Relative motion — 1D and 2D; river-boat problems: Visual walkthrough

1.1.21 · D2 · Physics › Measurement, Vectors & Kinematics › Relative motion — 1D and 2D; river-boat problems

Step 1 — Velocity arrow hota kya hai?

KYA HAI. Velocity ko ek arrow ki tarah draw kiya jaata hai. Uski length = kitni tez (metres per second, likha jaata hai m/s), uski direction = motion kis taraf jaati hai.

KYUN. Pehle hum do motions ko combine karein, humein ek motion ko draw karne ke tarike par agree karna hoga. Ek arrow honest picture hai: isme ek number (length) aur ek direction dono saath hote hain — aur yahi velocity hoti hai.

PICTURE. Neeche ki figure mein, ek arrow ek dot (boat) se start hota hai aur wahan point karta hai jahan boat jaati hai. Hum apna map bhi set up karte hain: $x$ points across the river (nazdiki bank se door wali bank ki taraf), $y$ points downstream (jis taraf paani behta hai). River ki width $d$ hai — dono banks ke beech ki seedhi doori.

Figure — Relative motion — 1D and 2D; river-boat problems

Upar wala chhota arrow () bas yeh kehta hai "yeh ek vector hai — iska ek direction hai, sirf ek size nahi."

Step 2 — Boat ki ground velocity SUM kyun hoti hai

KYA HAI. Boat ki ground velocity do arrows ko tip-to-tail rakhne se milti hai: $bank kya dekhta hai v_{b g} = boat through water v_{b w} + water over ground v_{w}$

KYUN yeh hai, subtraction kyun nahi? Subscripts ko ek chain ki tarah padho: $b w + w g = b g$ . Inner letter $w$ pehle wale ke right mein aur doosre ke left mein aata hai — yeh cancel ho jaata hai, $b g$ bachta hai. Yeh parent note ka chain rule of frames hai. Physically: boat paani ke through tairti hai, aur paani ki poori sheet ground ke upar slide karti hai — toh do motions genuinely stack hoti hain.

PICTURE. $v_{w}$ ki tail ko $v_{b w}$ ki tip par rakho. Bilkul shuru se bilkul akhir tak ka arrow $v_{b g}$ hai. Yeh vector addition hai — tip-to-tail.

Step 3 — Har arrow ko across + downstream mein split karo

KYA HAI. Har velocity ko do independent pieces mein todo: ek $x$ -piece (across) aur ek $y$ -piece (downstream). $x$ -axis se angle $θ$ par tilted, length $v$ wale arrow ke liye: $v_{x} = v cos θ, v_{y} = v sin θ$

Split kyun karein? Kyunki river cross karna sirf $x$ -piece ki parwah karta hai. Door wali bank ek fixed distance $d$ across par hai; tum us gap ko kitni tez close karte ho yeh sirf tumhari across-speed par depend karta hai. Downstream piece tumhe sideways slide karta hai lekin tumhe door wali bank ke paas kabhi nahi laata. Splitting se hum do directions ko do alag 1D problems ki tarah treat kar sakte hain — wahi trick jo projectile motion mein use hoti hai.

$cos$ aur $sin$ woh tools hain jo "is arrow ka kitna hissa $x$ ke along point karta hai" aur "kitna $y$ ke along" padhte hain. $cos θ$ adjacent-over-hypotenuse fraction hai (across share); $sin θ$ opposite-over-hypotenuse hai (downstream share).

PICTURE. Tilted boat arrow $x$ -axis par ek shadow daalta hai (uska across part) aur $y$ -axis par (uska downstream part). Yeh do shadows arrow ke right-triangle legs hain.

Step 4 — Case A: SEEDHA ACROSS aim karo (fastest crossing)

KYA HAI. $v_{b w}$ ko purely $+ x$ ke along point karo (across-direction se angle $0$ ). Tab boat ki poori engine speed across jaati hai: $v_{b g, x} = v_{b w}, v_{b g, y} = v_{w}$ Crossing time sirf across-speed use karta hai: $t_{m i n} = \frac{d}{v _{b w}}$

Term by term: $d$ = river width (across cover karne ki distance); $v_{b w}$ = poori boat speed, ab 100% across aim ki gayi; unka ratio hai time = distance ÷ speed.

KYUN yeh minimum hai. Current sirf ek $y$ -velocity add karta hai — yeh crossing ke perpendicular hai. Perpendicular motion $x$ -gap ko close karne ki speed ko speed up ya slow down nahi kar sakti. Toh engine ka 100% across-direction mein lagaana sabse fast possible crossing hai. $d / v_{b w}$ se kuch bhi beat nahi kar sakta.

Cost — drift. Drift ek downstream displacement hai, toh yeh $y$ -axis par rehta hai — ise $y_{drift}$ kaho. Jab tum $t_{m i n}$ time tak cross karte ho, current tumhe downstream le jaata hai: $y_{drift} = v_{w} t_{m i n} = \frac{v _{w} d}{v _{b w}}$ Yahan $v_{w}$ = current speed, $t_{m i n}$ = upar wali crossing time; multiply karo downstream ( $y$ ) distance get karne ke liye. (Yeh $y$ ke along hai, $x$ ke along nahi — $x$ -axis across define ki gayi thi, aur crossing us waqt finish hoti hai jab tum door wali bank tak pahuncho.)

PICTURE. Boat arrow dead across point karta hai; current arrow ek downstream nudge add karta hai; ground par actual path slanted resultant hai, jo tumhe seedhe-across se downstream land karata hai.

Step 5 — Case B: UPSTREAM aim karo taaki seedha opposite land karo

KYA HAI. $v_{b w}$ ko upstream angle $θ$ par tilt karo (seedhe-across direction se measure kiya) taaki uska upstream piece current ko exactly cancel kare. "Zero drift" ka matlab hai total downstream velocity $0$ hai: $boat ka upstream piece v_{b w} sin θ = current v_{w} \Rightarrow sin θ = \frac{v _{w}}{v _{b w}}$

Term by term: $v_{b w} sin θ$ woh slice hai jo boat ki velocity flow ke against point karta hai; ise $v_{w}$ ke equal set karne se dono cancel ho jaate hain toh net downstream velocity zero ho jaati hai. $θ$ ke liye solve karne par $sin θ = v_{w} / v_{b w}$ milta hai.

KYUN sirf $sin$ piece current se ladhta hai. Current purely $y$ ke along chalta hai. Sirf boat ka $y$ -piece usse oppose kar sakta hai, aur woh piece $v_{b w} sin θ$ hai. $x$ -piece ( $v_{b w} cos θ$ ) crossing karta hai aur kabhi us ladaai mein nahi aata.

Bacha hua across-speed. Current se ladhne ke liye $sin$ slice dene ke baad, jo bachta hai woh river cross karta hai: $v_{across} = v_{b w} cos θ = v_{b w}^{2} - v_{w}^{2}, t = \frac{d}{v _{b w}^{2} - v _{w}^{2}}$ Square root seedhe right triangle se aata hai: boat arrow (hypotenuse $v_{b w}$ ) ka ek upstream leg $v_{w}$ hai, toh across leg Pythagoras se $v_{b w}^{2} - v_{w}^{2}$ hai.

PICTURE. Boat arrow upstream jhukta hai; uska downstream shadow current arrow se exactly match karta hai aur dono annihilate ho jaate hain; sirf surviving motion seedha across hai — ground path door wali bank tak ek clean perpendicular line hai.

Step 6 — Do edge cases: exact tie, aur jab current jeet jaata hai

KYA HAI. "Land opposite" recipe ko $sin θ = v_{w} / v_{b w}$ chahiye. Lekin real angle ka $sin$ kabhi $1$ se zyada nahi ho sakta. Jaise $v_{w}$ , $v_{b w}$ ki taraf badhta hai, teen regimes nazar aate hain:

$opposite land karta hai, finite time v_{b w} > v_{w} tie: θ = 9 0^{\circ}, kabhi cross nahi karta v_{b w} = v_{w} koi real θ nahi, impossible v_{b w} < v_{w}$

Exact tie $v_{w} = v_{b w}$ (ise alag treat karo). Yahan $sin θ = 1$ , toh $θ = 9 0^{\circ}$ — tum fully upstream aim karte ho. Isse across-speed $v_{across} = v_{b w} cos 9 0^{\circ} = 0$ ho jaati hai. Ek real angle exist karta hai, lekin zero across-speed ke saath crossing time $t = \frac{d}{v _{b w}^{2} - v _{w}^{2}} = \frac{d}{0} \to \infty$ blow up ho jaata hai: tum perfectly upstream point karte ho, current ke against apni jagah exactly hold karte ho, aur kabhi actually door wali bank tak nahi pahunchte. Toh " $v_{b w} \geq v_{w}$ ke liye ek steering angle exist karta hai" technically sach hai lekin misleading hai — equality par tum start ke opposite position hold kar sakte ho lekin cross nahi kar sakte. Door wale point par genuine crossing ke liye strict inequality $v_{b w} > v_{w}$ chahiye.

Jab current jeette, $v_{w} > v_{b w}$ . Boat ki sabse zyada upstream push kabhi bhi uski poori speed $v_{b w}$ ho sakti hai (fully upstream aim karo, $θ = 9 0^{\circ}$ ). Agar current $v_{w}$ usse bada hai, toh full-upstream boat bhi downstream slip karti hai. Koi real angle $sin θ = v_{w} / v_{b w} > 1$ solve nahi karta — tum seedha opposite wala point reach hi nahi kar sakte.

PICTURE. Do panels side by side: left par ( $v_{b w} > v_{w}$ ) boat ka upstream reach current ko cover karta hai aur ek real triangle close hota hai. Right par ( $v_{b w} < v_{w}$ ) boat arrow current arrow ke across reach karne ke liye bahut chhota hai — triangle close nahi ho sakta, toh koi valid heading exist nahi karta. Tie exactly beech mein baithta hai: boat arrow vertical ho jaata hai (fully upstream), across-leg zero ho jaata hai.

Step 7 — Numbers lagao

Recall Quick self-test

Yahan fastest crossing time kya hai? ::: $20$ s (seedha across aim karo) Us fastest crossing mein downstream ( $y$ ) drift kitni hai? ::: $60$ m Seedha opposite land karne ke liye angle? ::: $3 7^{\circ}$ upstream from straight-across Zero-drift crossing time? ::: $25$ s Kya $v_{b w} = 2$ m/s wali boat yahan opposite land kar sakti hai? ::: Nahi — $2 < 3$ , toh $sin θ = 3/2 > 1$ , impossible Agar $v_{b w} = v_{w} = 3$ exactly ho toh? ::: Tum start ke opposite position hold kar sakte ho, lekin across-speed $0$ hai → infinite time, toh tum kabhi actually cross nahi karte

Ek-picture summary

Sab ek saath: ek boat arrow fixed length $v_{b w}$ ka, swing karne ke liye free. Seedha aim karo → fastest crossing, lekin downstream drift ho jaata hai ( $y$ ke along). Upstream aim karo → downstream shadow current arrow ko cancel karta hai, toh tum opposite land karte ho, lekin shorter across-leg ka matlab hai slower crossing. $v_{w}$ ko $v_{b w}$ tak push karo → across-leg vanish ho jaata hai (infinite time); usse aage triangle close hi nahi ho sakta.

Recall Feynman retelling — kisi dost ko batao

Socho tum ek moving carpet par tairo. Tumhare haath tumhe carpet ke relative ek fixed speed dete hain — yeh tumhara ek arrow hai, aur tum ise jahan chaaho point kar sakte ho. Carpet khud hamesha ek fixed rate par tumhe downstream slide karta hai. Door wali wall ki taraf seedha point karo: tum kam se kam possible time mein cross karte ho, kyunki tumhari saari mehnat crossing mein jaati hai — lekin carpet tumhe downstream drop karta hai. Carpet ke slide ke kuch against point karo: tumhari kuch mehnat ab carpet ki push cancel karti hai, toh tum exactly across land karte ho — lekin crossing ke liye kam mehnat bachti hai, toh zyada time lagta hai. Agar carpet bilkul utni hi tez slide kare jitna tum tairo: fully against point karte hue, tum apni jagah hold karte ho lekin across koi progress nahi — tum kabhi door wali wall tak nahi pahunchte. Agar carpet tumse tez slide kare: fully against point karte hue bhi, tum phir bhi peeche hote ho — tum kabhi seedha across land nahi kar sakte. Same picture, chaar answers, ek adjustable arrow.

Connections

Relative motion — 1D and 2D; river-boat problems — parent note jise yeh page visualise karta hai.
Vectors — addition, components, unit vectors — Steps 2–3 pure vector addition aur components hain.
Kinematics in 2D — projectile motion — same "independent perpendicular components" idea.
Frames of reference & Galilean transformation — kyun boat/water/ground velocities add hoti hain.
Rain-man umbrella problem — is problem ka twin, current ki jagah rain ke saath.