Breaking Vectors into Components

For a vector $\vec{v}$ at angle $\theta$ from the horizontal:

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

Magnitude from components: $v = \sqrt{v_x^2 + v_y^2}$

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

Sign Conventions

• $v_x > 0$: pointing right
• $v_x < 0$: pointing left
• $v_y > 0$: pointing up
• $v_y < 0$: pointing down

Independence of Motion

KEY PRINCIPLE: Horizontal and vertical motions are independent.

This means:

• $x$-direction motion doesn't affect $y$-direction motion
• $y$-direction motion doesn't affect $x$-direction motion
• We can analyze each direction separately!

Horizontal Direction

$x = x_0 + v_{0x}t + \frac{1}{2}a_x t^2$ $v_x = v_{0x} + a_x t$

Vertical Direction

$y = y_0 + v_{0y}t + \frac{1}{2}a_y t^2$ $v_y = v_{0y} + a_y t$

Important: Time $t$ is the same for both directions!

Position and Displacement Vectors

Position Vector

$\vec{r} = x\hat{i} + y\hat{j}$

Where $\hat{i}$ and $\hat{j}$ are unit vectors in $x$ and $y$ directions.

Displacement Vector

$\Delta \vec{r} = \Delta x \hat{i} + \Delta y \hat{j}$

$\Delta \vec{r} = (x_f - x_i)\hat{i} + (y_f - y_i)\hat{j}$

Magnitude of displacement: $|\Delta \vec{r}| = \sqrt{(\Delta x)^2 + (\Delta y)^2}$

Velocity Vectors

Average Velocity Vector

$\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{\Delta x}{\Delta t}\hat{i} + \frac{\Delta y}{\Delta t}\hat{j}$

$\vec{v}_{avg} = v_{avg,x}\hat{i} + v_{avg,y}\hat{j}$

Instantaneous Velocity Vector

$\vec{v} = \frac{d\vec{r}}{dt} = v_x\hat{i} + v_y\hat{j}$

Where: $v_x = \frac{dx}{dt}, \quad v_y = \frac{dy}{dt}$

Speed (magnitude of velocity): $v = |\vec{v}| = \sqrt{v_x^2 + v_y^2}$

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

Key fact: Velocity vector is always tangent to the path.

Acceleration Vectors

Acceleration Vector

$\vec{a} = a_x\hat{i} + a_y\hat{j}$

Where: $a_x = \frac{dv_x}{dt}, \quad a_y = \frac{dv_y}{dt}$

Magnitude: $a = |\vec{a}| = \sqrt{a_x^2 + a_y^2}$

Relative Velocity

The velocity of object A relative to object B:

$\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$

Example: Velocity of plane relative to ground = velocity of plane relative to air + velocity of air relative to ground (wind).

$\vec{v}_{plane/ground} = \vec{v}_{plane/air} + \vec{v}_{air/ground}$

Problem-Solving Strategy

1. Set up coordinate system ($x$-$y$ axes)
2. Break initial velocity into components using trig
3. Write separate equations for $x$ and $y$
4. Use the fact that $t$ is the same in both directions
5. Solve for unknowns
6. Combine components if asked for magnitude/direction

Common Scenarios

Motion on an Incline

• Rotate axes: one parallel to incline, one perpendicular
• Gravity component parallel: $g \sin \theta$
• Gravity component perpendicular: $g \cos \theta$

Circular Motion (Preview)

• Velocity is always tangent to circle
• Acceleration points toward center
• Speed can be constant, but velocity changes (direction changes)

Find: Average velocity vector and magnitude

Step 1: Find displacement components $\Delta x = x_f - x_i = 7 - 2 = 5 \text{ m}$ $\Delta y = y_f - y_i = 15 - 3 = 12 \text{ m}$

Step 2: Find average velocity components $v_{avg,x} = \frac{\Delta x}{\Delta t} = \frac{5}{4} = 1.25 \text{ m/s}$

$v_{avg,y} = \frac{\Delta y}{\Delta t} = \frac{12}{4} = 3.0 \text{ m/s}$

Step 3: Write vector $\vec{v}_{avg} = 1.25\hat{i} + 3.0\hat{j} \text{ m/s}$

Or: $\vec{v}_{avg} = (1.25, 3.0)$ m/s

Step 4: Find magnitude $v_{avg} = \sqrt{v_{avg,x}^2 + v_{avg,y}^2}$ $v_{avg} = \sqrt{(1.25)^2 + (3.0)^2}$ $v_{avg} = \sqrt{1.5625 + 9.0}$ $v_{avg} = \sqrt{10.5625}$ $v_{avg} \approx 3.25 \text{ m/s}$

Answers:

• Average velocity vector: $(1.25, 3.0)$ m/s or $1.25\hat{i} + 3.0\hat{j}$ m/s
• Magnitude: 3.25 m/s