Position vector, displacement, distance: Worked examples

1.1.13 · D3 · Physics › Measurement, Vectors & Kinematics › Position vector, displacement, distance

Yeh page ek practice arena hai. Pehle hum har tarah ke question list karte hain jo yeh topic throw kar sakta hai — ek "scenario matrix" — phir aise examples work karte hain jo milke us matrix ki har cell ko cover karti hain. Koi bhi case andhera mein nahi chodha: seedha ek-taraf walk, reversal, closed loops, 3D, negative signs, degenerate (zero) motion, aur ek word problem.

Definition Is page par use ki gayi Notation (ek baar padh lo, haath mein rakho)

Ek chhota glossary taaki yeh page apne aap mein complete ho. Parent page Position vector, displacement, distance mein poori pictures hain.

$r$ — ek position vector: chosen origin $O$ se ek point tak ka arrow. Hum iski start ko $r_{i}$ (initial) aur end ko $r_{f}$ (final) likhte hain.
$r_{A}$ — named point $A$ ka position vector: bilkul same idea, $O$ se us point tak ka arrow jo $A$ label hai. Toh $r_{A}, r_{B}, \dots$ bas position vectors hain jisme "initial/final" ki jagah point ka naam subscript mein hai.
$\hat{i}, \hat{j}, \hat{k}$ — unit vectors: length $1$ ke arrows jo $x$ , $y$ , $z$ axes ke along point karte hain. $3 \hat{i} + 4 \hat{j}$ likhne ka matlab hai " $x$ along $3$ jao, phir $y$ along $4$ jao." (Dekho Coordinate systems and unit vectors.)
$Δ r = r_{f} - r_{i}$ — displacement: start se end tak ka seedha arrow. Greek $Δ$ ("delta") ka matlab sirf "change in" hai. Yeh ek vector hai jo subtraction se banta hai.
$∣ v ∣$ — vector $v$ ki magnitude (length); hamesha $\geq 0$ . Toh $∣Δ r ∣$ = displacement arrow ki length, yaani start aur end ke beech seedha gap.
distance — actual path ki total length jo walk ki gayi; ek scalar, $\geq 0$ .
Ek headline fact jo hum har jagah test karte hain: $∣Δ r ∣ \leq distance$ , equality sirf seedhi, ek-taraf motion mein.

Recall Do parent "mistakes" jinka hum sahara lete hain (summarised taaki yeh page akeli khadi rahe)

Parent Mistake 2 — "Displacement origin par depend karta hai." Aisa hota nahi. Kyunki displacement ek difference $r_{f} - r_{i}$ hai, origin dono terms mein aata hai aur cancel ho jaata hai. Do observers alag-alag origins ke saath bhi $Δ r$ par agree karenge. (Example 7b mein live prove kiya gaya hai.) Parent Mistake 3 — "Displacement/distance negative ho sakta hai." 1D mein hum $- 5$ m jaisi cheezein likhte hain, lekin woh " $-$ " axis ke along ek direction hai, negative length nahi. Distance (ek path length) hamesha $\geq 0$ hoti hai. (Example 2 mein illustrate kiya gaya hai.)

The scenario matrix

Is topic ka har problem actually inhi case classes mein se ek hoti hai. Pehle yeh table padho — neeche har example us cell ke saath tagged hai jo woh fill karta hai.

Cell	Case class	Tricky kya hai	Example
A	Seedha ek-taraf (1D ya 2D)	distance $=	\Delta\vec r
B	Line par reversal (aage-peechhe-aadha)	distance $>	\Delta\vec r
C	Closed loop (start = end)	$Δ r = 0$ par distance $> 0$	Ex 3
D	2D bent path (corner turn)	Straight gap ke liye Pythagoras	Ex 4
E	Negative components / quadrant signs	$Δ r$ ka direction kisi bhi quadrant mein point kar sakta hai	Ex 5
F	3D motion	Magnitude ke liye teen squares chahiye	Ex 6
G	Degenerate / zero input	Bilkul motion nahi, ya origin-shift test	Ex 7
H	Word problem + exam twist	Words → vectors, phir average speed vs velocity	Ex 8

Ab hum Ex 1–8 ke through chalte hain. Milke yeh cells A, B, C, D, E, F, G, H ko touch karte hain — poora grid.

Example 1 — Cell A: seedha ek-taraf walk (equality case)

Forecast: Calculate karne se pehle — guess karo: displacement ki magnitude se distance badi hogi ya equal? (Path ek seedha segment hai...)

Displacement components. $Δ r = (8 - 2) \hat{i} + (9 - 1) \hat{j} = 6 \hat{i} + 8 \hat{j}$ m. Yeh step kyun? Displacement hai $r_{f} - r_{i}$ , coordinate by coordinate kiya jaata hai.
Magnitude. $∣Δ r ∣ = 6^{2} + 8^{2} = 36 + 64 = 100 = 10$ m. Yeh step kyun? Seedha-line gap woh right triangle ka hypotenuse hai jiske legs $6$ aur $8$ hain (Pythagoras theorem).
Distance. Path wahi seedha segment hai, toh distance $= 10$ m. Yeh step kyun? Koi turning nahi, koi reversing nahi — actual road aur straight arrow same line hain.

Figure 1 dekho (Ex1, Cell A). Caption: amber arrow jis par "|dr| = 10" likha hai displacement hai seedha $A (2, 1)$ se $B (8, 9)$ tak; do white dashed lines jine "leg 6" aur "leg 8" label kiya gaya hai us right triangle ki horizontal aur vertical legs hain jiska hypotenuse woh amber arrow hai. Kyunki cyclist ki road exactly wahi amber line hai, road ki length aur arrow ki length coincide karte hain (dono $10$ m).

Figure — Position vector, displacement, distance

Verify: Kyunki motion seedha & ek-taraf hai, hum equality expect karte hain: $∣Δ r ∣ = distance$ . Indeed $10 = 10$ ✓. Units: poore mein metres ✓. Yeh exactly $∣Δ r ∣ \leq distance$ ka equality case hai.

Example 2 — Cell B: line par reversal

Forecast: Displacement ka sign aur size guess karo. Guess karo ki distance $4$ , $10$ , ya $11$ m hai.

Path ko legs mein todho. Leg 1: $0 \to 7$ (length $7$ ). Leg 2: $7 \to 3$ (length $∣3 - 7∣ = 4$ ). Yeh step kyun? Distance har segment ki length ka sum karta hai, toh hum unhe list karte hain.
Distance. $7 + 4 = 11$ m. Yeh step kyun? Total path length = leg lengths ka sum, direction ignore karke.
Displacement. $Δ r = (x_{f} - x_{i}) \hat{i} = (3 - 0) \hat{i} = 3 \hat{i}$ m; magnitude $3$ m. Yeh step kyun? Displacement sirf do endpoints ( $0$ aur $3$ ) dekhta hai, $7$ tak ke detour ko nahi.
Sign ka matlab. $+ 3$ matlab hai net motion $+ x$ direction mein point karta hai. Agar woh $x = - 3$ par khatam hoti, toh hum $- 3 \hat{i}$ likhte — " $-$ " ek direction hai, negative distance nahi. (Yeh parent Mistake 3 hai, glossary mein summarise ki gayi hai: $-$ sign ek direction hai, path length kabhi negative nahi.) Yeh step kyun? Cell B woh jagah hai jahan signs aur reversals collide karte hain.

Figure 2 dekho (Ex2, Cell B). Caption: number line par, cyan arrow leg 1 show karta hai ( $0$ se $7$ tak), amber arrow leg 2 show karta hai ( $7$ se $3$ tak wapas); neeche white arrow, jis par "dr = +3 (net)" likha hai net displacement hai. Do colored arrows milke length mein $11$ hain jabki single white arrow sirf $3$ hai — yeh visible gap exactly "distance $>$ displacement" hai.

Verify: Inequality check karo: $∣Δ r ∣ = 3 \leq 11 = distance$ ✓ (strict, kyunki path reverse hua). Sanity: net eastward progress $3$ m hai chahe usne $11$ m road chali ho ✓.

Example 3 — Cell C: closed loop

Forecast: Ek route ke liye jo apne start par wapas aaye, displacement zaroor kya hona chahiye — chahे square kitna bhi bada ho?

Displacement. Yahan $r_{A}$ point $A$ ka position vector hai (origin $O$ se $A$ tak ka arrow), glossary mein define kiya gaya hai. Kyunki drone $A$ par start aur end karta hai, $r_{i} = r_{f} = r_{A}$ , toh $Δ r = r_{A} - r_{A} = 0$ . Yeh step kyun? End point start point ke equal hai, toh do position vectors identical hain aur unka difference zero vector hai.
Distance = perimeter. $A B + B C + C D + D A = 5 + 5 + 5 + 5 = 20$ m. Yeh step kyun? Distance har walked edge ko sum karta hai; side $5$ ka square chaar sides of $5$ rakhta hai.
Interpretation. $∣Δ r ∣ = 0$ par distance $= 20$ — inequality ki extreme: net kahin nahi, par road bahut hai. Yeh step kyun? Closed loops "Displacement = Dots, Distance = Dirt road" ka sabse sharp illustration hain.

Figure 3 dekho (Ex3, Cell C). Caption: cyan square poora flown path trace karta hai $A \to B \to C \to D \to A$ (har vertex apne coordinates ke saath labelled); amber text "dr = 0, distance = 20" centre mein hai jahan ek displacement arrow hota — lekin woh ek single point (length $0$ ) mein collapse ho gaya hai kyunki start aur end coincide karte hain. Chaar cyan edges phir bhi $20$ add up karte hain.

Verify: $∣Δ r ∣ = 0 \leq 20$ ✓. Koi bhi closed loop hamesha $Δ r = 0$ deta hai, loop ke size ya shape se independent ✓.

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

Forecast: Distance easy hai ( $3 + 4$ ). Lekin displacement magnitude — kya yeh $7$ , $5$ , ya kuch aur hai?

Distance. $P Q = ∣4 - 1∣ = 3$ m; $QR = ∣5 - 1∣ = 4$ m; total $= 3 + 4 = 7$ m. Yeh step kyun? Do straight legs; unki lengths sum karo.
Displacement components. $Δ r = (5 - 1) \hat{i} + (4 - 1) \hat{j} = 4 \hat{i} + 3 \hat{j}$ m. Yeh step kyun? $r_{f} - r_{i}$ start $P$ se end $R$ tak.
Displacement magnitude. $∣Δ r ∣ = 4^{2} + 3^{2} = 25 = 5$ m. Yeh step kyun? $P$ se $R$ tak seedha "bird's flight" us right triangle ka hypotenuse hai jiske legs do walked legs hain — Pythagoras theorem.

Figure 4 dekho (Ex4, Cell D). Caption: do cyan segments jine "3" aur "4" label kiya gaya hai walked legs hain ( $P$ se $Q$ tak $3$ upar, phir $Q$ se $R$ tak $4$ across, total $7$ ); amber arrow jis par "|dr| = 5" likha hai displacement ki tarah diagonally corner cut karta hai. Bent cyan road visibly longer hai straight amber shortcut se.

Verify: Inequality: $5 \leq 7$ ✓ (strict, kyunki path moda). Do legs $3, 4$ aur hypotenuse $5$ ek classic right triangle hain: $3^{2} + 4^{2} = 5^{2}$ ✓.

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

Forecast: Dono coordinates decrease karte hain. Guess karo: kya $Δ r$ ke dono components negative honge? Arrow kis quadrant mein point karta hai?

Components. $Δ r = (- 3 - 4) \hat{i} + (- 6 - 2) \hat{j} = - 7 \hat{i} - 8 \hat{j}$ m. Yeh step kyun? Badi initial se chhoti final subtract karne par negative results milte hain — yeh theek hai; sign har axis ke along direction hai.
Arrow ka Quadrant. Dono components negative ( $-, -$ ) matlab displacement arrow neeche-aur-left, third quadrant direction mein point karta hai. Yeh step kyun? Vector ka sign pair $(x, y)$ uska direction batata hai: $(+, +)$ =QI, $(-, +)$ =QII, $(-, -)$ =QIII, $(+, -)$ =QIV. Dekho Coordinate systems and unit vectors.
Magnitude. $∣Δ r ∣ = (- 7)^{2} + (- 8)^{2} = 49 + 64 = 113 \approx 10.63$ m. Yeh step kyun? Squares signs ko maar dete hain — magnitude hamesha $\geq 0$ hoti hai chahe quadrant koi bhi ho.

Figure 5 dekho (Ex5, Cell E). Caption: amber arrow $A (4, 2)$ se upar-right mein start hoke $B (- 3, - 6)$ par neeche-left mein land karta hai; woh clearly neeche-aur-left slope karta hai, third-quadrant (QIII) direction confirm karta hai (amber mein labelled). Chaar quadrants QI–QIV white mein labelled hain taaki aap sign pattern ek nazar mein padh sako.

Verify: $113 \approx 10.63$ m, aur yeh seedha-line motion hai toh distance $= ∣Δ r ∣ = 113$ m bhi (equality). Signs square ke under cancel hue: $(- 7)^{2} = 49, (- 8)^{2} = 64$ ✓.

Example 6 — Cell F: full 3D motion

Forecast: In numbers mein se do "3-4-5" form karte hain; teesra $12$ se jump karta hai. Calculate karne se pehle magnitude guess karo (hint: $5$ aur $12$ ...).

Components. $Δ r = (4 - 1) \hat{i} + (6 - 2) \hat{j} + (14 - 2) \hat{k} = 3 \hat{i} + 4 \hat{j} + 12 \hat{k}$ m. Yeh step kyun? Same subtraction rule, ab ek teesre axis $\hat{k}$ ke saath.
Magnitude in 3D. $∣Δ r ∣ = 3^{2} + 4^{2} + 1 2^{2} = 9 + 16 + 144 = 169 = 13$ m. Yeh step kyun? Pythagoras extend hota hai: space diagonal ka square = teen edge-squares ka sum.
Teen squares kyun? Pehle $3^{2} + 4^{2} = 5$ floor $(x, y)$ plane mein diagonal hai; phir woh $5$ aur height $12$ ek naya right triangle banate hain jo $5^{2} + 1 2^{2} = 13$ deta hai. Yeh step kyun? Yeh dikhata hai ki 3D Pythagoras bas 2D Pythagoras do baar apply kiya gaya hai.

(Koi figure nahi — yeh ek pure 3D arithmetic case hai; step 3 mein nested-triangle reasoning "picture in words" hai.)

Verify: $169 = 13$ ✓. Seedhi flight toh distance $= 13$ m bhi. Nested check: $3^{2} + 4^{2} = 5$ , phir $5^{2} + 1 2^{2} = 13$ ✓.

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

Forecast: Motionless object ke liye displacement aur distance guess karo. Part (b) ke liye, guess karo ki shift $Δ r$ ko change karta hai ya nahi.

(a) No motion. $r_{f} = r_{i} = 6 \hat{i} - 2 \hat{j}$ , toh $Δ r = 0$ aur distance $= 0$ m. Yeh step kyun? Degenerate case: start = end har waqt, koi path swept nahi. Note karo yeh closed loop (Ex 3) se alag hai: wahan distance positive thi; yahan zero hai kyunki kuch bhi kabhi move nahi hua.
(b) Observer 1. $Δ r_{1} = (4 - 1) \hat{i} + (6 - 2) \hat{j} = 3 \hat{i} + 4 \hat{j}$ m. Yeh step kyun? Original coordinates ke saath standard subtraction.
(b) Observer 2 ke coordinates explicitly banao. $(+ 10, + 10)$ ka shift matlab hai "har point ke har coordinate mein $10$ add karo." Toh point $A$ ke naye coordinates hain $A^{'} = (1 + 10, 2 + 10) = (11, 12)$ , aur point $B$ ke hain $B^{'} = (4 + 10, 6 + 10) = (14, 16)$ . Yeh step kyun? Hum spell out karte hain ki $A^{'} = (11, 12)$ aur $B^{'} = (14, 16)$ kahan se aate hain taaki shift transparent ho, magic nahi.
(b) Observer 2 ka displacement. $Δ r_{2} = (14 - 11) \hat{i} + (16 - 12) \hat{j} = 3 \hat{i} + 4 \hat{j}$ m. Yeh step kyun? Dono endpoints ne $(+ 10, + 10)$ gain kiya; difference us shift ko subtract karke hata deta hai — origin cancel ho jaata hai. Yeh parent Mistake 2 hai (glossary mein summarise ki gayi): displacement origin-independent hai.

(Koi figure nahi — yeh case numbers ke cancel hone ke baare mein hai, jo upar ke aligned subtractions mein sabse achha dikhta hai.)

Verify: (a) dono zero hain — sabse chhoti possible values, aur $0 \leq 0$ ✓. (b) $Δ r_{1} = Δ r_{2} = 3 \hat{i} + 4 \hat{j}$ ✓: displacement origin-independent hai.

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

Forecast: Guess karo ki average speed ka number average velocity magnitude se bada hoga ya nahi, aur kitna.

Axes set karo. East $= \hat{i}$ , North $= \hat{j}$ . Path: $A (0, 0) \to B (300, 0) \to C (300, 400)$ . Yeh step kyun? Compass words ko vector components mein translate karna (Coordinate systems and unit vectors).
(a) Distance. $300 + 400 = 700$ m. Yeh step kyun? Do walked straight legs ka sum.
(b) Displacement magnitude. $Δ r = 300 \hat{i} + 400 \hat{j}$ , toh $∣Δ r ∣ = 30 0^{2} + 40 0^{2} = 90000 + 160000 = 250000 = 500$ m. Yeh step kyun? Start $A$ se end $C$ tak seedha gap ek right triangle ka hypotenuse hai jiske legs $300$ aur $400$ hain (ek scaled 3-4-5) — Pythagoras theorem.
(c) Average speed $= \frac{distance}{time} = \frac{700}{100} = 7$ m/s. Yeh step kyun? Speed ek scalar hai jo distance se banta hai — dekho Average velocity and average speed.
(d) Average velocity magnitude $= \frac{∣Δ r ∣}{time} = \frac{500}{100} = 5$ m/s. Yeh step kyun? Velocity ek vector hai jo displacement se banta hai; uski magnitude $500$ m ka straight gap use karta hai, $700$ m ki road nahi.
(c) aur (d) kyun differ karte hain. Kyunki path moda, distance ( $700$ ) displacement ( $500$ ) se zyada hai, toh speed ( $7$ ) velocity magnitude ( $5$ ) se zyada hai. Yeh step kyun? Exam twist reward karta hai yeh jaanna ki kaun si length kaun si quantity feed karti hai: dirt road → speed, straight dots → velocity.

Figure 6 dekho (Ex8, Cell H). Caption: do cyan legs jine "East 300" aur "North 400" label kiya gaya hai woh runs hain jo paon actually karte hain (total $700$ m); amber arrow jis par "|dr| = 500" likha hai start $A$ se end $C$ tak seedha displacement hai. Average speed total cyan length se aata hai ( $700 \Rightarrow 7$ m/s), average velocity amber length se ( $500 \Rightarrow 5$ m/s) — figure literally dikhata hai kyun $7 > 5$ .

Verify: $∣Δ r ∣ \leq distance$ : $500 \leq 700$ ✓. Speed $\geq$ velocity magnitude: $7 \geq 5$ ✓. Units: metres per second ✓. Yeh Motion in a straight line se connect hota hai sirf tab jab path seedha hota (phir chaar agree karte).

Recall Self-test: example ko matrix cell se match karo

Kaun sa example prove karta hai ki 1D mein bhi distance displacement se zyada ho sakta hai? ::: Example 2 (Cell B, line par reversal). Kaun sa example $Δ r = 0$ deta hai positive distance ke saath? ::: Example 3 (Cell C, closed loop) — uski distance $20$ m hai, jabki Example 7a ki distance $0$ hai. Kaun sa example dono $Δ r = 0$ aur distance $= 0$ deta hai? ::: Example 7a (Cell G, bilkul motion nahi). Kaun sa example displacement arrow ko third quadrant mein dikhata hai? ::: Example 5 (Cell E, dono components negative). Kaun sa example prove karta hai ki displacement origin-independent hai? ::: Example 7b (Cell G, origin shift). Kaun sa example average speed ko average velocity se alag karta hai? ::: Example 8 (Cell H, word/exam twist).

Connections

Position vector, displacement, distance — woh parent jise yeh page drill karti hai.
Scalars and vectors — displacement (vector) vs distance (scalar) throughout.
Vector addition and subtraction — har $Δ r = r_{f} - r_{i}$ .
Pythagoras theorem — Ex 1, 4, 5, 6, 8 mein magnitudes.
Coordinate systems and unit vectors — quadrant signs (Ex 5), compass axes (Ex 8).
Average velocity and average speed — Ex 8 ka twist.
Motion in a straight line — equality case (Ex 1).