Vectors and Two-Dimensional Motion Study Guide

Scalar and Vector Fundamentals

• Physical Quantities Classification: All physical quantities encountered in kinematics and mechanics are classified as either scalars or vectors.

• Scalar Quantity: A physical quantity completely specified by magnitude (size) alone, along with appropriate physical units. Examples include time, mass, temperature, and speed.

• Vector Quantity: A physical quantity that possesses both a magnitude (size) and a direction in space, obeying specific rules of vector addition. Examples include displacement, velocity, acceleration, force, and momentum.

Vector Notation and Graphic Representation

• Handwritten Vector Notation: Represented by placing a rightward ray/arrow directly above the variable symbol (e.g., $\vec{A}$, $\vec{B}$, $\vec{\mathbf{r}}$).

• Printed Vector Notation: Displayed as boldface letters topped with a vector arrow (e.g., $\vec{\mathbf{A}}$, $\vec{\mathbf{B}}$, $\vec{\mathbf{R}}$).

• Vector Magnitude Notation: When referencing solely the non-negative magnitude (length) of a vector in printed text, standard italic letters without bolding or arrows are used (e.g., $A$, $B$, $R$).

• Graphical Representation: Vectors are drawn as directed line segments (arrows) in a coordinate frame:

  • The tail of the arrow represents the starting point/origin of the vector.

  • The tip (arrowhead) points in the direction of the vector.

  • The length of the arrow is proportional to the magnitude of the vector according to a chosen scale factor.

Fundamental Properties of Vectors

• Equality of Two Vectors: Two vectors $\vec{\mathbf{A}}$ and $\vec{\mathbf{B}}$ are equal ($\vec{\mathbf{A}} = \vec{\mathbf{B}}$) if and only if they have identical magnitudes ($A = B$) and point in the exact same direction, regardless of where their initial points (tails) are located in space.

• Parallel Transport Property: A vector can be moved parallel to its original orientation anywhere within a coordinate diagram without altering its value, magnitude, or direction.

• Negative Vectors: Two vectors $\vec{\mathbf{A}}$ and $\vec{\mathbf{B}}$ are defined as negative vectors relative to each other if they have equal magnitudes ($A = B$) but point in opposite directions ($180^\circ apart):\n\n\vec{\mathbf{A}} = -\vec{\mathbf{B}}\n\n\vec{\mathbf{A}} + (-\vec{\mathbf{A}}) = 0\n\n* **Resultant Vector:** The overall vector sum resulting from combining two or more individual vectors:\n\n\vec{\mathbf{R}} = \vec{\mathbf{A}} + \vec{\mathbf{B}}\n\n# Graphical Vector Operations\n\n* **General Rules for Vector Addition:**\n * Vector directions must explicitly be accounted for; simple algebraic addition of magnitudes is valid only when vectors are collinear and unidirectional.\n * All vectors being added together must possess identical physical units.\n* **Triangle or Polygon Method (Tip-to-Tail Construction):**\n * Step 1: Establish a standard coordinate system and choose a convenient spatial scale factor.\n * Step 2: Draw the first vector \vec{\mathbf{A}} with the correct length and orientation relative to the axes.\n * Step 3: Draw the second vector \vec{\mathbf{B}}$starting from the tip (arrowhead) of$\vec{\mathbf{A}}$, maintaining$\vec{\mathbf{B}}'s direction relative to a set of coordinate axes parallel to the main coordinate system.\n\n\n\n * Step 4: For additional vectors (\vec{\mathbf{C}}$,$\vec{\mathbf{D}}, etc.), repeat the tip-to-tail placement sequentially.\n\n\n\n * Step 5: Draw the resultant vector \vec{\mathbf{R}}$from the origin (tail) of the first vector$\vec{\mathbf{A}} directly to the tip (arrowhead) of the final vector.\n * Step 6: Measure the length of \vec{\mathbf{R}}$with a ruler and convert it to real units using the scale factor; measure its direction angle$\theta with a protractor.\n* **Commutative Law of Vector Addition:** The order in which vectors are added does not affect the final resultant vector:\n\n\vec{\mathbf{A}} + \vec{\mathbf{B}} = \vec{\mathbf{B}} + \vec{\mathbf{A}}\n\n\n\n\n\n* **Vector Subtraction:** Vector subtraction is defined as a special case of vector addition, where the negative of the vector being subtracted is added to the first vector:\n\n\vec{\mathbf{A}} - \vec{\mathbf{B}} = \vec{\mathbf{A}} + (-\vec{\mathbf{B}})\n\n\n\n* **Multiplication and Division of a Vector by a Scalar:**\n * Multiplying a vector \vec{\mathbf{A}}$by a scalar$m$yields a new vector$\vec{\mathbf{B}} = m \vec{\mathbf{A}}$whose magnitude is$|m|A.\n * If m > 0$, the resulting vector retains the same direction as$\vec{\mathbf{A}}.\n * If m < 0, the direction of the resulting vector is inverted ($180^\circ opposite to $\vec{\mathbf{A}}$).

Components of a Vector

• Rectangular Components: Projections of a vector along the orthogonal axes of a Cartesian coordinate system ($x$-axis and $y$-axis).

• Trigonometric Formulas for Components: For a vector $\vec{\mathbf{A}}$ making an angle $\theta$ measured counterclockwise from the positive horizontal $x$-axis:

  • Horizontal projection ($x$-component):

$A_x = A \cos(\theta)$

*   Vertical projection ($y$-component):

$A_y = A \sin(\theta)$

*   Vector expression in terms of component vectors:

$\vec{\mathbf{A}} = \vec{\mathbf{A}}_x + \vec{\mathbf{A}}_y$

• Reconstructing Magnitude and Direction from Components:

  • The magnitude $A$ is computed via the Pythagorean theorem:

$A = \sqrt{A_x^2 + A_y^2}$

*   The direction angle $\theta$ relative to the coordinate axis is determined using the inverse tangent function:

$\tan(\theta) = \frac{A_y}{A_x} \implies \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)$

• Quadrant Considerations for Angle Calculations:

  • The inverse tangent function $\tan^{-1}(A_y/A_x)$ yields standard polar angles in Quadrant I ($A_x > 0, A_y > 0$) and Quadrant IV ($A_x > 0, A_y < 0$).

  • If $\vec{\mathbf{A}}$ lies in Quadrant II ($A_x < 0, A_y > 0$) or Quadrant III ($A_x < 0, A_y < 0$), an angle correction of $180^\circ$ must be added to the calculated calculator value to obtain the true angle with respect to the positive $x$-axis.

Algebraic Vector Addition

• Step-by-Step Analytical Component Method:

  1. Establish an appropriate Cartesian coordinate frame ($x, y$) and sketch all vectors.

  2. Decompose each vector into its horizontal ($V_x$) and vertical ($V_y$) components using cosine and sine functions.

  3. Algebraically sum all horizontal components to find the total horizontal component $R_x$ of the resultant:

$R_x = \sum V_x$

4.  Algebraically sum all vertical components to find the total vertical component $R_y$ of the resultant:

$R_y = \sum V_y$

5.  Calculate the magnitude $R$ of the resultant vector using the Pythagorean theorem:

$R = \sqrt{R_x^2 + R_y^2}$

6.  Calculate the direction angle $\theta$ using the inverse tangent function, applying quadrant adjustments where necessary:

$\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

Worked Vector Examples

• Example 3.1: Resolution of Displacement Components

  • Problem: A motorist undergoes a displacement of $250\,\text{km}$ in a direction $30.0^\circ$ North of East. Resolve this displacement into components in the directions north and east.

*   *Calculation:*
    *   East component ($x$-direction): $d_x = (250\,\text{km}) \cos(30.0^\circ) = (250)(0.8660) = 216.51\,\text{km}$
    *   North component ($y$-direction): $d_y = (250\,\text{km}) \sin(30.0^\circ) = (250)(0.5000) = 125.00\,\text{km}$

• Example: Four-Path Walk Displacement

  • Problem: A person follows a path consisting of four straight-line movements: $100\,\text{m}$ East ($0^\circ$), $300\,\text{m}$ South ($-90^\circ$), $150\,\text{m}$ at $30.0^\circ$ South of West ($210^\circ$), and $200\,\text{m}$ at $60.0^\circ$ North of West ($120^\circ$). Find the resultant displacement from the starting point.

*   *Component Summation:*
    *   Horizontal Component $R_x$:

$R_x = 100 - 150 \cos(30.0^\circ) - 200 \cos(60.0^\circ)$

$R_x = 100 - 150(0.8660) - 200(0.5000) = 100 - 129.90 - 100 = -129.90\,\text{m}$

    *   Vertical Component $R_y$:

$R_y = 0 - 300 - 150 \sin(30.0^\circ) + 200 \sin(60.0^\circ)$

$R_y = -300 - 150(0.5000) + 200(0.8660) = -300 - 75 + 173.21 = -201.79\,\text{m}$

*   *Resultant Magnitude and Angle:*
    *   Magnitude $R$:

$R = \sqrt{(-129.90)^2 + (-201.79)^2} = \sqrt{16874.01 + 40719.20} = \sqrt{57593.21} = 239.99\,\text{m}$

    *   Reference Angle $\phi$ within Quadrant III:

$\phi = \tan^{-1}\left(\frac{-201.79}{-129.90}\right) = \tan^{-1}(1.5534) = 57.23^\circ$

    *   True Polar Angle $\theta$ from positive $x$-axis:

$\theta = 180^\circ + 57.23^\circ = 237.23^\circ$

*   *Conclusion:* The resultant displacement is $239.99\,\text{m}$ at an angle of $237.23^\circ$.

Motion in Two Dimensions

• Position Vector: The spatial position of an object in a 2D plane relative to an origin $O$ is denoted by the vector $\vec{\mathbf{r}}$.

• Two-Dimensional Displacement Vector: Defined as the change in the position vector during a time interval:

$\Delta \vec{\mathbf{r}} = \vec{\mathbf{r}}_f - \vec{\mathbf{r}}_i$

• Mechanisms of Acceleration in Two Dimensions: Acceleration vector $\vec{\mathbf{a}}$ occurs whenever velocity $\vec{\mathbf{v}}$ changes. An object accelerates if:

  1. The magnitude of the velocity (speed) changes while direction remains constant.

  2. The direction of the velocity changes while magnitude (speed) remains constant.

  3. Both the magnitude and the direction of the velocity change simultaneously.

Projectile Motion Principles

• Definition of Projectile Motion: Two-dimensional motion of an object moving under the sole influence of Earth's gravitational force. The physical trajectory executed by any projectile is a parabola.

• Independence of Motion Components: The horizontal ($x$) and vertical ($y$) motions of a projectile are completely independent of each other.

• Horizontal Motion Properties ($x$-axis):

  • No horizontal acceleration exists (neglecting air resistance): $a_x = 0$

  • Horizontal velocity component remains constant throughout the flight: $v_x = v_{x0} = v_0 \cos(\theta_0) = \text{constant}$

  • Operative horizontal position equation:

$x = x_0 + v_{x0} t$

• Vertical Motion Properties ($y$-axis):

  • Subject to constant downward gravitational acceleration: $a_y = -g = -9.80\,\text{m/s}^2$

  • Initial vertical velocity component: $v_{y0} = v_0 \sin(\theta_0)$

  • Executes standard 1D motion under constant acceleration (free fall).

• Stroboscopic Verification: Stroboscopic photographs of two falling balls—one dropped vertically from rest and one projected horizontally—demonstrate that their vertical positions remain identical at every instant, while the horizontal position of the projected ball increases linearly with time.

• Effect of Launch Angle on Range and Height:

  • Maximum range on flat terrain is achieved at a launch angle of $\theta_0 = 45^\circ$.

  • Complementary launch angles (e.g., $15^\circ$ and $75^\circ$, or $30^\circ$ and $60^\circ$) yield identical horizontal ranges, though the higher angle produces a greater peak height and longer time of flight.

Kinematic Equations for Projectile Motion

• Summary Table of Governing Kinematic Equations:

• Note on Sign Conventions: If downward is chosen as positive, $-g$ terms change to $+g$.

• Instantaneous Velocity Vector of a Projectile: At any point along the parabola, the instantaneous speed $v$ and direction angle $\theta$ are calculated as:

$v = \sqrt{v_x^2 + v_y^2}$

$\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)$

• At the peak (maximum height) of symmetrical projectile motion, $v_y = 0$, so the instantaneous velocity equals the horizontal component: $v = v_x = v_{x0}$.

Projectile Motion Worked Examples

• Example 3.1: Horizontal Launch Off a Cliff

  • Problem: A movie stunt driver on a motorcycle speeds horizontally off a $50.0\,\text{m}$ high cliff. How fast must the motorcycle leave the cliff-top to land on level ground below, $90.0\,\text{m}$ from the base of the cliff?

*   *Solution:*
    *   Vertical motion analysis ($y_0 = 0, y = -50.0\,\text{m}, v_{y0} = 0, a_y = -9.80\,\text{m/s}^2$):

$y = -\frac{1}{2} g t^2 \implies -50.0 = -\frac{1}{2}(9.80) t^2 \implies 4.90 t^2 = 50.0$

$t^2 = 10.204 \implies t = 3.19\,\text{s}$

    *   Horizontal motion analysis ($x = 90.0\,\text{m}, t = 3.19\,\text{s}$):

$x = v_{x0} t \implies 90.0 = v_{x0} (3.19) \implies v_{x0} = \frac{90.0}{3.19} = 28.2\,\text{m/s}$

• Example 3.2: Maximum Height of a Kicked Football

  • Problem: A football is kicked at an angle of $30.0^\circ$ with an initial speed of $20.0\,\text{m/s}$. Calculate the maximum height reached by the ball.

  • *Solution Steps (using peak conditions $v_y = 0$):

    • Time $t$ to reach maximum height using angle parameter $37.0^\circ$ as evaluated in the transcript slide:

$t = \frac{v_0 \sin(\theta)}{g} = \frac{20.0 \sin(37.0^\circ)}{9.80} = \frac{12.036}{9.80} = 1.22\,\text{s}$

    *   Maximum vertical height $y_{\max}$ calculation:

$y_{\max} = v_{y0} t - \frac{1}{2} g t^2$

$y_{\max} = (20.0 \sin(37.0^\circ))(1.22) - \frac{1}{2}(9.80)(1.22)^2$

$y_{\max} = (12.036)(1.22) - 4.90(1.4884) = 14.68 - 7.29 = 7.39\,\text{m}$

Relative Velocity

• Frame of Reference Concept: Measurements of position, velocity, and acceleration depend explicitly on the observer's chosen coordinate frame of reference.

• Relative Position Subscript Notation:

  • Let $E$ represent an observer stationary relative to Earth.

  • Let $A$ and $B$ represent two moving objects/frames.

  • $\vec{\mathbf{r}}_{AE}$ = position of object $A$ relative to Earth observer $E$

  • $\vec{\mathbf{r}}_{BE}$ = position of object $B$ relative to Earth observer $E$

  • $\vec{\mathbf{r}}_{AB}$ = position of object $A$ relative to object $B$

  • Relative Position Equation:

$\vec{\mathbf{r}}_{AB} = \vec{\mathbf{r}}_{AE} - \vec{\mathbf{r}}_{BE}$

• Relative Velocity Equation: Differentiating relative position vectors with respect to time yields the relative velocity relation:

$\vec{\mathbf{v}}_{AB} = \vec{\mathbf{v}}_{AE} - \vec{\mathbf{v}}_{BE}$

*   Alternatively written as:

$\vec{\mathbf{v}}_{AE} = \vec{\mathbf{v}}_{AB} + \vec{\mathbf{v}}_{BE}$

• Subscript Order Rules: The first subscript indicates the object being observed, while the second subscript indicates the reference frame of the observer. Inverting subscripts reverses vector direction:

$\vec{\mathbf{v}}_{AB} = -\vec{\mathbf{v}}_{BA}$

Relative Velocity Worked Examples

• Example 3.3: Heading Upstream to Cross a River

  • Problem: A boat's speed in still water is $v_{BW} = 2.00\,\text{m/s}$. The boat must travel directly across a river whose current speed is $v_{WS} = 1.50\,\text{m/s}$. At what upstream angle $\theta$ must the boat head?

*   *Solution:*
    *   Let $v_{BS}$ be the resultant velocity of the boat relative to the shore, pointing directly across the river (perpendicular to current).
    *   From the vector right triangle formed by $\vec{\mathbf{v}}_{BW}$, $\vec{\mathbf{v}}_{WS}$, and $\vec{\mathbf{v}}_{BS}$:

$\sin(\theta) = \frac{v_{WS}}{v_{BW}} = \frac{1.50\,\text{m/s}}{2.00\,\text{m/s}} = 0.750$

$\theta = \sin^{-1}(0.750) = 48.59^\circ$

    *   *Conclusion:* The boat must head upstream at an angle of $48.59^\circ$ relative to the line directly across the river.

• Follow-Up Example: Boat Steering Directly Across

  • Problem: The same boat ($v_{BW} = 2.00\,\text{m/s}$) aims directly across the river perpendicular to the current ($v_{WS} = 1.50\,\text{m/s}$).

    1. What is the velocity (magnitude and direction) of the boat relative to the shore?

    2. If the river is $110\,\text{m}$ wide, how long will it take to cross, and how far downstream will the boat land?

*   *Solution Part (a) - Resultant Velocity:*
    *   Magnitude $v_{BS}$ via Pythagorean theorem:

$v_{BS} = \sqrt{v_{BW}^2 + v_{WS}^2} = \sqrt{(2.00)^2 + (1.50)^2} = \sqrt{4.00 + 2.25} = \sqrt{6.25} = 2.50\,\text{m/s}$

*   *Solution Part (b) - Crossing Time and Drift Distance:*
    *   Width of river $d = 110\,\text{m}$. Time $t$ to cross depends solely on the perpendicular velocity component $v_{BW} = 2.00\,\text{m/s}$:

$t = \frac{d}{v_{BW}} = \frac{110\,\text{m}}{2.00\,\text{m/s}} = 55.0\,\text{s}$

    *   Downstream drift distance $x$ carried by river current:

$x = v_{WS} \times t = (1.50\,\text{m/s})(55.0\,\text{s}) = 82.5\,\text{m}$