Lesson 3 Notes: 2D Velocity & Vectors

Velocity Vectors and Independent Motions

Core idea: Motion in the x-direction (horizontal) and y-direction (vertical) are independent; the x-motion does not affect the y-motion and vice versa. This is illustrated repeatedly in the materials as a guiding principle for analyzing 2D motion.
Velocity vectors are decomposed into components along the x and y axes. The magnitude and direction of a velocity vector come from its components.
The material uses several concrete numbers to illustrate 1D and 2D velocity vectors, including constant velocities, instantaneous velocities, and how winds or currents modify the resultant motion.

1D Velocity Vectors

In 1D, velocity vectors lie along a single axis (typically the x-axis). An object may have a constant velocity: e.g., - $v = 2 \text{ m/s}$ along +x, or $v = -2 \text{ m/s}$ along -x depending on direction.

Instantaneous velocity: At any given time t, the velocity value can be $V(t)$ . A table shows $v = 2 \text{ m/s}$ at $t = 0 \text{ s}$ for a particular 1D motion.
If velocity is constant, the position x changes linearly with time: $x(t) = x_0 + v t$ . The content emphasizes instantaneous velocity (the velocity at a given moment) as a key concept in 1D motion.

2D Velocity Vectors: Components, Magnitude, and Direction

A 2D velocity vector can be described by its x-component $Vx$ and y-component $Vy$ , with the real velocity vector $V$ having magnitude and a direction angle $e$ (often denoted $\theta$ ).
Decomposition formulas (given a speed v and direction e):
$V_x = v \cos(e)$
$V_y = v \sin(e)$

Conversely, given components, the magnitude and direction are:
Magnitude: $v = \sqrt{Vx^2 + Vy^2}$
Direction: $\theta = \tan^{-1}\left(\frac{Vy}{Vx}\right)$

Example ( $60°$ angle with speed $v = 2 \text{ m/s}$ ):
$V_x = v \cos(60°) = 2 \times \tfrac{1}{2} = 1 \text{ m/s}$
$V_y = v \sin(60°) = 2 \times \tfrac{\sqrt{3}}{2} \approx 1.73 \text{ m/s} \approx 1.7 \text{ m/s}$
Magnitude: $v = \sqrt{Vx^2 + Vy^2} = \sqrt{1^2 + 1.73^2} \approx 2.0 \text{ m/s}$
Direction angle: $\theta = \tan^{-1}\left(\frac{Vy}{Vx}\right) = \tan^{-1}(1.73/1) \approx 60°$ . This aligns with the given example values ( $Vx \approx 1 \text{ m/s}, Vy \approx 1.7 \text{ m/s}$ ).

The material also shows mixed examples with $Vy = 1 \text{ m/s}$ and $Vx = 2 \text{ m/s}$ , yielding a magnitude ~2.24 m/s and angle ~26.6° (since $\tan^{-1}(1/2) \approx 26.565° = 26.6°$ ).

Positive and Negative Velocity Components

Components can be positive or negative depending on the chosen coordinate directions.
If +x is East and +y is North, then a velocity with $Vx = +1 \text{ m/s}$ and $Vy = +1.7 \text{ m/s}$ points northeast.
If $Vx = -1 \text{ m/s}$ and $Vy = +1.7 \text{ m/s}$ , the direction is northwest.
If $Vy$ is negative, the motion is downward along -y; if $Vx$ is negative, motion is to the west.

The magnitude remains always nonnegative, i.e., $|V| = \sqrt{Vx^2 + Vy^2} \ge 0$ .
Several slides illustrate right-angle (perpendicular) components and the resulting vector magnitude in different quadrants.

Adding Velocity Vectors (Vector Substitution)

When combining motions (e.g., the motion of an object and a wind or current), components add independently along each axis:
If an object has velocity components $Vx$ , $Vy$ and an external effect (wind) adds $V{\text{wind},x}$ and/or $V{\text{wind},y}$ , the total becomes:
$V{\text{tot},x} = Vx + V_{\text{wind},x}$
$V{\text{tot},y} = Vy + V_{\text{wind},y}$

Example from the slides: If the object has $Vx = 2 \text{ m/s}$ and $Vy = 1 \text{ m/s}$ , and a wind adds $V_{\text{wind}} = (1 \text{ m/s} \text{ in the x-direction})$ while not altering y, then:
$V_{\text{tot},x} = 2 + 1 = 3 \text{ m/s}$
$V_{\text{tot},y} = 1 \text{ m/s}$

Resultant magnitude and direction follow from the standard formulas: $v{\text{tot}} = \sqrt{V{\text{tot},x}^2 + V{\text{tot},y}^2}$ and $\theta{\text{tot}} = \tan^{-1}\left(\frac{V{\text{tot},y}}{V{\text{tot},x}}\right)$ .
The slides show a concrete set: $v = 2.2 \text{ m/s}, Vx = 2 \text{ m/s}, Vy = 1 \text{ m/s}, V{\text{wind}} = 1 \text{ m/s}$ . The resulting components are: $V{\text{tot},x} = 2 + 1 = 3\ \text{m/s},$ $V{\text{tot},y} = 1\ \text{m/s}.$ The magnitude is then $v{\text{tot}} = \sqrt{3^2 + 1^2} = \sqrt{10} \approx 3.16 \text{ m/s}$ and the angle is $\theta_{\text{tot}} = \tan^{-1}(1/3) \approx 18.4°$ . (Note: The slides sometimes present slightly different numerical examples; the method remains the same.)

The Boat Across the River (Independent Motions in x and y)

A classic multi-step problem is used to illustrate independence: the boat crosses the river (y-direction) while also moving downstream (x-direction) due to a current or given x-velocity.
Given data (one common configuration in the slides):
Boat velocity in the river frame: $v = 10\ \text{m/s}$ (across or diagonal depending on setup).
River current (wind-like) produces an x-velocity component: $V_x = 5\ \text{m/s}$ .
Across-river (y) motion has vertical component: you can assign $V_y = 8.66\ \text{m/s}$ , which corresponds to a vertical component chosen by the problem (often from a given angle, such as 60° with the horizontal for the boat’s intended path).
Across-river displacement to cross: $\Delta y = 40 \text{ m}$ .

Time to reach the opposite bank (y-direction):
Since $\Delta y = 40\ \text{m}$ and the vertical velocity is $V_y = 8.66\ \text{m/s}$ , the time to cross is
$t = \frac{\Delta y}{V_y} = \frac{40}{8.66} \approx 4.6\ \text{s}.$

How far downstream (x-direction) does it move in that time? If the horizontal velocity is $V_x = 5\ \text{m/s}$ , then the downstream distance is
$\Delta x = V_x t = 5 \times 4.6 \approx 23\ \text{m}.$

This problem explicitly demonstrates independence: the x-motion (downstream) does not affect the y-motion (across), and vice versa. The total resultant velocity can be found by combining components: $V{\text{tot}} = \sqrt{Vx^2 + Vy^2}$ and the direction by $\theta{\text{tot}} = \tan^{-1}\left(\frac{Vy}{Vx}\right)$ .

No Friction and Newton’s First Law (Curling Stone Example)

The slides mention a curling stone on ice with effectively no friction, so its horizontal velocity remains constant if no external force acts on it. This is a direct illustration of Newton's First Law: an object in motion will continue at a constant speed in a straight line unless acted on by an unbalanced force.
This concept underpins the idea that, in the absence of external forces (like drag or wind), velocity magnitude and direction remain fixed in 2D motion along a straight path.

Magnitude, Direction, and the “Pythagorean” Relationship

When given components, the magnitude v can be found by: $v = \sqrt{Vx^2 + Vy^2}$ . When the direction is given, or when it needs to be found: $\theta = \tan^{-1}\left(\frac{Vy}{Vx}\right)$ .
The slides show a key example where $Vx = 2 \text{ m/s}$ and $Vy = 1 \text{ m/s}$ , yielding:
Magnitude: $v = \sqrt{2^2 + 1^2} = \sqrt{5} \approx 2.24\ \text{m/s}.$
Angle: $\theta = \tan^{-1}(1/2) \approx 26.6°$ .

Another common example uses $Vy = 1.7 \text{ m/s}$ and $Vx = 1 \text{ m/s}$ , giving: $v = \sqrt{1^2 + 1.7^2} = \sqrt{1 + 2.89} = \sqrt{3.89} \approx 1.97\ \text{m/s},$ and $\theta = \tan^{-1}(1.7/1) \approx 60°$ . These examples appear repeatedly to illustrate how components determine both magnitude and direction.

Wind, Wind-Driven Velocity, and Vector Addition

In several slides, wind (or current) is introduced as an additional velocity component that adds vectorially to the object’s motion.
Key takeaways:
The total velocity is the vector sum of the object’s velocity and the wind/current velocity: $\vec{V}{\text{tot}} = \vec{V} + \vec{V}{\text{wind}}$ .
Components add independently: $V{\text{tot},x} = Vx + V{\text{wind},x},\quad V{\text{tot},y} = Vy + V{\text{wind},y}$ .
The resulting speed and direction follow from the standard magnitude and angle formulas:
$v{\text{tot}} = \sqrt{V{\text{tot},x}^2 + V_{\text{tot},y}^2}$
$\theta{\text{tot}} = \tan^{-1}\left(\frac{V{\text{tot},y}}{V_{\text{tot},x}}\right)$ .

Concrete example (from the notes): with $Vx = 2 \text{ m/s}, Vy = 1 \text{ m/s}$ , and wind $V{\text{wind}} = (1 \text{ m/s} \text{ in the x-direction})$ and no vertical wind, the total components become $V{\text{tot},x} = 3\ \text{m/s},\quad V{\text{tot},y} = 1\ \text{m/s}.$ The resulting speed is $v{\text{tot}} = \sqrt{3^2 + 1^2} = \sqrt{10} \approx 3.16\ \text{m/s},$ and the direction is $\theta_{\text{tot}} = \tan^{-1}(1/3) \approx 18.4°$ . (Numbers shown in the slides may vary slightly due to rounding, but the method remains the same.)
When wind changes, it often changes the resultant path, even if the original intended direction of motion is fixed.

Practical Application: Summary of Procedures

Decompose any 2D velocity into components:
Given speed v and direction e: $Vx = v \cos(e), \quad Vy = v \sin(e)$ .

If you know the components, compute magnitude and direction:
$v = \sqrt{Vx^2 + Vy^2}, \quad \theta = \tan^{-1}\left(\frac{Vy}{Vx}\right)$ .

For problems with additional velocity sources (wind/current):
Compute total components: $V{\text{tot},x} = Vx + V{\text{wind},x}, \quad V{\text{tot},y} = Vy + V{\text{wind},y}$ .
Then compute the total speed and direction as above.

Independence of x and y motions ensures that one can solve for the time to cover a vertical distance independently of the horizontal movement. If given $\Delta y$ and $Vy$ , time is $t = \frac{\Delta y}{Vy}$ . If you also know $Vx$ and t, the horizontal displacement is $\Delta x = Vx t$ . These relationships are demonstrated in the river-crossing problem.

Worked Examples (Key Numbers Reiterated)

Example A: 2D vector with $Vx = 2 \text{ m/s}, Vy = 1 \text{ m/s}$
Magnitude: $v = \sqrt{2^2 + 1^2} = \sqrt{5} \approx 2.24\,\text{m/s}$
Direction: $\theta = \tan^{-1}(1/2) \approx 26.6°$ .

Example B: 2D vector with $v = 2 \text{ m/s}$ at $60°$
$V_x = v \cos(60°) = 1\text{m/s},$
$V_y = v \sin(60°) = 1.73\text{m/s} \approx 1.7\text{m/s}.$
Magnitude: $v = \sqrt{1^2 + 1.73^2} \approx 2.0\text{m/s}.$

Example C: Boat across river ( $\Delta y = 40 \text{ m}, V_y = 8.66 \text{ m/s}$ )
Time to cross: $t = \frac{\Delta y}{V_y} = \frac{40}{8.66} \approx 4.6\text{s}.$
Downstream displacement: with $Vx = 5 \text{ m/s}$ , $\Delta x = Vx t \approx 5 \times 4.6 = 23\text{m}.$

Example D: Wind adds $V{\text{wind}} = 1 \text{ m/s}$ in x-direction; original velocity $V = (Vx, V_y) = (2, 1)$
Total components: $V{\text{tot},x} = 2 + 1 = 3\text{m/s},\quad V{\text{tot},y} = 1\text{m/s}.$
Total speed: $v_{\text{tot}} = \sqrt{3^2 + 1^2} = \sqrt{10} \approx 3.16\text{m/s},$
Direction: $\theta_{\text{tot}} = \tan^{-1}(1/3) \approx 18.4°$ .

Quick Reference Formulas (LaTeX)

Decomposition: $Vx = v \cos(e), \quad Vy = v \sin(e)$
Magnitude: $v = \sqrt{Vx^2 + Vy^2}$
Direction: $\theta = \tan^{-1}\left(\frac{Vy}{Vx}\right)$
Total components with wind/current: $V{\text{tot},x} = Vx + V{\text{wind},x}, \quad V{\text{tot},y} = Vy + V{\text{wind},y}$
Total speed and direction: $v{\text{tot}} = \sqrt{V{\text{tot},x}^2 + V{\text{tot},y}^2}, \quad \theta{\text{tot}} = \tan^{-1}\left(\frac{V{\text{tot},y}}{V{\text{tot},x}}\right)$
Time to cover vertical displacement: $t = \frac{\Delta y}{V_y}$
Horizontal displacement in that time: $\Delta x = V_x \cdot t$

Final Notes and Real-World Relevance

The independence of x and y motions is fundamental for analyzing projectiles, boats crossing rivers, and any motion in two dimensions. It allows us to treat horizontal and vertical motions separately and then combine results.
Velocity vectors and their components enable us to predict trajectories, crossing times, and drift under environmental influences like wind or currents.
The physical interpretation of velocity components helps in designing navigation strategies, sports trajectories (curling stone, curling paths), and any scenario where mixed motions occur.

Quick Recap of Key Points

Velocity is a vector; decompose into components: $Vx, Vy$ .
Magnitude and direction are given by: $v = \sqrt{Vx^2 + Vy^2}, \theta = \tan^{-1}(Vy/Vx)$ .
Environmental effects (wind/current) add vectorially to motion: $\vec{V}{\text{tot}} = \vec{V} + \vec{V}{\text{wind}}$ .
In problems with two perpendicular motions, solve each direction separately, then combine for the overall motion.
Numerical examples throughout illustrate how the formulas are applied in practice, with common values like $v = 2\,\text{m/s}, \ Vx = 1\,\text{m/s}, \ Vy = 1.7\,\text{m/s}$ , and angles such as $60°$ and $26.6°$ .

Appendix: Problem-Solving Checklist

Identify independent directions (x and y).
Determine what is given for each direction ( $Vx, Vy$ , speed v, angle e, distances $\Delta x, \Delta y$ ).
Compute components if needed using $Vx = v \cos(e)$ and $Vy = v \sin(e)$ .
If there is wind/current, add components: $V{\text{tot},x} = Vx + V{\text{wind},x}, \ V{\text{tot},y} = Vy + V{\text{wind},y}$ .
Compute total speed and direction: $v{\text{tot}} = \sqrt{V{\text{tot},x}^2 + V{\text{tot},y}^2}, \theta{\text{tot}} = \tan^{-1}\left(\frac{V{\text{tot},y}}{V{\text{tot},x}}\right)$ .
For crossing problems: use vertical displacement to find time, then horizontal displacement from the time.
Always check units and signs for directions; ensure consistency with the chosen coordinate axes.