Lesson 0: Vectors · First Principles Physics

Lesson 00 · Prerequisite

Vectors.

The math you need before Unit 1. Most students who struggle with kinematics are actually struggling with vectors. If any of this feels shaky, work through it before moving on — every unit after this one assumes you've got it cold.

Prerequisite for · All 8 units Estimated time · 60–90 minutes Difficulty · Foundation

The lesson

Why arrows are the secret language of physics

A car goes 60 mph. North or south? It matters enormously. Drive north for an hour and you end up somewhere completely different than driving south for an hour, even though both trips covered the same distance. The number alone — "60 mph" — doesn't tell you where you'll end up. You need direction too. That's a vector: a number with a direction attached.

Physics is largely the study of how things move and interact in space, which means it's largely the study of vectors. Position, velocity, acceleration, force, momentum — all vectors. Mass, time, temperature, energy — all scalars (just numbers). The first thing to learn about any physical quantity you meet is: vector or scalar? Because the math you use to combine them is completely different.

The single most important sentence in this lesson: vectors don't add like numbers do. Two forces of 3 N each don't necessarily add up to 6 N — they add up to anywhere from 0 N to 6 N depending on their directions. Until you understand this, every problem in AP Physics 1 will feel harder than it is.

Scalars vs. vectors

A scalar is a quantity that's fully described by a single number with units. Mass is 5 kg, end of story. Time is 30 seconds, done. Temperature is 295 K, complete information.

A vector is a quantity that requires two pieces of information: a magnitude (how big) and a direction (which way). Velocity is "30 m/s north" — both the speed and the direction matter. Force is "20 N at 30° above horizontal" — both the strength and the angle matter.

Here's a quick reference:

Scalars (just a number)	Vectors (number + direction)
Mass, Time	Position, Displacement
Speed, Distance	Velocity
Energy, Work	Acceleration
Temperature	Force
Density	Momentum

Notice some pairs that get confused: speed is a scalar (just how fast); velocity is a vector (how fast AND which way). Distance is a scalar (total path length); displacement is a vector (straight-line shortcut from start to end, with direction). The distinction sounds picky but matters constantly — a car driving in a circle has constant speed but constantly changing velocity (because direction changes).

How vectors are drawn and written

Vectors are usually drawn as arrows. The length of the arrow represents the magnitude (longer arrow = bigger vector). The direction the arrow points represents — surprise — the direction of the vector.

In writing, vectors are denoted with an arrow on top: $\vec{v}$ means "the vector v." The magnitude of that vector (just its size, no direction) is written either as $|\vec{v}|$ or simply $v$ without the arrow. So when you see $v = 5$ m/s in a physics problem, that's the magnitude of some velocity vector. When you see $\vec{v}$, that's the whole vector — magnitude AND direction.

A notation warning

Some textbooks use bold letters instead of arrows: v means the vector, while plain italic $v$ means the magnitude. Other books underline (v) or use a half-arrow. They all mean the same thing. AP Physics 1 questions typically use the arrow notation, but you'll see all of them. Don't let notation throw you off.

Two ways to describe the same vector

Any 2D vector can be described in two equivalent ways. You'll switch between them constantly in physics, so it's worth getting comfortable with both right now.

Magnitude-and-direction form: "5 m/s at 37° above the +x axis." This is intuitive — it matches how you'd describe a vector in words. Often the direction is given relative to a reference (north, east, the horizontal, etc.).

Component form: "(4 m/s in the +x direction, 3 m/s in the +y direction)." Or written more compactly: $v_x = 4$ m/s, $v_y = 3$ m/s. This breaks the vector into perpendicular pieces along the coordinate axes — usually horizontal ($x$) and vertical ($y$).

These two descriptions look different but represent the same arrow. The first description tells you where the tip of the arrow is using polar-coordinate language (a length and an angle). The second tells you where the tip is using rectangular coordinates (an x-coordinate and a y-coordinate).

Decomposing a vector into components

Given a vector with magnitude $v$ at angle $\theta$ above the +x axis, the components are:

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

This comes straight from right-triangle trigonometry. The vector is the hypotenuse, and the x and y components are the horizontal and vertical legs of a right triangle. The angle $\theta$ is measured from the +x axis going counterclockwise (the standard convention).

Worked example 1 · decomposition

A velocity at an angle

A ball moves at 10 m/s at 30° above the horizontal. Find its x and y components.

Apply the formulas

$v_x = (10)\cos 30° = 10(0.866) \approx 8.66$ m/s.

$v_y = (10)\sin 30° = 10(0.5) = 5.0$ m/s.

Sanity check

The angle is 30° — pretty shallow, so most of the motion should be horizontal. Indeed, $v_x = 8.66$ is bigger than $v_y = 5.0$. Good. Also, the components should always be smaller than the magnitude (they're legs of a right triangle, the magnitude is the hypotenuse). Both 8.66 and 5.0 are less than 10. Good.

$v_x \approx 8.66$ m/s, $v_y = 5.0$ m/s.

Going the other way: components → magnitude and direction

Given the components, you can recover the magnitude and direction using the Pythagorean theorem and inverse tangent:

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

$$\theta = \tan^{-1}\!\left(\dfrac{v_y}{v_x}\right)$$ angle above +x axis

The magnitude formula is just the Pythagorean theorem applied to the right triangle formed by the components. The angle formula is the inverse tangent of "opposite over adjacent."

A trap with inverse tangent

The $\tan^{-1}$ function on your calculator only returns angles between −90° and +90°. If your vector is in the second or third quadrant (where $v_x$ is negative), the calculator's answer will be wrong by 180°. Always draw the vector to make sure your computed angle puts it in the right quadrant. If it doesn't, add or subtract 180° to fix it.

Adding vectors: tip-to-tail (graphical)

To add two vectors graphically: draw the first vector. Then draw the second vector starting from the tip of the first vector. The sum (called the resultant) is the arrow drawn from the tail of the first to the tip of the second.

This works because vector addition represents physical "and-then" combinations. If you walk 3 m east and then 4 m north, your total displacement is the diagonal from start to finish — a vector of magnitude 5 m (by the Pythagorean theorem) at angle $\tan^{-1}(4/3) \approx 53°$ north of east.

The graphical method is intuitive but not always accurate. For real problems, use components instead.

Adding vectors: components (the right way)

The component method is the workhorse of vector addition, and you'll use it in every unit. The recipe:

Break each vector into components using $v_x = v\cos\theta$, $v_y = v\sin\theta$.
Add the x-components: $R_x = A_x + B_x + C_x + \ldots$.
Add the y-components separately: $R_y = A_y + B_y + C_y + \ldots$.
Recombine into magnitude and direction using $|\vec R| = \sqrt{R_x^2 + R_y^2}$ and $\theta = \tan^{-1}(R_y/R_x)$.

The key insight: x-components only add with x-components, y-components only with y-components. They don't mix. This is why decomposition is so powerful — it splits a hard 2D problem into two easy 1D problems.

Worked example 2 · vector addition

A walker changes direction

A person walks 5 m east, then 8 m at 60° north of east. Find their total displacement (magnitude and direction).

Break each leg into components

Leg 1 (5 m east): $A_x = 5$, $A_y = 0$.

Leg 2 (8 m at 60° N of E): $B_x = 8\cos 60° = 4$, $B_y = 8\sin 60° \approx 6.93$.

Add the components

$R_x = 5 + 4 = 9$ m. $R_y = 0 + 6.93 = 6.93$ m.

Recombine

$|\vec R| = \sqrt{81 + 48} = \sqrt{129} \approx 11.4$ m.

$\theta = \tan^{-1}(6.93/9) \approx 37.6°$ north of east.

Total displacement: 11.4 m at 37.6° north of east.

Subtracting vectors

Subtracting vectors is just adding the negative. The negative of a vector is the same magnitude but pointing in the opposite direction. So if $\vec{B}$ points east at 5 m/s, then $-\vec{B}$ points west at 5 m/s.

To compute $\vec{A} - \vec{B}$: first find $-\vec{B}$ by flipping its direction (or equivalently, by negating both its components). Then add $\vec{A} + (-\vec{B})$ using the component method.

The most important application is finding change in velocity: $\Delta\vec{v} = \vec{v}_f - \vec{v}_i$. This shows up constantly in Unit 1 (kinematics) and Unit 4 (momentum). Even when an object's speed doesn't change, if its direction changes, $\Delta\vec{v}$ is nonzero — and that nonzero $\Delta\vec{v}$ implies acceleration.

Multiplying a vector by a scalar

If you multiply a vector by a positive scalar $c$, you scale its magnitude by $c$ but keep the direction the same. If $\vec{v}$ is 5 m/s east, then $3\vec{v}$ is 15 m/s east. If you multiply by a negative scalar, you scale AND flip the direction. So $-2\vec{v}$ is 10 m/s west.

In component form: just multiply each component by the scalar. $c\vec{v} = (cv_x, cv_y)$.

Unit vectors (optional but useful)

Some textbooks write vectors using unit vectors: $\hat{i}$ is a vector of magnitude 1 pointing in the +x direction, $\hat{j}$ is magnitude 1 in the +y direction. A vector can be written as:

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

For example, "3 m/s east plus 4 m/s north" becomes $\vec{v} = 3\hat{i} + 4\hat{j}$ m/s. It's just compact notation for component form. AP Physics 1 doesn't require this notation, but you'll see it in textbooks and in AP Physics C if you take it later. Worth knowing.

Why this matters for AP Physics 1

Here's a quick preview of where vectors show up in each unit:

Unit 1 (Kinematics): position, velocity, acceleration are all vectors. Projectile motion is the canonical 2D vector problem — horizontal and vertical components handled separately.
Unit 2 (Forces): forces are vectors. To find net force, decompose every force into x and y components, add them separately, then recombine.
Unit 4 (Momentum): momentum is a vector. Collisions in 2D require treating x and y components independently.
Unit 5 (Rotation): torque, angular velocity, angular momentum — all vectors (though for AP Physics 1, mostly along an axis perpendicular to the page, so they reduce to scalars with a +/- sign).

Every problem you solve in this course will, at some level, involve breaking quantities into components, adding or subtracting them, and recombining. The faster you can do that, the faster you'll move through every problem.

Conceptual summary

Vectors are quantities with both magnitude and direction (as opposed to scalars, which only have magnitude). Any 2D vector can be described in two equivalent ways: magnitude-and-direction (a length and an angle) or component form ($v_x$ and $v_y$, the projections onto the x and y axes). The formulas $v_x = v\cos\theta$, $v_y = v\sin\theta$ convert between them. To add vectors, break them into components, add x-components together and y-components together separately, then recombine. To subtract, add the negative. To multiply by a scalar, scale the components. The component method is the foundation of essentially every 2D problem in AP Physics 1.

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Reference sheet

Every formula you need for vectors

★ Components from magnitude/angle

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

$v$: magnitude of the vector
$\theta$: angle measured from +x axis (counterclockwise positive)

Use when you have a vector's magnitude and direction and need its x, y components.

★ Magnitude from components

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

Pythagorean: theorem applied to components

Use when you have the components and need the magnitude (length of the vector).

★ Angle from components

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

Watch for: quadrant ambiguity — draw the vector to check

Use when you have the components and need the angle of the vector.

★ Vector addition (components)

$$R_x = A_x + B_x$$ $$R_y = A_y + B_y$$

Then: combine $R_x, R_y$ into magnitude and direction

Use when adding two or more vectors. Always work in components.

Vector subtraction

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

$-\vec B$: same magnitude, opposite direction

Use when finding "change in" something: $\Delta\vec v$, $\Delta\vec p$. Flip signs of $\vec B$'s components, then add.

Scalar multiplication

$$c\vec v = (cv_x,\, cv_y)$$

If $c > 0$: scales magnitude, keeps direction
If $c < 0$: scales AND flips direction

Use when scaling a vector by a number — e.g., $\vec F = m\vec a$ (mass times acceleration).

Unit vector form (optional)

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

$\hat i$: unit vector in +x direction
$\hat j$: unit vector in +y direction

Use when a textbook uses this notation. Same info as component form, just compact.

Trig values worth memorizing

$$\sin 0° = 0,\, \sin 30° = 0.5$$ $$\sin 45° \approx 0.71,\, \sin 60° \approx 0.87$$ $$\sin 90° = 1$$

Use when doing component math without a calculator. Cosines are just sines flipped: $\cos 60° = 0.5$, $\cos 30° \approx 0.87$, etc.

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Study guide

One page. Before you start Unit 1.

The 5 concepts that matter most

Vector vs. scalar. Vectors (force, velocity, displacement) have direction; scalars (mass, energy) don't.
Every 2D vector has TWO equivalent descriptions. Magnitude + angle, or x-component + y-component. Master the conversion.
Decompose: $v_x = v\cos\theta$, $v_y = v\sin\theta$. Recombine: $v = \sqrt{v_x^2 + v_y^2}$, $\theta = \tan^{-1}(v_y/v_x)$.
To add vectors, add components. X with x, y with y. Never add magnitudes directly unless they're along the same line.
Subtracting vectors = adding the negative. Negative means same magnitude, opposite direction.

Key equations

$v_x = v\cos\theta$, $v_y = v\sin\theta$
$|\vec v| = \sqrt{v_x^2 + v_y^2}$
$\theta = \tan^{-1}(v_y/v_x)$
$R_x = A_x + B_x$, $R_y = A_y + B_y$
$\vec A - \vec B = \vec A + (-\vec B)$
$c\vec v = (cv_x, cv_y)$

5 traps to watch for

Adding vector magnitudes directly when they're not along the same line.
Using sine when you mean cosine (or vice versa) — sine goes with the opposite side, cosine with the adjacent.
$\tan^{-1}$ returning the wrong quadrant — always draw the vector to verify.
Forgetting that distance and displacement are different — distance is scalar (path length), displacement is vector (start to end).
Mixing up speed and velocity — speed is scalar, velocity is vector with direction.

The vector-addition recipe (memorize this)

Draw a picture. Sketch all vectors with rough lengths and directions.
Set up x-y axes. Usually horizontal and vertical, but you can rotate if convenient (e.g., along an incline).
Decompose every vector into x and y components: $v_x = v\cos\theta$, $v_y = v\sin\theta$. Mind your signs.
Add components separately: $R_x = $ sum of all $v_x$; $R_y = $ sum of all $v_y$.
Recombine: $|\vec R| = \sqrt{R_x^2 + R_y^2}$, $\theta = \tan^{-1}(R_y/R_x)$. Check quadrant.

If you only remember one thing X-components only add with x-components. Y-components only add with y-components. That single fact lets you turn every 2D vector problem into two independent 1D problems, which are much easier to solve. The component method is the master tool of AP Physics 1.

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Practice problems

Twelve problems to test your fluency

If you can solve all twelve quickly and correctly, you're ready for Unit 1. If any of them give you trouble, re-read the relevant part of the lesson before moving on.

Foundation

Identify each quantity as vector or scalar: (a) mass, (b) temperature, (c) velocity, (d) energy, (e) force, (f) speed, (g) displacement, (h) time.

Foundation

A vector has magnitude 20 m/s and points at 45° above the +x axis. Find its x and y components.

Foundation

A vector has components $v_x = 6$ and $v_y = 8$. Find its magnitude.

Foundation

A force has magnitude 50 N and points at 60° above the horizontal. Find its horizontal and vertical components.

Moderate

A vector has components $v_x = -3$ and $v_y = 4$. Find both its magnitude and its angle (measured from the +x axis, counterclockwise positive). Be careful about the quadrant.

Moderate

A person walks 4 m east, then 3 m north. Find the magnitude and direction of their total displacement.

Moderate

Two forces act on an object: $\vec F_1$ = 30 N east, and $\vec F_2$ = 40 N north. Find the magnitude and direction of the net force.

Moderate

A ball has initial velocity $\vec v_i$ = 5 m/s east. After some time, its velocity is $\vec v_f$ = 5 m/s north. Find the change in velocity $\Delta\vec v = \vec v_f - \vec v_i$ (magnitude and direction).

Moderate

A boat's velocity relative to the water is 4 m/s north. The water's velocity relative to the ground is 3 m/s east. Find the boat's velocity relative to the ground (magnitude and direction).

Moderate

A vector $\vec A$ has magnitude 10 and points at 30° above the +x axis. A second vector $\vec B$ has magnitude 6 and points at 120° above the +x axis (i.e., 30° past straight up). Find $\vec A + \vec B$.

Moderate

Three vectors: $\vec A$ = 5 units east, $\vec B$ = 3 units at 60° N of E, $\vec C$ = 4 units west. Find their sum (magnitude and direction).

Moderate

A car drives 50 m east, then 30 m at 37° south of east. (a) What is the car's total distance traveled? (b) What is the magnitude of its displacement?

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Complete worked solutions

Every problem, every step

· identifying scalars vs. vectors

Scalars (no direction): (a) mass, (b) temperature, (d) energy, (f) speed, (h) time.

Vectors (have direction): (c) velocity, (e) force, (g) displacement.

5 scalars and 3 vectors.

Speed vs. velocity is the most commonly missed distinction. Speed = "how fast" (scalar). Velocity = "how fast AND which way" (vector). Same for distance (scalar) vs. displacement (vector).

· decomposing 20 m/s at 45°

$v_x = 20\cos 45° = 20(0.707) \approx 14.14$ m/s.

$v_y = 20\sin 45° = 20(0.707) \approx 14.14$ m/s.

$v_x \approx 14.14$ m/s, $v_y \approx 14.14$ m/s.

At 45°, the components are equal — this is the only angle where $\cos\theta = \sin\theta$. Useful sanity check for "halfway between horizontal and vertical" vectors.

· magnitude from (6, 8)

$|\vec v| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.

$|\vec v| = 10$.

(3, 4, 5) and (6, 8, 10) are classic right triangles you'll see all over physics. Memorize them — they save calculator time.

· force at 60° above horizontal

$F_x = 50\cos 60° = 50(0.5) = 25$ N.

$F_y = 50\sin 60° = 50(0.866) \approx 43.3$ N.

$F_x = 25$ N, $F_y \approx 43.3$ N.

60° is more vertical than horizontal, so the vertical component should be bigger. Yes — 43.3 > 25. Sanity check passes.

· components (-3, 4) — watch the quadrant

$|\vec v| = \sqrt{9 + 16} = \sqrt{25} = 5$.

$\tan^{-1}(4/-3)$ on a calculator gives $\approx -53.1°$ (or +126.9° depending on calculator mode). But which is correct?

The vector has $v_x < 0$ and $v_y > 0$, so it points into the second quadrant (upper-left). The correct angle is between 90° and 180°. The calculator's $-53.1°$ is wrong (it's in quadrant 4). The right answer is $180° - 53.1° = 126.9°$, measured counterclockwise from the +x axis.

$|\vec v| = 5$, $\theta \approx 126.9°$ (above +x axis, into the second quadrant).

This is the inverse-tangent trap I warned about in the lesson. ALWAYS draw the vector first to see which quadrant it's in, then adjust the calculator's answer if necessary.

· 4 m east + 3 m north

Components: $R_x = 4$, $R_y = 3$.

Magnitude: $R = \sqrt{16 + 9} = \sqrt{25} = 5$ m.

Direction: $\theta = \tan^{-1}(3/4) \approx 36.9°$ north of east.

5 m at 36.9° north of east.

The classic 3-4-5 right triangle. The diagonal across is 5 m, even though the walker covered 4 + 3 = 7 m of total path. Displacement (start-to-end vector) is less than distance (path length).

· two perpendicular forces

$R_x = 30$, $R_y = 40$. $R = \sqrt{900 + 1600} = \sqrt{2500} = 50$ N.

$\theta = \tan^{-1}(40/30) \approx 53.1°$ north of east.

50 N at 53.1° north of east.

· change in velocity (east → north)

Components: $\vec v_i = (5, 0)$, $\vec v_f = (0, 5)$.

$\Delta\vec v = \vec v_f - \vec v_i = (0-5, 5-0) = (-5, 5)$.

Magnitude: $|\Delta\vec v| = \sqrt{25 + 25} = \sqrt{50} \approx 7.07$ m/s.

Direction: $v_x < 0$, $v_y > 0$, so second quadrant. $\tan^{-1}(5/-5) = -45°$ on calculator, but actually 135° (above +x).

$\Delta\vec v \approx 7.07$ m/s at 135° from +x axis (i.e., northwest).

Even though the ball's speed didn't change (5 m/s both times), its velocity did change — by ~7.07 m/s — because direction changed. This is why circular motion has acceleration even at constant speed.

· boat + water current

Boat relative to water: $\vec v_{bw} = (0, 4)$ (north). Water relative to ground: $\vec v_{wg} = (3, 0)$ (east).

Boat relative to ground: $\vec v_{bg} = \vec v_{bw} + \vec v_{wg} = (3, 4)$.

Magnitude: $\sqrt{9 + 16} = 5$ m/s. Direction: $\tan^{-1}(4/3) \approx 53.1°$ north of east.

5 m/s at 53.1° north of east.

Relative-velocity problems are just vector addition. The boat thinks it's going north, but the current pushes it east, so the observed motion from the ground is the sum of both.

· A (10 at 30°) + B (6 at 120°)

$A_x = 10\cos 30° = 8.66$. $A_y = 10\sin 30° = 5.0$.

$B_x = 6\cos 120° = -3.0$. $B_y = 6\sin 120° = 5.20$.

$R_x = 8.66 - 3.0 = 5.66$. $R_y = 5.0 + 5.20 = 10.20$.

$|\vec R| = \sqrt{32.0 + 104.0} = \sqrt{136} \approx 11.7$.

$\theta = \tan^{-1}(10.2/5.66) \approx 61°$.

$\vec R \approx 11.7$ at 61° above +x axis.

Notice $\cos 120°$ is negative because 120° is in the second quadrant. Always trust what your calculator says about $\cos$ and $\sin$ of angles past 90° — they can be negative.

· three vectors

$A_x = 5$, $A_y = 0$.

$B_x = 3\cos 60° = 1.5$, $B_y = 3\sin 60° \approx 2.60$.

$C_x = -4$, $C_y = 0$.

$R_x = 5 + 1.5 - 4 = 2.5$. $R_y = 0 + 2.60 + 0 = 2.60$.

$|\vec R| = \sqrt{6.25 + 6.76} = \sqrt{13.01} \approx 3.61$.

$\theta = \tan^{-1}(2.60/2.5) \approx 46.1°$ above +x axis.

~3.61 units at ~46° above +x.

· distance vs. displacement

(a) Distance is the total path length, regardless of direction: $50 + 30 = 80$ m.

(b) Displacement: components of 50 m east are (50, 0). Components of 30 m at 37° south of east are $30\cos(-37°) \approx 24, 30\sin(-37°) \approx -18$. Note the negative y because "south" is in the −y direction.

$R_x = 50 + 24 = 74$. $R_y = 0 - 18 = -18$. Magnitude: $\sqrt{74^2 + 18^2} = \sqrt{5476 + 324} = \sqrt{5800} \approx 76.2$ m.

Distance = 80 m; displacement ≈ 76.2 m (slightly less, because the path bent).

Displacement is always less than or equal to distance traveled. Equal only when motion is in a straight line; less whenever the path curves or changes direction. This is the fundamental difference between the two quantities.

End of Lesson 0 · Vectors · Continue to Unit 1: Kinematics →