Scalars and vectors

Some quantities are fully described by one number and a unit.

The mass of a skateboard, the temperature of a gym, the time a song lasts: 3.2 kg, 21 °C, 3 minutes.

Others are not.

If a friend says "walk 200 m from the school entrance", you cannot follow the instruction until you know which way.

Physics separates these two kinds of quantity from the very first page, because they obey different arithmetic.

Two 3 kg masses always make 6 kg.

Two 3 m walks can leave you 6 m from where you started, 0 m from it, or anywhere in between, depending on their directions.

This section builds the language of vectors that every later unit uses: forces, momentum, torque and fields are all vectors, and the methods you learn here (arrows, components, unit vectors, adding by components) are the methods you will use for all of them.

On the exam. Unit 1 makes up about 10 to 15% of the multiple-choice section, and any unit can be examined in any of the four free-response questions: Mathematical Routines (MR, 10 points), Translation Between Representations (TBR, 12), Experimental Design and Analysis (LAB, 10) and Qualitative/Quantitative Translation (QQT, 8).

The AP Physics C: Mechanics exam gives you 42 multiple-choice questions in 85 minutes and the four free-response questions in 95 minutes, with a calculator allowed throughout.

The kinematics tools of this unit (vectors, derivatives and integrals of motion, graph readings) also run through free-response questions set mainly on later units, so they earn points far beyond this unit's own share.

You need to know the two kinds of quantity by these exact descriptions, because a Section I question may ask which kind a quantity is, and every later derivation depends on treating each kind correctly.

The magnitude of a vector is its size, always written as a positive number with units: "5.0 m" or "12 m/s".

The direction says which way it points: "north", "37° above the $+x$-axis", "down the incline".

A vector's symbol carries an arrow, $\vec{v}$, and its magnitude is written $|\vec{v}|$ or just $v$.

The pairs in the third term are worth knowing as pairs, because the exam likes to separate them:

Is the distinction only bookkeeping?

No: it changes answers.

Walk 300 m east and then 400 m north.

The distance traveled is 700 m, but you are only 500 m from where you started (a 3-4-5 triangle).

Speed and velocity part company in the same way, and in Unit 2 the fact that velocity is a vector is exactly why an object moving in a circle at constant speed needs a net force.

The second key term is a drawing rule, and the exam scores drawings by it.

An arrow's direction shows the vector's direction, and its length is proportional to the magnitude: on a grid where one square stands for 10 m/s, a 30 m/s velocity is three squares long.

Two arrows of equal magnitude must be drawn equal in length, and an arrow twice as large must be twice as long.

The arrows you draw later in the course (forces on a dot in Unit 2, momentum arrows on a grid in Unit 4) are scored on exactly this: the right direction, and relative lengths that match the physics.

Two vectors are equal when they have the same magnitude and the same direction, wherever they are drawn.

That is what lets you slide an arrow across a page, keeping its length and direction, to add it tip to tail.

Any vector in a plane can be split into two perpendicular pieces, its components.

In the diagram below, a vector $\vec{A}$ of magnitude $A$ makes an angle $\theta$ with the $+x$-axis.

Its components are the "shadows" it casts on the two axes:

These come straight from the right-triangle definitions $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ and $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$, with $A$ as the hypotenuse.

They hold only when $\theta$ is measured from the $x$-axis.

If an angle is measured from the vertical (a rope "30° from the vertical"), the cosine goes with the $y$-component instead.

Draw the triangle every time and read off which side is adjacent.

Going the other way, from components back to magnitude and direction, uses Pythagoras and the tangent:

Check the quadrant from the signs of the components before trusting a calculator's inverse tangent: $A_x = -3$, $A_y = 4$ and $A_x = 3$, $A_y = -4$ give the same ratio's size but point in different quadrants.

Unit vectors. A unit vector has magnitude exactly 1 and no units; its only job is to point.

The unit vectors along the $x$-, $y$- and $z$-axes are written $\hat{\imath}$, $\hat{\jmath}$ and $\hat{k}$.

Writing a vector as a sum of components times unit vectors packages magnitude and direction into one expression: a displacement of 4.0 m east and 3.0 m north (taking $+x$ east and $+y$ north) is $\Delta\vec{r} = (4.0\,\hat{\imath} + 3.0\,\hat{\jmath})$ m.

You need to know the course's own statement of this, with its sub-parts:

The position vector and $\hat{r}$. The position vector $\vec{r}$ of a point runs from the origin to the point.

Its magnitude $r$ is the point's distance from the origin, and the unit vector $\hat{r}$ points the same way with length 1, so $\vec{r} = r\,\hat{r}$ and $\hat{r} = \vec{r}/r$.

You will meet $\hat{r}$ again in Unit 2, where the gravitational force between two objects acts along the line joining them, and in Unit 5, where torque is built from $\vec{r}$.

A position in three dimensions needs all three unit vectors.

A drone hovering at $\vec{r} = (3.0\,\hat{\imath} + 4.0\,\hat{\jmath} + 12\,\hat{k})$ m from a launch pad is a distance $r = \sqrt{3.0^2 + 4.0^2 + 12^2} = \sqrt{169} = 13$ m away, and $\hat{r} = \tfrac{1}{13}(3.0\,\hat{\imath} + 4.0\,\hat{\jmath} + 12\,\hat{k})$: a dimensionless vector whose own magnitude is $\sqrt{(9 + 16 + 144)/169} = 1$, as a unit vector must be.

In this course you analyze motion quantitatively in two dimensions; three-dimensional motion is described only in words.

The third sub-part of the unit vector term is the rule for adding: add the components.

If $\vec{A} = A_x\hat{\imath} + A_y\hat{\jmath}$ and $\vec{B} = B_x\hat{\imath} + B_y\hat{\jmath}$, the resultant is

Why does this work?

Each unit vector direction is independent of the others: walking 2 m east never changes how far north you are.

So the east parts add like ordinary numbers, and so do the north parts.

The drawing method says the same thing geometrically: place the tail of $\vec{B}$ on the tip of $\vec{A}$, and the resultant runs from the tail of $\vec{A}$ to the tip of $\vec{B}$.

Both methods give the same answer; components are more precise, and the drawing tells you whether the answer is sensible.

Subtraction is addition of the opposite vector: $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$, where $-\vec{B}$ has the same magnitude as $\vec{B}$ and the opposite direction.

In components, subtract component by component.

You will use subtraction constantly from the next section on, because a change in a vector quantity, such as a displacement $\Delta\vec{r} = \vec{r}_2 - \vec{r}_1$ or a change in velocity $\Delta\vec{v}$, is a vector difference.

Is the size of a resultant sensible?

Its magnitude can never exceed the sum of the magnitudes, $|\vec{A} + \vec{B}| \le A + B$, with equality only when the two point the same way, and it can never be less than $|A - B|$, reached when they point opposite ways.

Any answer outside that range is wrong before you check the arithmetic.

When all motion is along one line, you do not need unit vectors.

You choose one direction as positive and write every vector quantity as a signed number.

So with "up" chosen as positive, a ball's velocity of $-6.0$ m/s means 6.0 m/s downward, and the acceleration due to gravity is $-9.8$ m/s².

The sign is a direction, not a size: $-6.0$ m/s is a faster motion than $+2.0$ m/s.

Choose the positive direction once, write it down ("up is positive"), and keep it for the whole problem.

A single sign error early in a chain often turns a correct method into a wrong final expression, and a stated convention makes every sign after it checkable.

The equation sheet (its full name is the AP Physics C: Mechanics Table of Information, and you have it for the whole exam) prints the unit's constant-acceleration equations, $\Delta x = \int v_x(t)\,dt$ and $\Delta v_x = \int a_x(t)\,dt$, plus vector rules, right-triangle trigonometry and a table of derivatives and integrals.

What it leaves out you carry yourself: the definitions $v_x = dx/dt$ and $a_x = dv_x/dt$, average velocity and acceleration, when each equation applies, and how to read motion graphs.

This section uses the vector part.

Used again in: every later unit.

Forces are added by components in Unit 2, work uses the dot product in Unit 3, momentum is conserved component by component in Unit 4, and torque uses the cross product in Unit 5.

The next worked example adds two displacements with a unit conversion.

The diagram below draws them to scale, tip to tail, so you can check the calculation by eye.

Section I can ask for two quantities at once and sets the options as a table, one row per option.

The next worked example is a question of that kind.

The full-credit drawing for part A is shown below.

It is worth comparing with two weaker answers.

A student who draws the second arrow from the origin, not from the tip of the first, has shown the two displacements but not added them, and earns Point A1 only if the directions are right.

A student who draws the second arrow twice as long as the first has broken the lengths-proportional rule and loses Point A2.

1. A student walks 3.0 m east and then 4.0 m north. Which row correctly gives the distance the student walks and the magnitude of the student's displacement?

Total [1]

2. A football leaves a kicker's foot with a velocity of 20 m/s at 53° above the horizontal. The horizontal component of the ball's initial velocity is most nearly

(A) 12 m/s (B) 16 m/s (C) 20 m/s (D) 33 m/s

Total [1]

3. Two vectors are $\vec{A} = a\,\hat{\imath} + 2a\,\hat{\jmath}$ and $\vec{B} = -a\,\hat{\imath} + a\,\hat{\jmath}$, where $a$ is a positive constant. What is the magnitude of $\vec{A} - \vec{B}$?

(A) $a$ (B) $\sqrt{5}\,a$ (C) $3a$ (D) $\sqrt{13}\,a$

Total [1]

4. A robotic lawn mower starts at point O on a flat lawn. It moves with displacement $\vec{d}_1 = (4.0\,\hat{\imath} + 2.0\,\hat{\jmath})$ m and then with displacement $\vec{d}_2 = (-1.0\,\hat{\imath} + 3.0\,\hat{\jmath})$ m, where $+x$ is east and $+y$ is north. Figure 1 is a grid on which each square is 1.0 m.

A. On the grid in Figure 1, draw arrows to represent $\vec{d}_1$, $\vec{d}_2$ placed tip to tail starting at O, and the resultant displacement $\vec{D}$.

B. Determine the magnitude and direction of $\vec{D}$.

C. Indicate whether the magnitude of $\vec{D}$ is greater than, less than, or equal to $|\vec{d}_1| + |\vec{d}_2|$ by writing one of the following.

• greater than

• less than

• equal to

Briefly justify your answer.

Total [6]

5. The minute hand of a wall clock is 15 cm long. Indicate whether the distance traveled by the tip of the minute hand between 12:00 and 12:30 is greater than, less than, or equal to the magnitude of its displacement over the same time by writing one of the following.

• greater than

• less than

• equal to

Briefly justify your answer.

Total [2]

6. A laser pointer is mounted at the origin of a lab coordinate system. A small target sits at position $\vec{r} = (30\,\hat{\imath} - 40\,\hat{\jmath} + 120\,\hat{k})$ cm. Calculate the distance from the origin to the target in meters, and write the unit vector $\hat{r}$ that gives the direction from the origin to the target.

Total [3]

7. A vector $\vec{v}$ has components $v_x = -6.0$ m/s and $v_y = 8.0$ m/s. Which of the following correctly describes $\vec{v}$?

(A) 10 m/s at 53° above the $+x$-axis (B) 10 m/s at 53° above the $-x$-axis (C) 14 m/s at 37° above the $-x$-axis (D) 2.0 m/s at 53° above the $-x$-axis

Total [1]