🧭 Vectors — Magnitude, Direction, and Components

Part 1 of 7 — Two-Dimensional Motion

So far we've studied motion along a straight line. Real-world motion often occurs in two dimensions — a ball flying through the air, a car turning a corner, a boat crossing a river. To handle 2D motion, we need vectors.

Scalars vs. Vectors

Representing Vectors

A vector can be described by:

1. Magnitude and direction: $v = 5$ m/s at $30°$ north of east
2. Components: $v_x = 4.33$ m/s, $v_y = 2.5$ m/s

Notation

• Vectors are written as $\vec{v}$, $\vec{a}$, $\vec{F}$ (arrow notation)
• Magnitude: $|\vec{v}| = v$ (no arrow, or absolute value bars)

Vector Components

Any 2D vector can be broken into perpendicular components:

$v_x = v \cos\theta$ $v_y = v \sin\theta$

where $\theta$ is measured from the positive x-axis (standard position).

Reconstructing from Components

$v = \sqrt{v_x^2 + v_y^2}$ $\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)$

Example

A velocity of $10$ m/s at $60°$ above the positive x-axis:

$v_x = 10 \cos 60° = 10(0.5) = 5 \text{ m/s}$ $v_y = 10 \sin 60° = 10(0.866) = 8.66 \text{ m/s}$

Check: $\sqrt{5^2 + 8.66^2} = \sqrt{25 + 75} = \sqrt{100} = 10$ m/s ✓

Unit Vectors

Unit vectors have magnitude 1 and point along a specific axis:

• $\hat{i}$ (or $\hat{x}$): points in the $+x$ direction
• $\hat{j}$ (or $\hat{y}$): points in the $+y$ direction

Any vector can be written as:

$\vec{v} = v_x \hat{i} + v_y \hat{j}$

Example

$\vec{v} = 3\hat{i} + 4\hat{j}$ m/s means $v_x = 3$ m/s and $v_y = 4$ m/s.

Magnitude: $v = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$ m/s

Direction: $\theta = \tan^{-1}(4/3) = 53.1°$ above the $+x$ axis

Concept Check — Vectors 🎯

Vector Component Practice 🧮

1. A force of 50 N acts at $37°$ above the positive x-axis. What is $F_x$? (in N, use $\cos 37° = 0.8$)

2. For the same force, what is $F_y$? (in N, use $\sin 37° = 0.6$)

3. A velocity vector has $v_x = 6$ m/s and $v_y = 8$ m/s. What is the magnitude? (in m/s)

Vector Basics Review 🔍

Exit Quiz — Vectors ✅

➕ Vector Addition

Part 2 of 7 — Two-Dimensional Motion

In physics, we constantly need to add vectors — combining displacements, adding velocities, summing forces. There are two methods: graphical (tip-to-tail) and component (algebraic).

Graphical Method: Tip-to-Tail

To add $\vec{A} + \vec{B}$:

1. Draw $\vec{A}$
2. Place the tail of $\vec{B}$ at the tip of $\vec{A}$
3. The resultant $\vec{R}$ goes from the tail of $\vec{A}$ to the tip of $\vec{B}$

Properties of Vector Addition

• Commutative: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$
• Associative: $(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C})$
• The resultant is generally NOT the arithmetic sum of the magnitudes

Special Cases

Component Method

The component method is more precise and works for any number of vectors.

Steps

1. Resolve each vector into $x$ and $y$ components
2. Add all $x$-components: $R_x = A_x + B_x + C_x + \cdots$
3. Add all $y$-components: $R_y = A_y + B_y + C_y + \cdots$
4. Find the resultant: $R = \sqrt{R_x^2 + R_y^2}$
5. Find the direction: $\theta = \tan^{-1}(R_y/R_x)$

Example

$\vec{A} = 3\hat{i} + 4\hat{j}$ and $\vec{B} = -1\hat{i} + 2\hat{j}$

$\vec{R} = \vec{A} + \vec{B} = (3-1)\hat{i} + (4+2)\hat{j} = 2\hat{i} + 6\hat{j}$

$R = \sqrt{4 + 36} = \sqrt{40} = 6.32 \text{ m}$

$\theta = \tan^{-1}(6/2) = \tan^{-1}(3) = 71.6°$

Vector Subtraction

$\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$

To subtract $\vec{B}$, reverse its direction and then add.

Using components: $\vec{A} - \vec{B} = (A_x - B_x)\hat{i} + (A_y - B_y)\hat{j}$

Important Application: $\Delta \vec{v}$

Change in velocity: $\Delta\vec{v} = \vec{v}_f - \vec{v}_i$

This is vector subtraction — you can't just subtract the magnitudes if the directions differ!

Concept Check — Vector Addition 🎯

Vector Addition Practice 🧮

1. $\vec{A} = 6\hat{i} + 2\hat{j}$ and $\vec{B} = -1\hat{i} + 4\hat{j}$. What is the x-component of $\vec{A} + \vec{B}$?

2. What is the y-component of $\vec{A} + \vec{B}$?

3. A hiker walks 5 km east, then 12 km north. What is the magnitude of the resultant displacement? (in km)

Vector Addition Review 🔍

Exit Quiz — Vector Addition ✅

🚤 Relative Motion and Reference Frames

Part 3 of 7 — Two-Dimensional Motion

Have you ever noticed that a person walking on a moving train appears to move at different speeds depending on whether you're on the train or on the ground? That's relative motion — and it's a core concept in physics.

Reference Frames

A reference frame is the perspective from which you observe motion. Different observers in different reference frames may measure different velocities for the same object.

Key Principle

The velocity of object A relative to observer C can be found by adding velocities:

$\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$

where:

• $\vec{v}_{AC}$ = velocity of A relative to C
• $\vec{v}_{AB}$ = velocity of A relative to B
• $\vec{v}_{BC}$ = velocity of B relative to C

Subscript Trick

The inner subscripts ($B$ and $B$) must match and "cancel":

$\vec{v}_{A\cancel{B}} + \vec{v}_{\cancel{B}C} = \vec{v}_{AC}$

Reversing Direction

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

The velocity of A relative to B is the negative of B relative to A.

1D Relative Motion Examples

Example 1: Train Passenger

A passenger walks at 2 m/s toward the front of a train moving at 30 m/s relative to the ground.

Velocity of passenger relative to ground: $v_{PG} = v_{PT} + v_{TG} = 2 + 30 = 32 \text{ m/s}$

Example 2: Opposing Motion

A person walks at 1.5 m/s toward the back of the same train:

$v_{PG} = -1.5 + 30 = 28.5 \text{ m/s}$

Example 3: River Crossing

A boat crosses a river. The boat's speed relative to water is $v_{BW} = 4$ m/s (perpendicular to bank). The river flows at $v_{WG} = 3$ m/s.

Boat's speed relative to ground: $v_{BG} = \sqrt{v_{BW}^2 + v_{WG}^2} = \sqrt{16 + 9} = 5 \text{ m/s}$

Concept Check — Relative Motion 🎯

Relative Motion Calculations 🧮

1. A plane flies at 200 m/s relative to the air. A tailwind blows at 50 m/s in the same direction. What is the plane's speed relative to the ground? (in m/s)

2. The same plane now flies into a 50 m/s headwind. What is the plane's speed relative to the ground? (in m/s)

3. A boat moves at 3 m/s across a river (perpendicular to the bank). The river flows at 4 m/s. What is the boat's speed relative to the ground? (in m/s)

Relative Motion Review 🔍

Exit Quiz — Relative Motion ✅

🎯 Independence of Horizontal and Vertical Motion

Part 4 of 7 — Two-Dimensional Motion

One of the most powerful principles in 2D kinematics is that horizontal and vertical motions are independent. This means you can analyze each direction separately, using its own set of kinematic equations.

The Independence Principle

When an object moves in two dimensions (like a projectile), the motion in the x-direction and the motion in the y-direction are completely independent of each other.

What This Means

• The horizontal velocity has no effect on the vertical motion
• The vertical velocity has no effect on the horizontal motion
• Gravity affects only the vertical component
• Each direction obeys its own kinematic equations

Separate Equations

What Connects Them?

Time ($t$) is the same for both directions. This is the link between horizontal and vertical motion.

The Classic Demonstration

Imagine two balls released at the same time:

• Ball A: dropped from rest
• Ball B: launched horizontally from the same height

Result

Both balls hit the ground at the same time! 🤯

Why? Because:

• Ball B has horizontal velocity, but that doesn't affect its vertical fall
• Both balls have the same initial vertical velocity ($v_{0y} = 0$) and the same vertical acceleration ($a_y = -g$)
• The vertical motion is identical for both

This proves independence!

The horizontal motion of Ball B is completely separate from its vertical free fall. The only thing that determines when it hits the ground is the height and $g$.

Concept Check — Independence of Motion 🎯

Independence of Motion Practice 🧮

A ball is launched horizontally at 15 m/s from the top of a 20 m tall building. Use $g = 10$ m/s².

1. How long does it take to reach the ground? (in seconds)

2. How far from the base of the building does it land? (in meters)

3. What is the vertical velocity just before hitting the ground? (in m/s, magnitude only)

Key Concepts Review 🔍

Exit Quiz — Independence of Motion ✅

🧮 Vector Practice Problems

Part 5 of 7 — Two-Dimensional Motion

Time to sharpen your vector skills! This lesson focuses on solving 2D motion problems step by step using vector decomposition and kinematic equations in each direction.

Problem-Solving Strategy for 2D Motion

Step-by-Step Method

1. Draw a diagram — sketch the situation with a coordinate system
2. Resolve into components — break initial velocity into $v_{0x}$ and $v_{0y}$
3. Write equations for each direction separately
4. Use time as the link between $x$ and $y$ equations
5. Solve and check units

Key Equations Summary

Finding the Resultant

When you need the final speed or direction:

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

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

Problem 1 — Horizontal Launch 🏀

A basketball is thrown horizontally at 8 m/s from a window 5 m above the ground. Use $g = 10$ m/s².

1. Time to hit the ground (in seconds)

2. Horizontal distance traveled (in meters)

3. Final speed just before hitting the ground (in m/s)

Problem 2 — Vector Addition ➕

Problem 3 — Vector Components 📐

A force of 50 N is applied at an angle of 53° above the horizontal.

Use $\cos 53° \approx 0.6$ and $\sin 53° \approx 0.8$.

1. What is the horizontal component (in N)?

2. What is the vertical component (in N)?

Problem 4 — Relative Motion 🚗

Problem 5 — Adding Vectors by Components ➕

Two displacement vectors: $\vec{A} = 6$ m east and $\vec{B} = 8$ m north.

1. Magnitude of $\vec{A} + \vec{B}$ (in meters)

2. Direction of $\vec{A} + \vec{B}$ above east (in degrees, use $\tan^{-1}(4/3) \approx 53°$)

Exit Quiz — Vector Practice ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Two-Dimensional Motion

This workshop walks through multi-step 2D motion problems that combine vector decomposition, kinematic equations, and the independence principle. These are the kinds of problems you'll see on the AP exam!

AP Problem-Solving Framework

DVAT Approach for 2D

Create two separate DVAT tables — one for each direction:

Common Mistakes to Avoid

• ❌ Using the total speed where a component is needed
• ❌ Forgetting that $a_x = 0$ (not $g$!)
• ❌ Using different times for $x$ and $y$
• ❌ Mixing up $\sin$ and $\cos$ for components

Worked Example

A ball is kicked at 20 m/s at 37° above horizontal from ground level. Use $g = 10$ m/s², $\cos 37° = 0.8$, $\sin 37° = 0.6$.

Step 1: Resolve Components

$v_{0x} = 20\cos 37° = 20(0.8) = 16 \text{ m/s}$

$v_{0y} = 20\sin 37° = 20(0.6) = 12 \text{ m/s}$

Step 2: Find Time of Flight

At landing, $\Delta y = 0$ (returns to ground):

$0 = v_{0y}t - \frac{1}{2}gt^2 = 12t - 5t^2 = t(12 - 5t)$

$t = 0$ (launch) or $t = 2.4$ s (landing)

Step 3: Find Range

$\Delta x = v_{0x} \cdot t = 16(2.4) = 38.4 \text{ m}$

Step 4: Find Maximum Height

At max height, $v_y = 0$:

$v_y^2 = v_{0y}^2 - 2g\Delta y_{max}$ $0 = 144 - 20\Delta y_{max}$ $\Delta y_{max} = 7.2 \text{ m}$

Guided Problem 1 — Cliff Launch 🏔️

A stone is thrown horizontally at 12 m/s from a 45 m cliff. Use $g = 10$ m/s².

1. Time to reach the ground (in seconds)

2. Horizontal distance from the cliff base (in meters)

3. Vertical velocity at impact, magnitude (in m/s)

Guided Problem 2 — Angled Launch ⚽

A soccer ball is kicked at 25 m/s at 53° above horizontal from ground level. Use $g = 10$ m/s², $\cos 53° = 0.6$, $\sin 53° = 0.8$.

1. $v_{0x}$ (in m/s)

2. $v_{0y}$ (in m/s)

3. Time of flight (in seconds)

4. Range (in meters)

Concept Checks 🔍

Challenge Problems 🏆

Exit Problem ✅

A ball is launched from the ground at 40 m/s at 37° above horizontal. Use $g = 10$ m/s², $\cos 37° = 0.8$, $\sin 37° = 0.6$.

1. Maximum height (in meters)

2. Total time of flight (in seconds)

Round all answers to 3 significant figures.

🎓 Synthesis & AP Review

Part 7 of 7 — Two-Dimensional Motion

Congratulations on completing the Two-Dimensional Motion unit! This final lesson ties all the concepts together and prepares you with AP-style questions covering vectors, components, relative motion, and the independence principle.

Unit Summary

Vectors

• Vectors have magnitude and direction
• Components: $A_x = A\cos\theta$, $A_y = A\sin\theta$
• Magnitude: $A = \sqrt{A_x^2 + A_y^2}$
• Direction: $\theta = \tan^{-1}(A_y / A_x)$

Vector Addition

• Graphical: tip-to-tail method
• Component: add $x$-components, add $y$-components, then find resultant

Independence Principle

• Horizontal and vertical motions are independent
• $a_x = 0$, $a_y = -g$ (for projectiles)
• Time links the two directions

Relative Motion

• $\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}$ — add velocities in the subscript chain
• Always specify the reference frame

AP-Style Multiple Choice — Set 1 🎯

AP Calculation Problem 🧮

A cannon fires a shell at 50 m/s at 53° above horizontal from ground level. Use $g = 10$ m/s², $\cos 53° = 0.6$, $\sin 53° = 0.8$.

1. Horizontal component of initial velocity (in m/s)

2. Vertical component of initial velocity (in m/s)

3. Maximum height (in meters)

4. Total time of flight (in seconds)

5. Horizontal range (in meters)

Conceptual Review 🔍

AP-Style Multiple Choice — Set 2 🎯

AP Free-Response Style 📝

A rescue helicopter hovers at 80 m altitude. It drops a supply package while a person on the ground is 120 m away horizontally. The package must be dropped with a horizontal velocity to reach the person. Use $g = 10$ m/s².

1. Time for the package to fall 80 m (in seconds)

2. Required horizontal velocity (in m/s)

Final Assessment ✅