Position vector, displacement, distance: Question bank

1.1.13 · D5 · Physics › Measurement, Vectors & Kinematics › Position vector, displacement, distance

Vocabulary jo tumhare paas parent note se pehle se honi chahiye: position vector $r$ (origin O se point tak ka arrow), displacement $Δ r = r_{f} - r_{i}$ (start position $r_{i}$ se end position $r_{f}$ tak ka arrow), distance (actual path ki length jo chali gayi).

Shuru karne se pehle ek picture apne dimaag mein rakho. Ek curved/looping path aur uska straight shortcut, side by side:

Figure — Position vector, displacement, distance

Green feet-path (distance) hamesha blue bird-arrow (displacement) se kam se kam utni lambi hoti hai. Neeche har trap usi ka variation hai — question actually dono mein se kaunsa pooch raha hai.

True or false — justify karo

Ek closed loop mein displacement hamesha zero hota hai.

True — end position, start position ke barabar hoti hai, yaani

r_{f} = r_{i}

, toh

Δ r = r_{f} - r_{i} = 0

, chahe loop kitna bhi lamba ya twisty kyun na ho.

Zero displacement force karta hai zero distance.

False — ek round trip mein

Δ r = 0

hota hai lekin distance badi hoti hai; sirf endpoints mile, pair toh phir bhi chale.

Agar distance, displacement ke magnitude ke barabar hai, toh motion ek straight line thi bina reverse kiye.

True — straight segment sabse chota unique path hota hai, toh

∣Δ r ∣ \leq distance

mein equality path ko usi straight line pe pin kar deti hai, ek baar traverse kiya, koi backtracking nahi.

Displacement magnitude mein distance se bada ho sakta hai.

False — straight line sabse chota route hoti hai, toh

∣Δ r ∣

zyada se zyada distance ke barabar ho sakta hai, kabhi exceed nahi kar sakta.

Origin ko move karne se displacement change hota hai.

False —

Δ r = r_{f} - r_{i}

; origin dono

r_{f}

aur

r_{i}

ke andar hota hai aur subtraction mein cancel ho jaata hai. Neeche ki figure ye cancellation graphically dikhati hai. Dekho Vector addition and subtraction.

Origin ko move karne se position vector change hota hai.

True —

r

origin se measure hota hai, toh O ko relocate karne par arrow redraw ho jaata hai. Yahi reason hai ki position frame pe depend karti hai lekin displacement nahi.

1D motion mein distance negative ho sakti hai.

False — 1D mein jo "

-

" likha jaata hai woh displacement pe ek direction encode karta hai; distance ek path length hai aur hamesha

\geq 0

rehti hai.

Displacement aur distance dono scalars hain.

False — displacement ek vector hai (direction carry karta hai), distance ek scalar hai (ek plain non-negative number).

Ek hi journey ke liye do log alag coordinate systems use karke same displacement report karenge.

True — displacement origin-independent hota hai, toh ye ek physical quantity hai jis pe dono observers agree karte hain, chahe unke position vectors alag hon.

Position vector ka magnitude woh distance hai jo object ne travel ki hai.

False —

∣ r ∣

abhi origin se point tak ka straight gap hai; distance poore path ki length hai jo kahin bhi pahunchne ke liye li gayi.

Origin kyun cancel hota hai — dekho, sirf kaho mat. Do observers do alag origins $O_{1}$ aur $O_{2}$ choose karte hain usi journey A→B ke liye:

Dono alag position arrows draw karte hain, lekin difference arrow (red displacement A→B) identical hai — origin-to-origin shift add aur subtract ho jaata hai, kuch nahi bachta.

Error dhundho

Neeche ki mislabelled walk sabse common trap hai. Sketch study karo, phir corrections padho:

"Runner ne 400 m track ka ek lap kiya, toh uska displacement 400 m hai."

Galat — ek full lap start par wapas le aata hai, toh displacement

0

hai; 400 m toh distance hai.

"Woh 3 m east chali phir 4 m north, toh uski distance

3^{2} + 4^{2} = 5

m hai."

Galat — 5 m displacement magnitude hai (figure mein straight A→C hypotenuse). Distance

3 + 4 = 7

m hai, do green legs ka sum jo chali gayi.

"Displacement

- 5

m hai, toh object zero distance se 5 m neeche hai."

Galat — "

-

" axis pe ek direction hai, signed distance nahi. Distance

5

m hai; negative distance jaisi koi cheez nahi hoti.

"Kyunki

∣ r_{i} ∣ = ∣ r_{f} ∣ = 5

m, displacement zero hona chahiye."

Galat — equal magnitudes sirf ye matlab hai ki dono points O se same distance pe hain (radius 5 ke ek circle pe). Displacement

r_{f} - r_{i}

hai, jo ek lamba chord ho sakta hai jab tak points actually coincide na karein.

"Displacement dhundne ke liye main position vectors add karta hoon:

r_{f} + r_{i}

Galat — displacement difference hai

r_{f} - r_{i}

. Addition ek meaningless origin-dependent arrow dega, start→end change nahi.

"Path bend karta hai, lekin kyunki woh same height pe start aur end hota hai, distance = displacement."

Galat — equality ke liye single-direction straight line chahiye. Ek bending path already "shortest route" tod deta hai, toh distance strictly

∣Δ r ∣

se zyada hogi.

"Distance ek vector hai kyunki ye road pe kisi bhi direction mein point kar sakta hai."

Galat — road bend karti hai, toh koi single direction ise describe nahi kar sakta; distance direction bilkul discard kar deta hai aur scalar hota hai.

Why questions

Origin displacement mein kyun cancel hota hai lekin position mein nahi?

Position ek arrow hai O se, toh O ise define karta hai; displacement do aise arrows subtract karta hai aur shared O term drop ho jaata hai (

r_{f} - r_{i}

mein koi O nahi bachta) — bilkul upar ki two-origin figure mein identical red arrow ki tarah.

Straight-line path do points ke beech hamesha shortest kyun hoti hai?

Koi bhi detour sideways travel add karta hai jo baad mein undo karna padta hai, toh total length sirf badhti hai — straight segment woh ek path hai jismein koi wasted motion nahi, giving

∣Δ r ∣ \leq distance

; neeche ki detour figure wasted sideways bulge dikhati hai.

Hum ek reference point se bother kyun karte hain?

Ek raw coordinate jaise "3 par" ka koi matlab nahi jab tak ye na batao ki 3 kahaan se, kis direction mein; origin aur axes (Coordinate systems and unit vectors) woh anchor supply karte hain.

Do bahut alag journeys same displacement kyun share kar sakti hain?

Displacement sirf do endpoints yaad rakhta hai, toh same start aur end wale koi bhi paths identical

Δ r

denge, chahe pair kitne bhi alag chale hon.

Average velocity displacement kyun use karta hai jabki average speed distance use karti hai?

Velocity ek vector hai jo net directed change (displacement) se time ke saath banta hai; speed ek scalar hai jo total road covered (distance) se time ke saath banta hai, toh ek round trip ki average velocity zero hogi lekin average speed nonzero.

∣Δ r ∣

Pythagoras theorem se kyun compute hota hai, components directly add karke nahi?

Components

Δ x, Δ y, Δ z

perpendicular legs hain; straight arrow unki hypotenuse hai, toh

∣Δ r ∣ = Δ x^{2} + Δ y^{2} + Δ z^{2}

Δ x + Δ y

nahi.

Detours kyun lambe hote hain — sideways-bulge picture. Same do dots ke beech ek straight shortcut versus ek bulging detour:

Detour sideways jaane mein length spend karta hai (orange) jo straight route kabhi pay nahi karta; sideways bulge ki har unit eventually wapas walk karni padti hai, toh green path sirf lambi ho sakti hai.

Edge cases

Ek object ek ghante tak bilkul still rehta hai. Uski distance aur displacement kya hain?

Dono zero hain — koi path nahi chala (distance

= 0

) aur start equals end (displacement

= 0

Ek object 5 m east chalta hai, phir 5 m west wapas start par. Distance aur displacement?

Distance

= 10

m (dono legs count hote hain); displacement

= 0

(endpoints coincide karte hain). Ye sabse clean case hai jahan dono maximally disagree karte hain.

1D mein ek purely straight one-way walk mein, distance aur

∣Δ r ∣

kaise compare karte hain?

Ye bilkul equal hain — koi bending nahi, koi reversing nahi, toh inequality

∣Δ r ∣ \leq distance

equality ban jaati hai.

Ek object ek lambi journey ke baad origin pe end hota hai — kya uska final position vector zero hai?

Haan,

r_{f} = 0

(woh origin pe baitha hai); uski distance travelled nonzero hai, aur uska displacement

r_{f} - r_{i}

nonzero hai jab tak start $r_{i}$ bhi origin par na ho, us case mein

Δ r = 0

bhi.

Kya displacement nonzero ho sakta hai jabki har individual coordinate change ek axis mein "cancelled" lagta hai?

Haan — e.g. same

x

par wapas aao lekin alag

y

pe:

Δ x = 0

lekin

Δ y \neq = 0

, toh

Δ r = (Δ y) \hat{j} \neq = 0

. Ek axis mein cancellation pura vector zero nahi karta.

Ek drone full 3D mein

(1, 0, 2)

(1, 3, 6)

m tak fly karta hai. Uska displacement magnitude kya hai?

Teeno axes use karo:

Δ r = 0 \hat{i} + 3 \hat{j} + 4 \hat{k}

, toh

∣Δ r ∣ = 0^{2} + 3^{2} + 4^{2} = 5

m — Pythagoras theorem third dimension mein bhi unchanged extend hota hai.

Polar coordinates mein ek point radius

r = 5

, angle

60°

pe hai. Kya displacement ko care hai ki tumne endpoints describe karne ke liye kaunsa coordinate system use kiya?

Nahi — displacement physical start→end arrow hai; Cartesian ya polar sirf same do points ke liye do alag languages hain, toh

Δ r

(aur uski length) dono tarah same niklerti hai.

Ek particle radius

R

ke semicircle par ek end se dusre end tak (diameter ke) move karta hai. Distance vs displacement?

Distance

= π R \approx 3.14 R

(arc length); displacement

= 2 R

(straight diameter). Kyunki

π R > 2 R

, inequality strictly hold karti hai, jaisa ki curved path ke liye hona chahiye.

Kisi given nonzero displacement ke liye sabse choti possible distance kya hai?

Exactly

∣Δ r ∣

hi — sirf direct straight line chalne se achieve hota hai; koi bhi aur path lambi hoti hai.