* **Yellowstone Biology Student Walking Example**:\n * *Problem*: A student collects water samples across three walking legs:\n 1. 120\,\text{m} due East.\n 2. 100\,\text{m}$at$37^\circ North of West.\n 3. 150\,\text{m}$at$53^\circ South of East.\n        Find the total displacement in component form and as magnitude/direction.\n * *Leg Decompositions*:\n * **Leg 1**: \n            +120\,\text{m}\,\hat{x} + 0\,\text{m}\,\hat{y}\n * **Leg 2**: \n            -100\cos(37^\circ)\,\text{m}\,\hat{x} + 100\sin(37^\circ)\,\text{m}\,\hat{y} = -80\,\text{m}\,\hat{x} + 60\,\text{m}\,\hat{y}\n * **Leg 3**: \n            +150\cos(53^\circ)\,\text{m}\,\hat{x} - 150\sin(53^\circ)\,\text{m}\,\hat{y} = +90\,\text{m}\,\hat{x} - 120\,\text{m}\,\hat{y}\n * *Sum of Components*:\n * x_{\text{total}} = +120\,\text{m} - 80\,\text{m} + 90\,\text{m} = +130\,\text{m}\,\hat{x}\n * y_{\text{total}} = 0\,\text{m} + 60\,\text{m} - 120\,\text{m} = -60\,\text{m}\,\hat{y}\n * **Component Result**: \n            +130\,\text{m}\,\hat{x} - 60\,\text{m}\,\hat{y}\n * *Magnitude and Direction*:\n * R = \sqrt{(130)^2 + (-60)^2} = \sqrt{16900 + 3600} = \sqrt{20500} \approx 143.18\,\text{m} \approx 143\,\text{m}\n * \theta = \tan^{-1}\left(\frac{-60}{130}\right) \approx -24.78^\circ \approx 25^\circ\quad\text{(below East / } +\hat{x} \text{ axis)}\n\n# Curved Motion, Turning Acceleration, and Path Analysis\n\n\n\n* **Three Fundamental Ways to Accelerate**:\n 1. **Speeding Up**: Acceleration vector \vec{a}$is parallel to velocity$\vec{v}.\n 2. **Slowing Down**: Acceleration vector \vec{a}$is anti-parallel (opposite) to velocity$\vec{v}.\n 3. **Turning**: Acceleration vector \vec{a}$is perpendicular to velocity$\vec{v}.\n\n* **Velocity Vector Properties in Curved Motion**:\n * Velocity is **always tangent** to the circular or curved path at every point.\n\n* **Decomposition of Acceleration in 2D**:\n * **Parallel Component (\vec{a}{\parallel})**: Changes the magnitude (speed) of the velocity vector.\n * **Perpendicular Component (\vec{a}{\perp})**: Changes the direction of the velocity vector without changing speed. Always points toward the inside center of the curve.\n\n* **Angles Between Velocity and Total Acceleration**:\n * **Speeding Up and Turning**: Angle \theta$between$\vec{v}$and$\vec{a}$is **less than$90^\circ$** ($\theta < 90^\circ).\n * **Turning Only (Constant Speed)**: Angle \theta$between$\vec{v}$and$\vec{a}$is **exactly$90^\circ$** ($\theta = 90^\circ).\n * **Slowing Down and Turning**: Angle \theta$between$\vec{v}$and$\vec{a}$is **greater than$90^\circ$** ($\theta > 90^\circ).\n\n* **Physical Factors Influencing Turning Acceleration**:\n * **Speed Effect**: More turning acceleration is required when moving at higher speeds around a curve.\n * **Tightness Effect**: More turning acceleration is required when navigating tighter (sharper) curves.\n\n* **Change in Velocity Vector for a 90^\circ Turn**:\n * *Problem*: A car enters a turn at \vec{v}i = 20\,\text{m/s}$and exits at$\vec{v}_f = 20\,\text{m/s}$rotated by$90^\circ$. What is the change in velocity vector$\Delta \vec{v}{i,f}?\n * *Vector Equation*: \n        \vec{v}i + \Delta \vec{v}{i,f} = \vec{v}f \implies \Delta \vec{v}{i,f} = \vec{v}f - \vec{v}_i\n * *Magnitude*:\n        |\Delta \vec{v}{i,f}| = \sqrt{(20\,\text{m/s})^2 + (20\,\text{m/s})^2} = \sqrt{800} \approx 28.3\,\text{m/s}\n * *Direction*: Points downward / inward toward the center of curvature.\n\n* **Pendulum Motion Analysis**:\n * **Velocity**: Directed tangent to the circular arc. Zero at turnaround points (extremes), maximum at the bottom of the swing.\n * **Acceleration**: Consists of a tangential component (\vec{a}{\parallel}$) driving speed changes and a radial/centripetal component ($\vec{a}{\perp}) pointing up along the string toward the pivot to redirect velocity.\n\n\n\n* **Track Path Motion Analysis (Points A through H)**:\n * **Point A**: Object starts at rest (v = 0$$).

  • Segment A to C: Object continuously speeds up until reaching point C.

  • Segment C to D: Object navigates a curve while moving.

  • Segment D to E: Object maintains constant speed through point E and just beyond.

  • Beyond E to F: Object begins slowing down.

  • Segment F to G: Object travels along a straight path.

  • Segment G to H: Object continues slowing down along the path until coming to a complete stop at point H.