Two-Dimensional Motion

Vectors, components, and motion in a plane

Two-Dimensional Motion

Introduction to Vectors

Vectors have both magnitude and direction. In 2D motion, we need to track both xx and yy components.

Vector Notation

  • Vector: v\vec{v} or v (bold)
  • Magnitude: v|\vec{v}| or vv (no arrow/bold)
  • Components: vxv_x (horizontal), vyv_y (vertical)

Breaking Vectors into Components

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

vx=vcosθv_x = v \cos \theta vy=vsinθv_y = v \sin \theta

Magnitude from components: v=vx2+vy2v = \sqrt{v_x^2 + v_y^2}

Angle from components: θ=tan1(vyvx)\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)

Sign Conventions

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

Independence of Motion

KEY PRINCIPLE: Horizontal and vertical motions are independent.

This means:

  • xx-direction motion doesn't affect yy-direction motion
  • yy-direction motion doesn't affect xx-direction motion
  • We can analyze each direction separately!

Horizontal Direction

x=x0+v0xt+12axt2x = x_0 + v_{0x}t + \frac{1}{2}a_x t^2 vx=v0x+axtv_x = v_{0x} + a_x t

Vertical Direction

y=y0+v0yt+12ayt2y = y_0 + v_{0y}t + \frac{1}{2}a_y t^2 vy=v0y+aytv_y = v_{0y} + a_y t

Important: Time tt is the same for both directions!

Position and Displacement Vectors

Position Vector

r=xi^+yj^\vec{r} = x\hat{i} + y\hat{j}

Where i^\hat{i} and j^\hat{j} are unit vectors in xx and yy directions.

Displacement Vector

Δr=Δxi^+Δyj^\Delta \vec{r} = \Delta x \hat{i} + \Delta y \hat{j}

Δr=(xfxi)i^+(yfyi)j^\Delta \vec{r} = (x_f - x_i)\hat{i} + (y_f - y_i)\hat{j}

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

Velocity Vectors

Average Velocity Vector

vavg=ΔrΔt=ΔxΔti^+ΔyΔtj^\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{\Delta x}{\Delta t}\hat{i} + \frac{\Delta y}{\Delta t}\hat{j}

vavg=vavg,xi^+vavg,yj^\vec{v}_{avg} = v_{avg,x}\hat{i} + v_{avg,y}\hat{j}

Instantaneous Velocity Vector

v=drdt=vxi^+vyj^\vec{v} = \frac{d\vec{r}}{dt} = v_x\hat{i} + v_y\hat{j}

Where: vx=dxdt,vy=dydtv_x = \frac{dx}{dt}, \quad v_y = \frac{dy}{dt}

Speed (magnitude of velocity): v=v=vx2+vy2v = |\vec{v}| = \sqrt{v_x^2 + v_y^2}

Direction of velocity: θ=tan1(vyvx)\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)

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

Acceleration Vectors

Acceleration Vector

a=axi^+ayj^\vec{a} = a_x\hat{i} + a_y\hat{j}

Where: ax=dvxdt,ay=dvydta_x = \frac{dv_x}{dt}, \quad a_y = \frac{dv_y}{dt}

Magnitude: a=a=ax2+ay2a = |\vec{a}| = \sqrt{a_x^2 + a_y^2}

Relative Velocity

The velocity of object A relative to object B:

vA/B=vAvB\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).

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

Problem-Solving Strategy

  1. Set up coordinate system (xx-yy axes)
  2. Break initial velocity into components using trig
  3. Write separate equations for xx and yy
  4. Use the fact that tt 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: gsinθg \sin \theta
  • Gravity component perpendicular: gcosθg \cos \theta

Circular Motion (Preview)

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

📚 Practice Problems

1Problem 1easy

Question:

A velocity vector has components vx=6v_x = 6 m/s and vy=8v_y = 8 m/s. Find the magnitude and direction of the velocity.

💡 Show Solution

Given:

  • Horizontal component: vx=6v_x = 6 m/s
  • Vertical component: vy=8v_y = 8 m/s

Find:

  1. Magnitude vv
  2. Direction θ\theta (angle from horizontal)

Part 1: Magnitude Use the Pythagorean theorem:

v=vx2+vy2v = \sqrt{v_x^2 + v_y^2} v=62+82v = \sqrt{6^2 + 8^2} v=36+64v = \sqrt{36 + 64} v=100v = \sqrt{100} v=10 m/sv = 10 \text{ m/s}

Part 2: Direction Use inverse tangent:

θ=tan1(vyvx)\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) θ=tan1(86)\theta = \tan^{-1}\left(\frac{8}{6}\right) θ=tan1(1.333)\theta = \tan^{-1}(1.333) θ53.1°\theta \approx 53.1°

Answers:

  • Magnitude: 10 m/s
  • Direction: 53.1° above the horizontal (or from the positive xx-axis)

Note: This is a 3-4-5 right triangle scaled by 2!

2Problem 2medium

Question:

An object moves from position (2,3)(2, 3) m to (7,15)(7, 15) m in 44 seconds. Find the average velocity vector and its magnitude.

💡 Show Solution

Given:

  • Initial position: (xi,yi)=(2,3)(x_i, y_i) = (2, 3) m
  • Final position: (xf,yf)=(7,15)(x_f, y_f) = (7, 15) m
  • Time interval: Δt=4\Delta t = 4 s

Find: Average velocity vector and magnitude

Step 1: Find displacement components Δx=xfxi=72=5 m\Delta x = x_f - x_i = 7 - 2 = 5 \text{ m} Δy=yfyi=153=12 m\Delta y = y_f - y_i = 15 - 3 = 12 \text{ m}

Step 2: Find average velocity components vavg,x=ΔxΔt=54=1.25 m/sv_{avg,x} = \frac{\Delta x}{\Delta t} = \frac{5}{4} = 1.25 \text{ m/s}

vavg,y=ΔyΔt=124=3.0 m/sv_{avg,y} = \frac{\Delta y}{\Delta t} = \frac{12}{4} = 3.0 \text{ m/s}

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

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

Step 4: Find magnitude vavg=vavg,x2+vavg,y2v_{avg} = \sqrt{v_{avg,x}^2 + v_{avg,y}^2} vavg=(1.25)2+(3.0)2v_{avg} = \sqrt{(1.25)^2 + (3.0)^2} vavg=1.5625+9.0v_{avg} = \sqrt{1.5625 + 9.0} vavg=10.5625v_{avg} = \sqrt{10.5625} vavg3.25 m/sv_{avg} \approx 3.25 \text{ m/s}

Answers:

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

3Problem 3hard

Question:

A boat can travel at 55 m/s in still water. It heads due north across a river that flows east at 33 m/s. What is the boat's velocity relative to the shore (magnitude and direction)?

💡 Show Solution

Given:

  • Boat velocity relative to water: vboat/water=5\vec{v}_{boat/water} = 5 m/s north
  • Water velocity relative to shore: vwater/shore=3\vec{v}_{water/shore} = 3 m/s east

Find: Boat velocity relative to shore

Set up components: Let east be +x+x and north be +y+y.

Boat relative to water:

  • vboat/water,x=0v_{boat/water, x} = 0 m/s
  • vboat/water,y=5v_{boat/water, y} = 5 m/s

Water relative to shore:

  • vwater/shore,x=3v_{water/shore, x} = 3 m/s
  • vwater/shore,y=0v_{water/shore, y} = 0 m/s

Apply relative velocity formula: vboat/shore=vboat/water+vwater/shore\vec{v}_{boat/shore} = \vec{v}_{boat/water} + \vec{v}_{water/shore}

Components: vboat/shore,x=0+3=3 m/sv_{boat/shore, x} = 0 + 3 = 3 \text{ m/s} vboat/shore,y=5+0=5 m/sv_{boat/shore, y} = 5 + 0 = 5 \text{ m/s}

Magnitude: vboat/shore=32+52=9+25=345.83 m/sv_{boat/shore} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83 \text{ m/s}

Direction: θ=tan1(53)59.0°\theta = \tan^{-1}\left(\frac{5}{3}\right) \approx 59.0°

This angle is measured from east (the positive xx-axis).

Answers:

  • Velocity relative to shore: 5.83 m/s
  • Direction: 59.0° north of east (or 31.0° east of north)

Physical interpretation: The current pushes the boat downstream (east) even though it's trying to go north!