Exercises — Position vector, displacement, distance

WHAT — we want the length of the position arrow, $|\vec{r}|$. WHY Pythagoras — the components $6$ and $8$ are the two perpendicular sides of a right triangle whose slanted side (hypotenuse) is the arrow itself. The tool that turns two perpendicular sides into the slanted length is Pythagoras theorem — that's exactly the question "how long is the diagonal?" $|\vec{r}|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{ m}.$

• (a) Displacement — vector (it has a direction: which way from start to end).
• (b) Distance — scalar (just a road length, no direction).
• (c) Position vector — vector (it points from $O$ to the point).
• (d) $|\Delta \vec{r}|$ — scalar. The bars $|\cdot |$ strip off the direction and keep only the length, so the result is a plain number. (See Scalars and vectors.)

Equal. The inequality $|\Delta \vec{r}|\le \text{distance}$ becomes an equality precisely when the path is a straight line travelled in one direction — no turning, no doubling back. Here the cat does exactly that, so distance $=|\Delta \vec{r}|=|7-2|=5$ m. This is the special case handled in Motion in a straight line.

WHAT — subtract the start arrow from the end arrow, component by component (Vector addition and subtraction): $\Delta \vec{r}=(5-1)\hat{i}+(-1-2)\hat{j}=4\hat{i}-3\hat{j}\text{ m}.$ WHY subtract — displacement answers "what arrow, added to the start, lands me on the end?" That arrow is ${\vec{r}}_f-{\vec{r}}_i$. Magnitude (Pythagoras on the sides $4$ and $3$; the sign of $-3$ does not matter because squaring erases it): $|\Delta \vec{r}|=\sqrt{4^2+(-3{)}^2}=\sqrt{16+9}=\sqrt{25}=5\text{ m}.$

Look at the figure: two segments on the same line, the second one reversing.

(a) Distance = sum of walked lengths: $AB=|9-0|=9,BC=|4-9|=5,\text{distance}=9+5=14\text{ m}.$ We use absolute values for each leg because a road length is never negative. (b) Displacement uses only the endpoints $A(0,0)$ and $C(0,4)$: $\Delta \vec{r}=(0-0)\hat{i}+(4-0)\hat{j}=4\hat{j}\text{ m}.$ (c) Magnitude $=|4|=4$ m. Sanity check: $4\le 14$ ✓ — the reversal is exactly why distance ($14$) pulls ahead of $|\Delta \vec{r}|$ ($4$).

$\Delta \vec{r}={\vec{r}}_f-{\vec{r}}_i=(-3-(-3))\hat{i}+(5-5)\hat{j}=0\hat{i}+0\hat{j}=\vec{0}.$ The zero vector — an arrow of length $0$ with no direction. Degenerate case: start $=$ end $\Rightarrow$ displacement is exactly zero, no matter where the origin sits.

Trace the L-shaped path in the figure.

Distance (walk the two legs): $OP=3,PQ=4,\text{distance}=3+4=7\text{ m}.$ Displacement (straight arrow from $O$ to $Q$): $\Delta \vec{r}=3\hat{i}+4\hat{j},|\Delta \vec{r}|=\sqrt{3^2+4^2}=5\text{ m}.$ The gap: $5<7$. The straight red arrow (the hypotenuse) is the shortest route; the black L-path bends, so it is longer. This is the geometric reason $|\Delta \vec{r}|\le \text{distance}$: the diagonal of a right triangle is always shorter than the two legs added.

Distance $=12+5=17$ km (add the two straight legs). Displacement magnitude — the two legs are perpendicular, so they are the sides of a right triangle: $|\Delta \vec{r}|=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{ km}.$ Direction: East-and-a-bit-North (into the first quadrant, since both components are positive: $12\hat{i}+5\hat{j}$). Check: $13\le 17$ ✓.

Same answer: $\Delta \vec{r}=6\hat{i}+8\hat{j}$, magnitude $10$ m. WHY — displacement is a difference ${\vec{r}}_f-{\vec{r}}_i$. Moving the origin adds the same shift vector $\vec{s}$ to both ${\vec{r}}_f$ and ${\vec{r}}_i$: $({\vec{r}}_f+\vec{s})-({\vec{r}}_i+\vec{s})={\vec{r}}_f-{\vec{r}}_i.$ The shift cancels. That is precisely why displacement is called physical: observers who disagree about the origin still agree on the displacement.

See the closed loop in the figure.

(a) Distance — walk all three legs: $AB=4,BC=3,CA=\sqrt{4^2+3^2}=5,\text{distance}=4+3+5=12\text{ m}.$ (b) Displacement — start $=$ end $=A$, so: $\Delta \vec{r}={\vec{r}}_A-{\vec{r}}_A=\vec{0}\text{ m},|\Delta \vec{r}|=0.$ (c) Check: $0\le 12$ ✓. A full loop is the extreme case: zero net arrow, non-zero road.

(a) $\Delta \vec{r}=(8-2)\hat{i}+(11-3)\hat{j}=6\hat{i}+8\hat{j}$, so $|\Delta \vec{r}|=\sqrt{6^2+8^2}=\sqrt{100}=10$ m. (b) Straight line, one direction $\Rightarrow$ distance $=|\Delta \vec{r}|=10$ m (equality case). (c) Time elapsed $=3-1=2$ s. Average speed $=\frac{\text{distance}}{\text{time}}=\frac{10}{2}=5$ m/s. (d) Magnitude of average velocity $=\frac{|\Delta \vec{r}|}{\text{time}}=\frac{10}{2}=5$ m/s. Here speed $=$ velocity-magnitude because the path is straight and undirectional — the tie-in to Average velocity and average speed.

Set-up. First leg: $6\hat{i}$. Second leg has length $6$ but variable direction. The distance is fixed at $6+6=12$ m; we are steering the net arrow. Smallest $|\Delta \vec{r}|$: send the second leg straight back (West), $-6\hat{i}$. Then $\Delta \vec{r}=6\hat{i}-6\hat{i}=\vec{0},|\Delta \vec{r}|=0\text{ m}.$ Reversing cancels the first leg — the minimum any two equal legs can give. Largest $|\Delta \vec{r}|$: send the second leg the same way (East again), $+6\hat{i}$. Then $\Delta \vec{r}=6\hat{i}+6\hat{i}=12\hat{i},|\Delta \vec{r}|=12\text{ m}.$ This hits the equality $|\Delta \vec{r}|=\text{distance}=12$, because the motion never turned direction. Range: $0\le |\Delta \vec{r}|\le 12$ m — the inequality's two extremes, made concrete.

Let the legs be vectors $\vec{a}$ and $\vec{b}$ with lengths $a=|\vec{a}|$, $b=|\vec{b}|$. The net displacement is the sum $\Delta \vec{r}=\vec{a}+\vec{b}$ (Vector addition and subtraction). The triangle inequality for vectors states: $|\vec{a}+\vec{b}|\le |\vec{a}|+|\vec{b}|=a+b.$ The left side is $|\Delta \vec{r}|$; the right side is the distance. Hence $|\Delta \vec{r}|\le \text{distance}$. ∎ Equality holds exactly when $\vec{a}$ and $\vec{b}$ point the same direction (no turning) — then the two arrows lie tip-to-tail on one straight line and their lengths simply add. Any turn strictly shortens the net arrow. This is the general proof behind Motion in a straight line being the equality case.

Break the motion into legs and take absolute values for lengths: $0\to 10:|10-0|=10,10\to 4:|4-10|=6,4\to 7:|7-4|=3.$ Distance $=10+6+3=19$ m. Net displacement — only endpoints matter, start $x=0$, end $x=7$: $\Delta \vec{r}=(7-0)\hat{i}=7\hat{i}\text{ m},|\Delta \vec{r}|=7\text{ m}.$ Check: $7\le 19$ ✓. Every reversal (the $10\to 4$ backtrack) adds to distance but not to the net arrow.

One-way straight line $\Rightarrow$ distance $=$ displacement magnitude? ::: Yes — the equality case. Closed loop $\Rightarrow$ displacement? ::: Zero vector (distance still positive). How to get each leg's length for distance in 1D? ::: Take the absolute value $|x_f-x_i|$ and add all legs. Why is $|\Delta \vec{r}|\le$ distance always true? ::: Triangle inequality — the straight arrow is the shortest route between endpoints.