Worked examples — Position vector, displacement, distance

This page is a practice arena. We first lay out every kind of question this topic can throw at you — a "scenario matrix" — then work through examples that together hit every cell of that matrix. No case is left in the dark: forward-only walks, reversals, closed loops, 3D, negative signs, degenerate (zero) motion, and a word problem.

Recall Two parent "mistakes" we lean on (summarised so this page stands alone)

Parent Mistake 2 — "Displacement depends on the origin." It does not. Because displacement is a difference ${\vec{r}}_f-{\vec{r}}_i$, the origin appears in both terms and cancels. Two observers with different origins still agree on $\Delta \vec{r}$. (Proven live in Example 7b.) Parent Mistake 3 — "Displacement/distance can be negative." In 1D we write things like $-5$ m, but that "$-$" is a direction along the axis, not a negative length. Distance (a path length) is always $\ge 0$. (Illustrated in Example 2.)

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The scenario matrix

Every problem in this topic is really one of these case classes. Read this table first — each example below is tagged with the cell it fills.

Cell	Case class	What makes it tricky	Example
A	Straight one-way (1D or 2D)	distance $=	\Delta\vec r
B	Reversal on a line (there-and-back-partway)	distance $>	\Delta\vec r
C	Closed loop (start = end)	$\Delta \vec{r}=\vec{0}$ but distance $>0$	Ex 3
D	2D bent path (corner turn)	Pythagoras for the straight gap	Ex 4
E	Negative components / quadrant signs	direction of $\Delta \vec{r}$ can point into any quadrant	Ex 5
F	3D motion	magnitude needs three squares	Ex 6
G	Degenerate / zero input	no motion at all, or origin-shift test	Ex 7
H	Word problem + exam twist	translate words → vectors, then average speed vs velocity	Ex 8

We will now walk through Ex 1–8. Together they touch cells A, B, C, D, E, F, G, H — the whole grid.

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Example 1 — Cell A: straight one-way walk (the equality case)

Forecast: Before computing — guess: will distance be bigger than or equal to the magnitude of displacement? (The path is a single straight segment...)

1. Displacement components. $\Delta \vec{r}=(8-2)\hat{i}+(9-1)\hat{j}=6\hat{i}+8\hat{j}$ m. Why this step? Displacement is ${\vec{r}}_f-{\vec{r}}_i$, done coordinate by coordinate.
2. Magnitude. $|\Delta \vec{r}|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$ m. Why this step? The straight-line gap is the hypotenuse of the right triangle with legs $6$ and $8$ (Pythagoras theorem).
3. Distance. The path is that straight segment, so distance $=10$ m. Why this step? No turning, no reversing — the actual road and the straight arrow are the same line.

See Figure 1 (Ex1, Cell A). Caption: the amber arrow labelled "|dr| = 10" is the displacement straight from $A(2,1)$ to $B(8,9)$; the two white dashed lines labelled "leg 6" and "leg 8" are the horizontal and vertical legs of the right triangle whose hypotenuse is that amber arrow. Because the cyclist's road is exactly that amber line, the road length and the arrow length coincide (both $10$ m).

Verify: Because motion is straight & one-way, we expect equality: $|\Delta \vec{r}|=\text{distance}$. Indeed $10=10$ ✓. Units: metres throughout ✓. This is exactly the equality case of $|\Delta \vec{r}|\le \text{distance}$.

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Example 2 — Cell B: reversal on a line

Forecast: Guess the displacement's sign and size. Guess whether distance is $4$, $10$, or $11$ m.

1. Break the path into legs. Leg 1: $0\to 7$ (length $7$). Leg 2: $7\to 3$ (length $|3-7|=4$). Why this step? Distance sums the length of every segment, so we list them.
2. Distance. $7+4=11$ m. Why this step? Total path length = sum of leg lengths, ignoring direction.
3. Displacement. $\Delta \vec{r}=(x_f-x_i)\hat{i}=(3-0)\hat{i}=3\hat{i}$ m; magnitude $3$ m. Why this step? Displacement only sees the two endpoints ($0$ and $3$), not the detour to $7$.
4. Sign meaning. The $+3$ means the net motion points in the $+x$ direction. If she had ended at $x=-3$, we'd write $-3\hat{i}$ — the "$-$" is a direction, not a negative distance. (This is parent Mistake 3, summarised in the glossary above: a $-$ sign is a direction, path length is never negative.) Why this step? Cell B is where signs and reversals collide.

See Figure 2 (Ex2, Cell B). Caption: on the number line, the cyan arrow shows leg 1 (out from $0$ to $7$), the amber arrow shows leg 2 (back from $7$ to $3$); the white arrow underneath, labelled "dr = +3 (net)" is the net displacement. The two coloured arrows total $11$ in length while the single white arrow is only $3$ — that visible gap is exactly "distance $>$ displacement."

Verify: Check the inequality: $|\Delta \vec{r}|=3\le 11=\text{distance}$ ✓ (strict, because the path reversed). Sanity: net eastward progress is $3$ m even though she walked $11$ m of road ✓.

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Example 3 — Cell C: closed loop

Forecast: For a route that returns to its start, what must the displacement be — regardless of how big the square is?

1. Displacement. Here ${\vec{r}}_A$ is the position vector of point $A$ (the arrow from origin $O$ to $A$), defined in the glossary. Since the drone starts and ends at $A$, ${\vec{r}}_i={\vec{r}}_f={\vec{r}}_A$, so $\Delta \vec{r}={\vec{r}}_A-{\vec{r}}_A=\vec{0}$. Why this step? End point equals start point, so the two position vectors are identical and their difference is the zero vector.
2. Distance = perimeter. $AB+BC+CD+DA=5+5+5+5=20$ m. Why this step? Distance sums every walked edge; a square of side $5$ has four sides of $5$.
3. Interpretation. $|\Delta \vec{r}|=0$ but distance $=20$ — the extreme of the inequality: net nowhere, but plenty of road. Why this step? Closed loops are the sharpest illustration of "Displacement = Dots, Distance = Dirt road."

See Figure 3 (Ex3, Cell C). Caption: the cyan square traces the full flown path $A\to B\to C\to D\to A$ (each vertex labelled with its coordinates); the amber text "dr = 0, distance = 20" sits at the centre where a displacement arrow would be — but it has collapsed to a single point (length $0$) because start and end coincide. The four cyan edges still add up to $20$.

Verify: $|\Delta \vec{r}|=0\le 20$ ✓. Any closed loop always gives $\Delta \vec{r}=\vec{0}$, independent of loop size or shape ✓.

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Example 4 — Cell D: bent 2D path (corner turn)

Forecast: Distance is easy ($3+4$). But the displacement magnitude — is it $7$, $5$, or something else?

1. Distance. $PQ=|4-1|=3$ m; $QR=|5-1|=4$ m; total $=3+4=7$ m. Why this step? Two straight legs; sum their lengths.
2. Displacement components. $\Delta \vec{r}=(5-1)\hat{i}+(4-1)\hat{j}=4\hat{i}+3\hat{j}$ m. Why this step? ${\vec{r}}_f-{\vec{r}}_i$ from start $P$ to end $R$.
3. Displacement magnitude. $|\Delta \vec{r}|=\sqrt{4^2+3^2}=\sqrt{25}=5$ m. Why this step? The straight "bird's flight" from $P$ to $R$ is the hypotenuse of the right triangle whose legs are the two walked legs — Pythagoras theorem.

See Figure 4 (Ex4, Cell D). Caption: the two cyan segments labelled "3" and "4" are the walked legs ($3$ up from $P$ to $Q$, then $4$ across from $Q$ to $R$, total $7$); the amber arrow labelled "|dr| = 5" cuts the corner diagonally as the displacement. The bent cyan road is visibly longer than the straight amber shortcut.

Verify: Inequality: $5\le 7$ ✓ (strict, because the path bends). The two legs $3,4$ and hypotenuse $5$ are a classic right triangle: $3^2+4^2=5^2$ ✓.

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Example 5 — Cell E: negative components (quadrant signs)

Forecast: Both coordinates decrease. Guess: will both components of $\Delta \vec{r}$ be negative? Into which quadrant does the arrow point?

1. Components. $\Delta \vec{r}=(-3-4)\hat{i}+(-6-2)\hat{j}=-7\hat{i}-8\hat{j}$ m. Why this step? Subtracting a larger initial from a smaller final gives negative results — that's fine; the sign is the direction along each axis.
2. Quadrant of the arrow. Both components negative ($-,-$) means the displacement arrow points down-and-left, into the third quadrant direction. Why this step? The sign pair $(x,y)$ of a vector tells you its direction: $(+,+)$=QI, $(-,+)$=QII, $(-,-)$=QIII, $(+,-)$=QIV. See Coordinate systems and unit vectors.
3. Magnitude. $|\Delta \vec{r}|=\sqrt{(-7{)}^2+(-8{)}^2}=\sqrt{49+64}=\sqrt{113}\approx 10.63$ m. Why this step? Squares kill the signs — magnitude is always $\ge 0$ regardless of quadrant.

See Figure 5 (Ex5, Cell E). Caption: the amber arrow starts at $A(4,2)$ in the upper-right and lands at $B(-3,-6)$ in the lower-left; it clearly slopes down-and-left, confirming the third-quadrant (QIII) direction (labelled in amber). The four quadrants QI–QIV are labelled in white so you can read off the sign pattern at a glance.

Verify: $\sqrt{113}\approx 10.63$ m, and it is straight-line motion so distance $=|\Delta \vec{r}|=\sqrt{113}$ m too (equality). Signs cancelled under the square: $(-7{)}^2=49,(-8{)}^2=64$ ✓.

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Example 6 — Cell F: full 3D motion

Forecast: Two of these numbers form a "3-4-5"; the third jumps by $12$. Guess the magnitude before computing (hint: $5$ and $12$...).

1. Components. $\Delta \vec{r}=(4-1)\hat{i}+(6-2)\hat{j}+(14-2)\hat{k}=3\hat{i}+4\hat{j}+12\hat{k}$ m. Why this step? Same subtraction rule, now with a third axis $\hat{k}$.
2. Magnitude in 3D. $|\Delta \vec{r}|=\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}=\sqrt{169}=13$ m. Why this step? Pythagoras extends: the space diagonal squared = sum of the three edge-squares.
3. Why three squares? First $\sqrt{3^2+4^2}=5$ is the diagonal in the floor $(x,y)$ plane; then that $5$ and the height $12$ form a new right triangle giving $\sqrt{5^2+12^2}=13$. Why this step? It shows 3D Pythagoras is just 2D Pythagoras applied twice.

(No figure — this is a pure 3D arithmetic case; the nested-triangle reasoning in step 3 is the "picture in words".)

Verify: $\sqrt{169}=13$ ✓. Straight flight so distance $=13$ m too. Nested check: $\sqrt{3^2+4^2}=5$, then $\sqrt{5^2+12^2}=13$ ✓.

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Example 7 — Cell G: degenerate / zero input + origin test

Forecast: For a motionless object, guess displacement and distance. For part (b), guess whether the shift changes $\Delta \vec{r}$.

1. (a) No motion. ${\vec{r}}_f={\vec{r}}_i=6\hat{i}-2\hat{j}$, so $\Delta \vec{r}=\vec{0}$ and distance $=0$ m. Why this step? Degenerate case: start = end at all times, no path swept. Note this differs from a closed loop (Ex 3): there distance was positive; here it is zero because nothing ever moved.
2. (b) Observer 1. $\Delta {\vec{r}}_1=(4-1)\hat{i}+(6-2)\hat{j}=3\hat{i}+4\hat{j}$ m. Why this step? Standard subtraction with the original coordinates.
3. (b) Build observer 2's coordinates explicitly. A shift of $(+10,+10)$ means "add $10$ to each coordinate of every point." So point $A$'s new coordinates are $A^{\prime }=(1+10,2+10)=(11,12)$, and point $B$'s are $B^{\prime }=(4+10,6+10)=(14,16)$. Why this step? We spell out where $A^{\prime }=(11,12)$ and $B^{\prime }=(14,16)$ come from so the shift is transparent, not magic.
4. (b) Observer 2's displacement. $\Delta {\vec{r}}_2=(14-11)\hat{i}+(16-12)\hat{j}=3\hat{i}+4\hat{j}$ m. Why this step? Both endpoints gained $(+10,+10)$; the difference subtracts that shift away — the origin cancels. This is parent Mistake 2 (summarised in the glossary): displacement is origin-independent.

(No figure — this case is about numbers cancelling, best seen in the aligned subtractions above.)

Verify: (a) both are zero — smallest possible values, and $0\le 0$ ✓. (b) $\Delta {\vec{r}}_1=\Delta {\vec{r}}_2=3\hat{i}+4\hat{j}$ ✓: displacement is origin-independent.

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Example 8 — Cell H: word problem + exam twist (velocity vs speed)

Forecast: Guess whether the average speed number is bigger than the average velocity magnitude, and by how much.

1. Set axes. East $=\hat{i}$, North $=\hat{j}$. Path: $A(0,0)\to B(300,0)\to C(300,400)$. Why this step? Translating compass words into vector components (Coordinate systems and unit vectors).
2. (a) Distance. $300+400=700$ m. Why this step? Sum of the two straight legs walked.
3. (b) Displacement magnitude. $\Delta \vec{r}=300\hat{i}+400\hat{j}$, so $|\Delta \vec{r}|=\sqrt{300^2+400^2}=\sqrt{90000+160000}=\sqrt{250000}=500$ m. Why this step? The straight gap from start $A$ to end $C$ is the hypotenuse of a right triangle with legs $300$ and $400$ (a scaled 3-4-5) — Pythagoras theorem.
4. (c) Average speed $=\frac{\text{distance}}{\text{time}}=\frac{700}{100}=7$ m/s. Why this step? Speed is a scalar built from distance — see Average velocity and average speed.
5. (d) Average velocity magnitude $=\frac{|\Delta \vec{r}|}{\text{time}}=\frac{500}{100}=5$ m/s. Why this step? Velocity is a vector built from displacement; its magnitude uses the $500$ m straight gap, not the $700$ m of road.
6. Why (c) and (d) differ. Because the path bends, distance ($700$) exceeds displacement ($500$), so speed ($7$) exceeds velocity magnitude ($5$). Why this step? The exam twist rewards knowing which length feeds which quantity: dirt road → speed, straight dots → velocity.

See Figure 6 (Ex8, Cell H). Caption: the two cyan legs labelled "East 300" and "North 400" are the runs the feet actually make (total $700$ m); the amber arrow labelled "|dr| = 500" is the straight displacement from start $A$ to end $C$. Average speed comes from the total cyan length ($700\Rightarrow 7$ m/s), average velocity from the amber length ($500\Rightarrow 5$ m/s) — the figure literally shows why $7>5$.

Verify: $|\Delta \vec{r}|\le \text{distance}$: $500\le 700$ ✓. Speed $\ge$ velocity magnitude: $7\ge 5$ ✓. Units: metres per second ✓. This connects to Motion in a straight line only if the path were straight (then all four would agree).

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Recall Self-test: match example to matrix cell

Which example proves distance can exceed displacement even in 1D? ::: Example 2 (Cell B, reversal on a line). Which example gives $\Delta \vec{r}=\vec{0}$ with positive distance? ::: Example 3 (Cell C, closed loop) — its distance is $20$ m, whereas Example 7a has distance $0$. Which example gives both $\Delta \vec{r}=\vec{0}$ and distance $=0$? ::: Example 7a (Cell G, no motion at all). Which example shows the displacement arrow in the third quadrant? ::: Example 5 (Cell E, both components negative). Which example proves displacement is origin-independent? ::: Example 7b (Cell G, origin shift). Which example separates average speed from average velocity? ::: Example 8 (Cell H, word/exam twist).

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Connections

• Position vector, displacement, distance — the parent this page drills.
• Scalars and vectors — displacement (vector) vs distance (scalar) throughout.
• Vector addition and subtraction — every $\Delta \vec{r}={\vec{r}}_f-{\vec{r}}_i$.
• Pythagoras theorem — magnitudes in Ex 1, 4, 5, 6, 8.
• Coordinate systems and unit vectors — quadrant signs (Ex 5), compass axes (Ex 8).
• Average velocity and average speed — the Ex 8 twist.
• Motion in a straight line — the equality case (Ex 1).