Exercises — Position vector, displacement, distance

WHAT — hum position arrow ki length chahte hain, $|\vec{r}|$. WHY Pythagoras — components $6$ aur $8$ ek right triangle ki do perpendicular sides hain jiske slanted side (hypotenuse) arrow khud hai. Woh tool jo do perpendicular sides ko slanted length mein convert karta hai woh hai Pythagoras theorem — yahi exactly woh sawaal hai "diagonal kitna lamba hai?" $|\vec{r}|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{ m}.$

• (a) Displacement — vector (iska ek direction hai: start se end tak kaun si taraf).
• (b) Distance — scalar (sirf ek road length, koi direction nahi).
• (c) Position vector — vector (yeh $O$ se point ki taraf point karta hai).
• (d) $|\Delta \vec{r}|$ — scalar. Bars $|\cdot |$ direction hataa dete hain aur sirf length rakhte hain, toh result ek plain number hai. (Dekho Scalars and vectors.)

Equal. Inequality $|\Delta \vec{r}|\le \text{distance}$ exactly tab equality ban jaati hai jab path ek seedhi line ho jo ek direction mein travel ki gayi ho — koi turning nahi, koi doubling back nahi. Yahan billi exactly wahi karti hai, toh distance $=|\Delta \vec{r}|=|7-2|=5$ m. Yeh woh special case hai jo Motion in a straight line mein handle kiya jaata hai.

WHAT — start arrow ko end arrow se component by component subtract karo (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 yeh jawaab deta hai "kaunsa arrow, start mein add karo toh end par pahuncho?" Woh arrow hai ${\vec{r}}_f-{\vec{r}}_i$. Magnitude (sides $4$ aur $3$ par Pythagoras; $-3$ ka sign koi matter nahi karta kyunki squaring use mitaa deta hai): $|\Delta \vec{r}|=\sqrt{4^2+(-3{)}^2}=\sqrt{16+9}=\sqrt{25}=5\text{ m}.$

Figure dekhte hain: ek hi line par do segments, doosra wala reversing.

(a) Distance = chali gayi lengths ka sum: $AB=|9-0|=9,BC=|4-9|=5,\text{distance}=9+5=14\text{ m}.$ Hum har leg ke liye absolute values use karte hain kyunki road length kabhi negative nahi hoti. (b) Displacement sirf endpoints $A(0,0)$ aur $C(0,4)$ use karta hai: $\Delta \vec{r}=(0-0)\hat{i}+(4-0)\hat{j}=4\hat{j}\text{ m}.$ (c) Magnitude $=|4|=4$ m. Sanity check: $4\le 14$ ✓ — reversal exactly wahi reason hai kyun distance ($14$) $|\Delta \vec{r}|$ ($4$) se aage nikal jaati hai.

$\Delta \vec{r}={\vec{r}}_f-{\vec{r}}_i=(-3-(-3))\hat{i}+(5-5)\hat{j}=0\hat{i}+0\hat{j}=\vec{0}.$ Zero vector — length $0$ ka arrow jiska koi direction nahi. Degenerate case: start $=$ end $\Rightarrow$ displacement exactly zero hai, chahe origin kahin bhi ho.

Figure mein L-shaped path trace karo.

Distance (do legs chalte hain): $OP=3,PQ=4,\text{distance}=3+4=7\text{ m}.$ Displacement ($O$ se $Q$ tak seedha arrow): $\Delta \vec{r}=3\hat{i}+4\hat{j},|\Delta \vec{r}|=\sqrt{3^2+4^2}=5\text{ m}.$ Gap: $5<7$. Seedha red arrow (hypotenuse) shortest route hai; black L-path bend karti hai, toh woh longer hai. Yahi geometric reason hai $|\Delta \vec{r}|\le \text{distance}$ ka: right triangle ka diagonal hamesha do legs ke sum se chota hota hai.

Distance $=12+5=17$ km (do straight legs add karo). Displacement magnitude — do legs perpendicular hain, toh yeh ek right triangle ki sides hain: $|\Delta \vec{r}|=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{ km}.$ Direction: East-aur-thoda-North (first quadrant mein, kyunki dono components positive hain: $12\hat{i}+5\hat{j}$). Check: $13\le 17$ ✓.

Same answer: $\Delta \vec{r}=6\hat{i}+8\hat{j}$, magnitude $10$ m. WHY — displacement ek difference ${\vec{r}}_f-{\vec{r}}_i$ hai. Origin ko move karne se ${\vec{r}}_f$ aur ${\vec{r}}_i$ dono mein same shift vector $\vec{s}$ add ho jaata hai: $({\vec{r}}_f+\vec{s})-({\vec{r}}_i+\vec{s})={\vec{r}}_f-{\vec{r}}_i.$ Shift cancel ho jaata hai. Yahi precisely woh reason hai kyun displacement physical kehlaata hai: jo observers origin ke baare mein disagree karte hain, woh bhi displacement par agree karte hain.

Figure mein closed loop dekhte hain.

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

(a) $\Delta \vec{r}=(8-2)\hat{i}+(11-3)\hat{j}=6\hat{i}+8\hat{j}$, toh $|\Delta \vec{r}|=\sqrt{6^2+8^2}=\sqrt{100}=10$ m. (b) Straight line, ek 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) Average velocity ka magnitude $=\frac{|\Delta \vec{r}|}{\text{time}}=\frac{10}{2}=5$ m/s. Yahan speed $=$ velocity-magnitude kyunki path straight aur unidirectional hai — yeh tie-in hai Average velocity and average speed se.

Set-up. Pehli leg: $6\hat{i}$. Doosri leg ki length $6$ hai lekin direction variable hai. Distance $6+6=12$ m fixed hai; hum net arrow ko steer kar rahe hain. Sabse chhota $|\Delta \vec{r}|$: doosri leg seedha wapas (West) bhejo, $-6\hat{i}$. Tab $\Delta \vec{r}=6\hat{i}-6\hat{i}=\vec{0},|\Delta \vec{r}|=0\text{ m}.$ Reverse karna pehli leg ko cancel kar deta hai — minimum jo do equal legs de sakti hain. Sabse bada $|\Delta \vec{r}|$: doosri leg usi direction mein bhejo (phir se East), $+6\hat{i}$. Tab $\Delta \vec{r}=6\hat{i}+6\hat{i}=12\hat{i},|\Delta \vec{r}|=12\text{ m}.$ Yeh equality $|\Delta \vec{r}|=\text{distance}=12$ hit karta hai, kyunki motion ne kabhi direction nahi badli. Range: $0\le |\Delta \vec{r}|\le 12$ m — inequality ke do extremes, concrete banaye.

Legs ko vectors $\vec{a}$ aur $\vec{b}$ maano jinka lengths $a=|\vec{a}|$, $b=|\vec{b}|$ hain. Net displacement sum hai $\Delta \vec{r}=\vec{a}+\vec{b}$ (Vector addition and subtraction). Vectors ke liye triangle inequality kehti hai: $|\vec{a}+\vec{b}|\le |\vec{a}|+|\vec{b}|=a+b.$ Left side $|\Delta \vec{r}|$ hai; right side distance hai. Isliye $|\Delta \vec{r}|\le \text{distance}$. ∎ Equality exactly tab hold karta hai jab $\vec{a}$ aur $\vec{b}$ ek hi direction mein point karein (koi turning nahi) — tab do arrows ek seedhi line par tip-to-tail lagte hain aur unki lengths simply add ho jaati hain. Koi bhi turn net arrow ko strictly chota kar deta hai. Yeh general proof hai jiske peeche Motion in a straight line equality case hai.

Motion ko legs mein todto aur lengths ke liye absolute values lo: $0\to 10:|10-0|=10,10\to 4:|4-10|=6,4\to 7:|7-4|=3.$ Distance $=10+6+3=19$ m. Net displacement — sirf endpoints matter karte hain, 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: $7\le 19$ ✓. Har reversal ($10\to 4$ backtrack) distance mein add karta hai lekin net arrow mein nahi.

One-way straight line $\Rightarrow$ distance $=$ displacement magnitude? ::: Haan — equality case. Closed loop $\Rightarrow$ displacement? ::: Zero vector (distance abhi bhi positive hai). 1D mein distance ke liye har leg ki length kaise nikalein? ::: Absolute value $|x_f-x_i|$ lo aur saari legs add karo. $|\Delta \vec{r}|\le$ distance hamesha kyun sach hai? ::: Triangle inequality — seedha arrow endpoints ke beech shortest route hai.