Kinematics in Two Dimensions; Vectors

Chapter 3: Kinematics in Two Dimensions; Vectors

Contents of Chapter 3

• Vectors and Scalars

• Addition of Vectors—Graphical Methods

• Subtraction of Vectors and Multiplication of a Vector by a Scalar

• Adding Vectors by Components

• Projectile Motion

• Solving Projectile Motion Problems

• Projectile Motion Is Parabolic

• Relative Velocity

3-1 Vectors and Scalars

• Definition of a Vector:

  • A vector has magnitude and direction.

  • Examples of vector quantities: displacement, velocity, force, momentum.

• Definition of a Scalar:

  • A scalar has only a magnitude.

  • Examples of scalar quantities: mass, time, temperature.

Scalars and Vectors

• A scalar quantity is described by a single number (with a unit).

• A vector quantity has both a size (How far? or How fast?) and a direction (Which way?).

• The size or length of a vector is called its magnitude.

• Vectors are graphically represented as arrows, where the length indicates magnitude and the arrow points in the direction of the vector.

Displacement Vectors

• The displacement vector indicates the distance and direction of an object’s motion.

• It is drawn from the initial position to the final position, regardless of the actual path followed.

Tactics Box 1.3: Adding Vectors

• To find the net displacement for a trip with two legs, add the two displacement vectors as follows:

  1. Draw vector A.

  2. Place the tail of vector B at the tip of vector A.

  3. Draw an arrow from the tail of A to the tip of B, representing vector A + B.

  • Refer to: PhET: Vectors and Motion; Text page 20.

Vectors and Trigonometry

• Trigonometry is utilized to calculate lengths and angles of triangles, which is essential when working with displacement vectors and other vectors.

3-2 Addition of Vectors—Graphical Methods

• For vectors in one dimension, straightforward addition and subtraction are sufficient, with careful attention to signs.

• In two dimensions, the addition of vectors becomes complex if they travel along non-overlapping paths.

  • The Pythagorean Theorem is applied for calculative displacement.

Graphical Methods for Vector Addition

• Adding vectors in different orders yields the same result (known as the parallelogram method).

• Vectors not at right angles can be added using the “tail-to-tip” method.

• Parallelogram Method: Place vectors tail-to-tip to find resultant vector magnitudes and angles.

QuickCheck 1.7

• Quick checks for students to understand how to calculate P + Q visually and conceptually.

Example 1.7: How Far Away Is Anna?

1. Anna walks 90 m due east and then 50 m due north. To calculate her displacement, follow the structured approach outlined below:

   • STRATEGIZE: Utilize trigonometry; sketch to form a right triangle.

   • PREPARE: Set the origin at Anna’s starting position, with subsequent vectors formulated as d₁ and d₂.

   • SOLVE: Use the right triangle formed by her movement, where 90 m is adjacent, and 50 m is opposite.

2. The hypotenuse (net displacement) can be calculated using the Pythagorean theorem:

   • $d<em>{net}^2 = d</em>1^2 + d_2^2$

   • $d_{net} = ext{hypotenuse} = ext{sqrt}( (90 ext{ m})^2 + (50 ext{ m})^2) = 103.1 ext{ m}$

   • Rounded appropriately gives 100 m.

3. To assess the angle ($θ$):

   • $θ = an^{-1}\bigg( rac{50 ext{ m}}{90 ext{ m}}\bigg)$

   • Calculates angle north of east approximately 29°.

4. ASSESS: The derived displacement and angle are validated against geometric expectations.

Velocity Vectors

• The velocity vector points in the direction of motion, and its magnitude is equivalent to the speed of the object.

3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar

• Subtraction of Vectors: Defines the negative of a vector, which has the same magnitude but points in the opposite direction. This negative vector is then added.

Tactics Box 3.1: Subtracting Vectors

• To subtract B from A:

  1. Draw A.

  2. Place the tail of vector -B at the tip of vector A.

  3. Draw an arrow from A’s tail to -B’s tip; this represents vector A - B.

  • Refer to: Text page 74.

3-4 Adding Vectors by Components

• Any vector can be expressed as the sum of two perpendicular vectors (components).

Component Vectors

• For vector A in an xy-coordinate system, define two new component vectors parallel to the axes, such that: $A = A<em>x + A</em>y$

Tactics Box 3.2: Determining the Components of a Vector

1. The magnitude of the x-component, $A_x$, is the absolute value respective to direction.

2. The sign of $A_x$ indicates the direction along the x-axis.

3. The y-component $A_y$ follows similar principles regarding direction.

Finding the Components of a Vector

• Components can be derived via trigonometry:

  • $A_x = A \cdot cos(θ)$

  • $A_y = A \cdot sin(θ)$

• The full magnitude of vector A can be computed: $A = ext{sqrt}( A<em>x^2 + A</em>y^2 )$

• The angle can also be determined using tangent: $θ = an^{-1}\bigg( rac{A<em>y}{A</em>x}\bigg)$

3-5 Projectile Motion

• Definition: A projectile is an object moving in two dimensions under the influence of Earth’s gravity; its trajectory is parabolic.

Understanding Projectile Motion

• Analyze horizontal and vertical motions independently.

• The x-direction velocity remains constant, while in the y-direction, acceleration due to gravity (≈ $g$ = 9.8 m/s²) occurs.

• Two balls dropped simultaneously depict that the vertical positions are identical over time, while one maintains a constant horizontal position.

Initial Angle θ₀

• When launched at angle $θ_0$, consider that the initial velocity consists of vertical and horizontal components.

3-6 Solving Projectile Motion Problems

• Steps to Follow:

  1. Read the problem.

  2. Draw a diagram.

  3. Choose coordinate systems.

  4. Determine the time interval for both dimensions.

  5. Analyze x and y motions separately.

  6. Identify known and unknown variables, particularly velocity at the apex of the trajectory.

  7. Use proper kinematic equations and proceed logically to solve for unknowns.

Problem-Solving Approach 3.1

• STRATEGIZE: Address horizontal and vertical motions as interrelated yet distinct problems.

• PREPARE:

  • Assume ideal projectile conditions.

  • Draw a comprehensive visual representation.

  • Create a coordinate system with defined axes where horizontal acceleration is zero ( $a<em>x = 0$ ) and vertical acceleration is free fall ( $a</em>y = -g$ ).

Example 3.9: Dock Jumping

1. Problem Layout: A dog runs off a dock at 8.5 m/s with a height of 0.61 m.

2. PREPARE: Identify both horizontal and vertical component velocities.

   • $v<em>{x</em>i}= 8.5 ext{ m/s}$ and $v<em>{y</em>i}= 0 ext{ m/s}$

3. Since these motions are independent, estimate how long the dog remains in free fall.

4. Use vertical motion equations to find $\Delta t$ for the dog to drop 0.61 m.

   • Vertical equation: $y<em>f = y</em>i + v<em>{y</em>i} \cdot \Delta t - rac{1}{2}g (\Delta t)^2$; solve to find $\Delta t ≈ 0.35 ext{ s}$

5. Calculate horizontal distance using the uniform motion equation:

   • $x<em>f = x</em>i + v_{x} \cdot \Delta t = 8.5 ext{ m/s} \cdot 0.35 ext{ s} = 3.0 ext{ m}$

6. ASSESS: A horizontal distance of 3.0 m aligns with expected behavior for this scenario.

3-7 Projectile Motion Is Parabolic

• The equation of projectile motion can be resolved into the form of a parabola, where: $y = Ax - Bx^2$

3-8 Relative Velocity

• Definition: Velocity as measured relative to different reference frames involves adding or subtracting vector quantities.

• Each velocity is denoted with corresponding objects and reference frames.

Relative Motion Example

• A runner moves at differing perceived speeds relative to observers.

• Specific formulae for relative velocities are derived based on their positional relationships.

Example 3.14: Finding the Ground Speed of an Airplane

1. Context: A plane traveling 500 mph towards the east confronts a wind of 100 mph blowing south.

2. STRATEGIZE: Frame relative velocities via vectors.

   • $v<em>{pg} = v</em>{pa} + v_{ag}$

3. PREPARE: with visual figures to represent motion.

4. SOLVE: Use right triangle relationships to find resultant velocities:

   • $v_{pg} = ext{sqrt}( (500 ext{ mph})^2 + (100 ext{ mph})^2) = 510 ext{ mph}$

   • Angle: $θ= an^{-1}\bigg( rac{100 ext{ mph}}{500 ext{ mph}}\bigg) = 11°$

5. Result: $v_{pg} = (510 ext{ mph}, 11° ext{ south of east})$

Summary of Chapter 3

• Understanding vectors vs scalars:

• Vector addition through graphical means or component analysis.

• Projectile motion defined with key gravitational effects.

Summary: General Principles

• Projectile characteristics under gravity and free-fall acceleration interactions.

• Derived kinematic equations for path prediction and analysis.

Summary: Important Concepts

• Decomposition of vectors into their components and understanding directionality and signs.

Summary: Applications

• Motion dynamics across ramps and relative motion understandings regarding reference points.