Visual walkthrough — Position vector, displacement, distance

1.1.13 · D2 · Physics › Measurement, Vectors & Kinematics › Position vector, displacement, distance

Hum Scalars and vectors (ek number vs ek number-with-direction), Coordinate systems and unit vectors (kisi point ko pin karna), Vector addition and subtraction (arrows kaise milte hain), aur Pythagoras theorem (seedhi length kaise maapte hain) ke ideas par bharosa karenge. Baaki sab hum khud se develop karenge.

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Step 1 — Ek reference dot se point ko pin karo

KYA. Hum board par ek dot rakhte hain aur use $O$ kehte hain, yaani origin — hamaara agreed "home base." Phir hum ek object ko kisi point $P$ par rakhte hain aur $O$ se $P$ tak ek seedha arrow kheenchte hain. Woh arrow position vector hai, likha jaata hai $\vec{r}$.

KYUN. "Object 3 par hai" jaisi bare location meaningless hai — 3 kya, kahaan se measure kiya, kis direction mein? Board par kisi bhi cheez ki location tab tak nahi hoti jab tak hum ek home base choose nahi karte. Arrow $\vec{r}$ do facts ko ek saath bundle karta hai: kitna door (uski length) aur kis taraf (woh kahan point karta hai).

PICTURE. Yellow dot $O$ hai. Blue arrow $\vec{r}$ hai, jiska tail $O$ par fixed hai aur head $P$ par resting hai. Uska horizontal shadow $x$ hai, vertical shadow $y$ hai.

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Step 2 — Do points, do arrows

KYA. Ab object move karta hai. Iske start ko $A$ aur end ko $B$ mark karo. Usi origin $O$ se do position arrows kheencho: ${\vec{r}}_i$ (start tak) aur ${\vec{r}}_f$ (end tak). Subscripts ka matlab hai $i$ = initial, $f$ = final.

KYUN. Motion ke baare mein baat karne ke liye humein ek before aur ek after chahiye. Dono arrows ka tail ek hi $O$ share karte hain isliye dono ek hi ruler se measure hote hain — yeh shared origin hi woh trick hai jo agla step itna neat bana deti hai.

PICTURE. Blue arrow ${\vec{r}}_i$ $A$ ki taraf point karta hai; pink arrow ${\vec{r}}_f$ $B$ ki taraf point karta hai. Dono yellow origin dot se nikalte hain.

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Step 3 — Bridging arrow: displacement

KYA. Hum ek sawaal poochte hain: woh kaun sa ek arrow hai jo ${\vec{r}}_i$ ki tip par rakh do aur ${\vec{r}}_f$ ki tip tak pahuncha de? Us bridging arrow ko $\Delta \vec{r}$ kehte hain (symbol $\Delta$, "delta," matlab hai "mein change").

KYUN. Yeh bridging arrow physical cheez hai — yeh seedha wahan se point karta hai jahan tum sach mein the, straight wahan tak jahan tum sach mein pahunche. Yeh origin ko bilkul ignore karta hai, aur yahi hum chahte hain: do alag home-bases use karne wale do logon ko phir bhi agree karna chahiye ki object kaisa move kiya.

PICTURE. Green bridging arrow tail-at-$A$, head-at-$B$ jaata hai. Notice karo ki yeh do position arrows ke saath ek triangle complete karta hai: ${\vec{r}}_i$, phir $\Delta \vec{r}$, wahi pahunchta hai jahan ${\vec{r}}_f$ pahunchta hai.

Picture ko ek arrow equation ki tarah padho (yeh Vector addition and subtraction in action hai): $\underset{\text{start tak jao}}{\underbrace{{\vec{r}}_i}}+\underset{\text{bridge karo}}{\underbrace{\Delta \vec{r}}}=\underset{\text{end par pahuncho}}{\underbrace{{\vec{r}}_f}}$

Bridge ke liye solve karo, ${\vec{r}}_i$ dono sides se hata do: $\boxed{\Delta \vec{r}={\vec{r}}_f-{\vec{r}}_i}$

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Step 4 — Woh bridge kitna lamba hai? (Pythagoras aata hai)

KYA. Bridge $\Delta \vec{r}$ ki ek length hai. Hum ise $|\Delta \vec{r}|$ kehte hain — bars ka matlab hai "size, direction bhool jao." Ise find karne ke liye hum horizontal leg $(x_f-x_i)$ aur vertical leg $(y_f-y_i)$ drop karte hain aur unhe right angle se jodte hain.

KYUN Pythagoras aur kuch nahi? Do legs $90^{\circ }$ par milte hain (right-step aur up-step construction se perpendicular hain). Woh ek tool jo do perpendicular legs ko seedhi hypotenuse mein badalta hai woh Pythagoras theorem hai — yahi exactly woh sawaal hai "is right triangle ka diagonal kitna lamba hai?"

PICTURE. Green hypotenuse $=|\Delta \vec{r}|$; yellow horizontal leg; blue vertical leg; chhota square right angle mark karta hai.

$|\Delta \vec{r}|=\sqrt{\underset{{\text{horizontal leg}}^2}{\underbrace{(x_f-x_i{)}^2}}+\underset{{\text{vertical leg}}^2}{\underbrace{(y_f-y_i{)}^2}}}$

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Step 5 — Woh sadak jo pair sach mein chale

KYA. Displacement ne journey bhool gayi. Ab hum use yaad karte hain. Distance asli path ki total length hai, choti choti seedhi segments mein kaat ke add ki gayi. Yeh ek scalar hai (dekho Scalars and vectors) — sirf ek non-negative number, koi direction nahi.

KYUN. Asli pair teleport nahi karte bridge ke saath; woh sadakon par chalte hain jo modd leti hain. Distance woh hai jo odometer padhta hai. Bendy path maapne ke liye hum use bahut saari chhoti seedhi pieces se approximate karte hain aur unki lengths add karte hain.

PICTURE. $A$ se $B$ tak bhatakti pink sadak, choti chalk segments $s_1,s_2,s_3,\dots$ mein kaati gayi; seedha green bridge $\Delta \vec{r}$ comparison ke liye neeche se jaata hai.

$\text{distance}=\underset{\text{har chhota step add karo}}{\underbrace{s_1+s_2+s_3+\cdots }}\ge 0$

Har $s_k$ ek length hai, isliye kabhi negative nahi ho sakti, isliye total kabhi negative nahi ho sakti.

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Step 6 — Kyun bridge sadak se hamesha haarta hai

KYA. Sirf do road pieces lo, $\vec{a}$ phir $\vec{b}$, jo milke $A$ se $B$ tak jaate hain. Bridge hai $\vec{a}+\vec{b}$. Triangle inequality kehti hai direct side kabhi bhi do-side detour se lamba nahi hota: $|\vec{a}+\vec{b}|\le |\vec{a}|+|\vec{b}|$

KYUN. Kisi bhi triangle mein, ek side akele dono sides ke sum se aage nahi ja sakti — warna triangle close hi nahi hota. Is fact ko saari chhoti road pieces par chain karo aur poori sadak har baar jeetti hai. Geometrically: straight line do points ke beech ka shortest route hoti hai.

PICTURE. Ek triangle jisme green direct side $\vec{a}+\vec{b}$ aur do pink detour sides $\vec{a}$, $\vec{b}$ hain — detour visibly longer hai.

Har segment par chain karne se headline result milta hai: $\boxed{|\Delta \vec{r}|\le \text{distance}}$

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Step 7 — Har case, sneaky wale bhi

KYA / KYUN / PICTURE — poora case sweep. Inequality ka ek equality case hai aur kaafi degenerate cases hain. Reader ko koi aisa scenario nahi milna chahiye jise humne skip kiya ho.

• Case A — seedha, ek direction (equality). Sadak hi bridge hai, toh $|\Delta \vec{r}|=\text{distance}$. Yeh ek hi equality case hai (the Motion in a straight line special case).
• Case B — path bend karta hai. Sadak bridge se lambi: $|\Delta \vec{r}|<\text{distance}$.
• Case C — aage phir wapas (partial reversal). Bridge shrink hota hai jabki sadak badhti rehti hai; gap badh jaata hai.
• Case D — closed loop, $A=B$. Yahan ${\vec{r}}_f={\vec{r}}_i$, isliye $\Delta \vec{r}=\vec{0}$ aur $|\Delta \vec{r}|=0$, phir bhi distance $>0$. Chidiya kahin nahi gayi; pair phir bhi thak gaye.
• Case E — bilkul nahi hila. Path length $0$ aur displacement $\vec{0}$: dono sides $0$, equality trivially hold karti hai.

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Ek-picture summary

Upar sab kuch ek single board mein compress kiya: origin $O$, do position arrows, green seedha bridge $\Delta \vec{r}$, bhatakti pink sadak, bridge maapne wala right triangle, aur neeche stamped inequality.

Recall Feynman retelling — poori walk plain words mein

Zameen par ek home dot giraao; woh origin hai. Home se tumhare tak ka arrow tumhara position hai — yeh batata hai kitna door aur kis taraf. Ab kahin chalo: apne start tak ek home-arrow kheencho aur end tak doosra. Start-to-end ko bridge karne wala single arrow tumhara displacement hai — aur kyunki dono home-arrows ek hi home se measure hote hain, home-dot cancel ho jaata hai, isliye displacement same rehta hai chahe koi apna home kahaan bhi rakhe. Us bridge ki length maapne ke liye, uske sideways aur upward shadows giraa do, right triangle banao, aur Pythagoras use karo. Iske saath saath, apni asli winding sadak ko chhote chhote steps mein kato aur sab add karo — woh distance hai, aur kyunki har step ek length hai isliye kabhi negative nahi hoti. Aakhir mein, kisi bhi chhote triangle mein seedha side do bent sides ko beat nahi kar sakti, aur poori sadak par woh fact stack karne se prove hota hai ki seedha bridge sadak se kabhi lamba nahi hota: $|\Delta \vec{r}|\le \text{distance}$. Woh sirf tab tie karte hain jab tum seedha march karo bina kabhi wapas mude — aur agar loop karte hue waapas ghar aao, toh bridge zero hai jabki tumhare pair thak gaye hain.

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Connections

• Position vector, displacement, distance — woh parent jise yeh walkthrough visualise karta hai.
• Scalars and vectors — displacement ek vector hai, distance ek scalar.
• Vector addition and subtraction — Step 3 mein ${\vec{r}}_i+\Delta \vec{r}={\vec{r}}_f$ solve hota hai.
• Coordinate systems and unit vectors — Steps 1–2 mein $\hat{i},\hat{j}$ aur coordinates.
• Pythagoras theorem — Step 4 mein bridge length maapi jaati hai.
• Motion in a straight line — Case A, ek hi equality case.
• Average velocity and average speed — velocity bridge use karti hai, speed sadak use karti hai.