Relative motion — 1D and 2D; river-boat problems: Worked examples

1.1.21 · D3 · Physics › Measurement, Vectors & Kinematics › Relative motion — 1D and 2D; river-boat problems

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Scenario matrix

Is topic ka har problem exactly inhi cells mein se ek mein aata hai. Last column uss worked example ka naam deta hai jo use cover karta hai.

#	Cell (case class)	Tricky bit	Covered by
C1	1D, dono positive (same direction)	closing/opening speed = difference	Ex 1
C2	1D, opposite signs (toward/away)	negative subtract karna ⇒ speeds add ho jaati hain	Ex 2
C3	2D min-time crossing	current perpendicular hai ⇒ time ke liye ignore karo	Ex 3
C4	2D zero-drift crossing (opposite land karo)	upstream aim karo, $sin θ = v_{w} / v_{b w}$	Ex 4
C5	Degenerate: current ≥ boat speed	$sin θ \geq 1$ ⇒ opposite land karna impossible	Ex 5
C6	Zero current ( $v_{w} = 0$ )	dono routes ek mein collapse ho jaate hain	Ex 6
C7	General aim angle (koi bhi special route nahi)	full vector add, ek heading se drift + time	Ex 7
C8	Real-world word twist (rain–man cousin)	moving frame mein jaao wind/rain "dekhne" ke liye	Ex 8
C9	Exam twist: minimum drift jab $v_{w} > v_{b w}$	opposite nahi pahunch sakte, lekin drift minimize karo	Ex 9

Hum C1–C9 cover karenge. Cells C3, C4, C7, C9 ke figures hain kyunki unki geometry hi saari baat hai.

Ex 1 — C1: 1D, same direction

Worked example Do trains parallel tracks pe

Train A $+ 30$ m/s pe chal rahi hai, Train B $+ 18$ m/s pe, dono right ki taraf. B mein baitha passenger A ko kitni speed se move karta hua dekhta hai?

Forecast: padhne se pehle andaaza lagao — kya yeh $48$ , $12$ , ya kuch aur hoga, aur kis direction mein?

Step 1 — Sign convention chuno. Right $= +$ . Dono velocities pehle se sign ki hui hain: $v_{A} = + 30$ , $v_{B} = + 18$ . Yeh step kyun? 1D mein relative motion sirf signed arithmetic hai; bina fixed "positive direction" ke signs bekar hain.

Step 2 — "Subtract the observer" apply karo. Observer B hai, toh $v_{A B} = v_{A} - v_{B} = 30 - 18 = + 12 m/s .$ Yeh step kyun? Rule $v_{A B} = v_{A} - v_{B}$ B ki apni motion remove karta hai — hum poochhte hain "B ke saath chalna band karein toh A mein kya bachta hai?"

Step 3 — Sign padho. $+ 12$ matlab A abhi bhi B se dekha toh right ki taraf drift kar raha hai, dheere. Yeh step kyun? 1D mein sign ek direction hai; positive answer matlab "hamare +axis ke same direction mein".

Verify: B ki khidki se, A har second $12$ m aage khisakta hai — yeh everyday feeling se match karta hai ki thodi zyada fast car slowly aage nikaiti hai. Units: m/s − m/s = m/s ✓.

Ex 2 — C2: 1D, opposite directions (negative subtract karna)

Worked example Head-on approach

Car A: $+ 25$ m/s. Car B, A ki taraf $20$ m/s pe chal rahi hai, toh $v_{B} = - 20$ m/s. $v_{A B}$ nikalo (B ko A kitni speed se aata dikhta hai).

Forecast: kya subtract karne se formula ke andar $20$ ko $+ 20$ milega ya $- 20$ ?

Step 1 — Velocities ko sign karo. Right $= +$ . A right move karta hai: $v_{A} = + 25$ . B left ki taraf A ki taraf: $v_{B} = - 20$ . Yeh step kyun? "A ki taraf" ko subtract karne se pehle ek signed number mein convert karna zaroori hai, warna direction kho jaata hai.

Step 2 — Observer B ko subtract karo. $v_{A B} = v_{A} - v_{B} = 25 - (- 20) = 25 + 20 = + 45 m/s .$ Yeh step kyun? Negative subtract karne se add hota hai — yahi reason hai ki ek doosre ki taraf aane wali objects ki closing speed, speeds ka sum hoti hai.

Step 3 — Interpret karo. B ko A $45$ m/s pe aata dikh raha hai. Mirror check: $v_{B A} = v_{B} - v_{A} = - 20 - 25 = - 45$ , aur actually $v_{A B} = - v_{B A}$ .

Verify: Agar woh $450$ m apart se shuru karein toh $450/45 = 10$ s mein milenge — sensible. Sign/units consistent ✓.

Ex 3 — C3: minimum-time crossing (figure)

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

Worked example Fastest crossing

River width $d = 120$ m. Boat-through-water speed $v_{b w} = 6$ m/s. Current $v_{w} = 4$ m/s. Boat ko seedha across point karo. Crossing time aur downstream drift nikalo.

Forecast: kya $4$ m/s current crossing ko no-current time $120/6 = 20$ s se zyada lamba banata hai, ya bilkul same?

Step 1 — Fixed axes use karo. Jaise upar define kiya, $x$ = across, $y$ = downstream (figure dekho). "Seedha across" matlab $θ = 0$ , toh boat ka engine poora $+ x$ pe point karta hai. Yeh axes kyun? Velocity ko "across" aur "downstream" mein split karne se dono motions independent ho jaati hain — same trick jaise Kinematics in 2D — projectile motion mein projectile motion.

Step 2 — Crossing sirf across-component use karta hai. Seedha across aim karne se poori engine speed $x$ pe aati hai: $v_{x} = v_{b w} cos 0 = v_{b w} = 6$ m/s. Current ka koi $x$ -part nahi hai (woh $y$ ke saath hai). $t = \frac{d}{v _{x}} = \frac{120}{6} = 20 s .$ Yeh step kyun? Orange $x$ -velocity across-distance ko khaati hai; current (magenta, $y$ ke saath) perpendicular hai aur crossing ko chhoo nahi sakti — figure mein dekho ki magenta arrow path ke sideways hai.

Step 3 — Drift. Current poore $20$ s downstream push karta hai: $x_{drift} = v_{w} t = 4 \times 20 = 80 m .$ Yeh step kyun? $y$ -motion $v_{w}$ pe uniform hai; distance = speed × time.

Verify: Crossing time equals current-free time $120/6 = 20$ s — confirm karta hai ki current min-time crossing ko affect nahi karta. Ground speed $= 6^{2} + 4^{2} = 52 \approx 7.21$ m/s, downstream angled, lekin crossing abhi bhi 20 s ✓.

Ex 4 — C4: zero-drift crossing, opposite land karo (figure)

Worked example Seedha opposite land karo

Same river: $d = 120$ m, $v_{b w} = 6$ m/s, $v_{w} = 4$ m/s. Ab upstream aim karo taaki boat seedha across land kare (koi drift nahi). Heading angle aur crossing time nikalo.

Forecast: kya yeh crossing Ex 3 ke 20 s se jaldi hogi ya deri se?

Step 1 — Zero downstream velocity demand karo. Heading angle $θ$ use karke (upstream from straight-across, jaise upar define kiya), boat ka upstream component current ko exactly cancel kare: $v_{b w} sin θ = v_{w} .$ Yeh step kyun? Agar koi bhi $y$ -velocity bachi, boat drift karegi. Sirf boat ka upstream part $v_{b w} sin θ$ current se lad sakta hai, toh hum use equal set karte hain.

Step 2 — Angle solve karo. $sin θ = \frac{v _{w}}{v _{b w}} = \frac{4}{6} = 0.6667 \Rightarrow θ \approx 41. 8^{\circ} upstream .$ Yeh step kyun? $sin$ angle-of-aim ko engine speed ka woh fraction banata hai jo upstream jaata hai; $arcsin$ batata hai "kaun sa angle woh fraction deta hai?"

Step 3 — Across-speed wahi hai jo bach jaati hai. Remaining component river cross karta hai: $v_{across} = v_{b w} cos θ = v_{b w}^{2} - v_{w}^{2} = 36 - 16 = 20 \approx 4.472 m/s .$ Kyun $v_{b w}^{2} - v_{w}^{2}$ ? Engine vector, uska upstream part $v_{w}$ , aur across part ek right triangle banate hain (figure) — Pythagoras across leg deta hai.

Step 4 — Crossing time. $t = \frac{d}{v _{across}} = \frac{120}{20} \approx 26.83 s .$ Yeh step kyun? Sirf across-component $v_{across}$ river ki width $d$ ko khaata hai; time = distance ÷ that across-speed, exactly Ex 3 jaisa lekin reduced (cosine-cut) speed ke saath.

Verify: $26.83 > 20$ — zero drift time cost karta hai, exactly jaise parent ka trade-off predict karta hai. Aur yahan ground velocity purely across hai ( $4.472$ m/s, $0$ downstream) ✓.

Ex 5 — C5: degenerate case, current boat ko beat karta hai

Worked example Jeet nahi sakte

Ek weak swimmer: $v_{b w} = 2$ m/s. Fast current $v_{w} = 3$ m/s, river width $d = 50$ m. Kya woh seedha opposite land kar sakte hain? Nahi toh kya hoga?

Forecast: kya koi bhi angle zero drift deta hai?

Step 1 — Zero-drift condition try karo. $θ$ upstream se straight-across measure kiya, hamesha ki tarah: $sin θ = \frac{v _{w}}{v _{b w}} = \frac{3}{2} = 1.5.$ Yeh step kyun? Yahi ek condition hai jo drift hatati hai; agar yeh poori nahi ho sakti, toh goal impossible hai.

Step 2 — Feasibility check karo. $sin θ$ kabhi $1$ se zyada nahi ho sakta (kisi bhi real angle ka sine $[- 1, 1]$ mein hota hai). Kyunki $1.5 > 1$ , koi real angle $θ$ exist nahi karta. Yeh step kyun? Swimmer jo sabse zyada upstream push kar sakta hai woh uski poori speed $v_{b w} = 2$ m/s hai (dead upstream aim karo, $θ = 9 0^{\circ}$ ), jo abhi bhi $3$ m/s current se kam hai.

Step 3 — Physical consequence. Chahe heading kaisi bhi ho, kuch downstream velocity hamesha bachti hai → swimmer hamesha downstream carry hota hai. Woh far bank tak pahunch sakta hai, bas kabhi opposite point pe nahi. Yeh step kyun? "No real $θ$ " algebra ko translate karna hai ki paani mein actually kya hota hai — impossible equation matlab target unreachable hai, motion ruk nahi jaati. Physical outcome ka naam dena hi maths ko trustworthy banata hai.

Boundary $v_{w} = v_{b w}$ (exactly tied). Agar current swimmer speed ke equal ho, toh $sin θ = v_{w} / v_{b w} = 1$ , toh $θ = 9 0^{\circ}$ : swimmer ko dead upstream aim karna hoga. Tab across-component $v_{b w} cos 9 0^{\circ} = 0$ hai — swimmer across mein zero progress karta hai aur crossing time $t = d /0$ infinite hai. Toh $v_{w} = v_{b w}$ knife-edge hai: tie pe tum drift barely rok sakte ho lekin actually cross nahi karte; opposite point finite time mein reach karne ke liye $v_{b w} > v_{w}$ strictly chahiye.

Verify: Opposite reach karne ki condition hai $v_{b w} > v_{w}$ (strict); yahan $2 < 3$ fail hai, aur tie $v_{w} = v_{b w}$ bhi $v_{across} = 0$ deta hai ✓. (Ex 9 dikhata hai ki swimmer ko actually kya karna chahiye.)

Ex 6 — C6: zero current (doosri taraf se degenerate)

Worked example Still water

Lake, koi current nahi: $v_{w} = 0$ . Boat $v_{b w} = 5$ m/s, width $d = 80$ m. Min-time aur zero-drift routes compare karo.

Forecast: kya dono routes alag answers denge jab koi current nahi hai?

Step 1 — Min-time route. Seedha across aim karo ( $θ = 0$ ): $t = d / v_{b w} = 80/5 = 16$ s, drift $= v_{w} t = 0 \times 16 = 0$ . Yeh step kyun? Straight-across hamesha time minimize karta hai; current nahi hone se drift term apne aap zero ho jaata hai.

Step 2 — Zero-drift route. $sin θ = v_{w} / v_{b w} = 0/5 = 0 \Rightarrow θ = 0^{\circ}$ (seedha across aim karo). Tab $t = d / v_{b w}^{2} - v_{w}^{2} = 80/ 25 - 0 = 80/5 = 16$ s. Yeh step kyun? $v_{w} = 0$ ke saath dono conditions same heading maangti hain — routes fuse ho jaate hain.

Step 3 — Conclusion. Koi current nahi ⇒ min-time aur zero-drift identical straight-across trip hai, $16$ s, zero drift. Yeh step kyun? Check karna ki dono formulas $v_{w} = 0$ pe same answer deti hain, hamare sanity limit ki tarah kaam karta hai: koi bhi correct river formula ordinary "no-current" crossing mein reduce honi chahiye jab current off ho.

Verify: Dono formulas $16$ s return karti hain; drift $0$ ✓. Yeh har pehle waale formula ka sensible limit hai jab $v_{w} \to 0$ .

Ex 7 — C7: general heading (koi bhi special route nahi) (figure)

Worked example Arbitrary aim

Ek boat $v_{b w} = 8$ m/s ke saath straight-across se $θ = 3 0^{\circ}$ upstream aim karti hai. Current $v_{w} = 3$ m/s, width $d = 90$ m. Crossing time, downstream drift, aur ground speed nikalo.

Forecast: kya drift hogi, aur kya woh upstream hogi ya downstream?

Step 1 — Axes yaad karo aur engine velocity split karo. Jaise upar fix kiya, $x$ = across, $y$ = downstream ( $+$ ), aur $θ$ $+ x$ (straight-across) direction se upstream measure kiya gaya hai. Figure yeh heading dikhata hai. Engine velocity ko components mein split karo: $v_{x} = v_{b w} cos θ = 8 cos 3 0^{\circ} = 8 (0.8660) = 6.928 m/s (across, + x),$ $v_{b y} = - v_{b w} sin θ = - 8 sin 3 0^{\circ} = - 8 (0.5) = - 4 m/s (upstream, so - y) .$ Yeh step kyun? Axes restate karna har sign ko honest rakhta hai; components across aur downstream ko alag handle karne dete hain, aur minus mark karta hai ki aim current ke against hai. Figure mein orange arrow yeh engine vector hai, upstream tilted.

Step 2 — Current add karo $y$ mein ground velocity pane ke liye. Figure mein magenta arrow current hai, tip-to-tail add kiya gaya: $v_{g y} = v_{b y} + v_{w} = - 4 + 3 = - 1 m/s (net upstream) .$ Yeh step kyun? Chain rule $v_{b g} = v_{b w} + v_{w}$ component-wise kaam karta hai; current sirf $y$ ko touch karta hai. Figure mein violet resultant $v_{b g}$ hai.

Step 3 — Crossing time (sirf $v_{x}$ use karta hai). $t = \frac{d}{v _{x}} = \frac{90}{6.928} \approx 12.99 s .$ Yeh step kyun? Crossing sirf $x$ ka kaam hai; aim ka across-component $v_{x}$ (current se unchanged) width $d$ khaata hai, toh time = width ÷ $v_{x}$ .

Step 4 — Drift. Signed drift $= v_{g y} t = (- 1) (12.99) \approx - 12.99$ m — yani lagbhag $13$ m upstream start se. Yeh step kyun? Drift woh hai jitna net downstream velocity boat ko crossing time mein le jaati hai; kyunki $v_{g y}$ negative hai (net upstream), boat opposite se upstream land karti hai — figure mein violet resultant thoda upstream tilt karta hua match karta hai. Aim ne current se zyada lada.

Step 5 — Ground speed. $∣ v_{b g} ∣ = v_{x}^{2} + v_{g y}^{2} = 6.92 8^{2} + 1^{2} = 48 + 1 = 49 = 7 m/s .$ Yeh step kyun? Do perpendicular ground-velocity components ( $v_{x}$ across, $v_{g y}$ along-stream) ek right triangle banate hain; Pythagoras woh true speed deta hai jo bank observer measure karta hai — figure mein violet arrow ki length.

Verify: Negative drift upstream landing confirm karta hai (aim ne current ko beat kiya). $49 = 7$ exactly ✓. Full vector addition use karta hai — dekho Vectors — addition, components, unit vectors.

Ex 8 — C8: real-world word twist (rain-man cousin)

Worked example Cyclist pe wind

Ek cyclist due north $6$ m/s pe ride kart hai. Usse wind north-west se aati lagti hai, seedha $4 5^{\circ}$ uske face mein, $62$ m/s pe. Wind ki ground ke upar true velocity kya hai?

Forecast: kya real wind jo woh feel kati hai usse tez hai ya dheemi?

Step 1 — Frames naam karo. $c$ = cyclist, $g$ = ground, $W$ = wind. Woh $v_{W c}$ measure karti hai (wind relative to her). Hume $v_{W g}$ chahiye. Yeh step kyun? "Wind jo woh feel karti hai" ek relative velocity hai; pole pe laga jhanda ground wind dikhata hai. Dono uski motion se alag hain — exactly Rain-man umbrella problem setup.

Step 2 — Chain rule use karo. $v_{W g} = v_{W c} + v_{c g} .$ Yeh step kyun? Inner index $c$ cancel hota hai, deta hai wind-over-ground: wind jo woh feel karti hai plus uski apni motion.

Step 3 — Axes chuno aur components daalo. East $= + x$ , north $= + y$ ho. "Wind from the north-west" matlab woh south-east ki taraf blow karti hai, yani $+ x$ (east) axis se $- 4 5^{\circ}$ pe. Uske components hain: $v_{W c} = 62 (cos (- 4 5^{\circ}), sin (- 4 5^{\circ})) = 62 (\frac{1}{2}, - \frac{1}{2}) = (+ 6, - 6) m/s,$ yani $+ 6$ east, $- 6$ south (uske face mein aur usse push back karti hai). Uski apni velocity due north hai: $v_{c g} = (0, + 6)$ m/s. Yeh step kyun? "North-west se $4 5^{\circ}$ pe" words ko signed components mein turn karna hi ek tarika hai dono velocities ko cleanly add karne ka.

Step 4 — Vectors add karo. $v_{W g} = v_{W c} + v_{c g} = (6, - 6) + (0, 6) = (6, 0) m/s .$ Yeh step kyun? Uska northward $+ 6$ add karna felt southward $- 6$ cancel kar deta hai, sirf eastward vector bachta hai — uski forward motion "head-on" feeling manufacture kar rahi thi.

Step 5 — Interpret karo aur magnitude nikalo. True wind due east blow karta hai: $∣ v_{W g} ∣ = 6^{2} + 0^{2} = 6 m/s .$ Yeh step kyun? Component answer $(6, 0)$ ko plain language mein convert karna zaroori hai — due-east wind of $6$ m/s — kyunki yahi "wind's true velocity" maanga tha.

Verify: True wind ( $6$ m/s) felt $62 \approx 8.49$ m/s se dheema hai ✓ — motion hamesha jo wind tum feel karte ho use inflate karta hai, everyday cycling experience se match karta hai.

Ex 9 — C9: exam twist — minimize drift jab tum opposite land nahi kar sakte (figure)

Worked example Strong current ke against best kya kar sakte ho

Wapas Ex 5 ke swimmer pe: $v_{b w} = 2$ m/s, $v_{w} = 3$ m/s, width $d = 60$ m. Opposite landing impossible hai. Kaun sa heading sabse kam drift deta hai, aur kitna drift hoga?

Forecast: kya swimmer ko jitna ho sake utna upstream aim karna chahiye, ya kuch beech mein?

Step 1 — Drift ko heading ke function ke roop mein set up karo. Angle $θ$ upstream aim karo straight-across se (same baseline hamesha ki tarah). Across-speed $= v_{b w} cos θ$ , net downstream speed $= v_{w} - v_{b w} sin θ$ . Crossing time $t = d / (v_{b w} cos θ)$ , toh $drift (θ) = (v_{w} - v_{b w} sin θ) \cdot \frac{d}{v _{b w} c o s θ} .$ Yeh step kyun? Hum drift ko zero force nahi kar sakte, toh ise ek quantity ki tarah treat karte hain jise hum minimize karein apni ek free choice — heading $θ$ — par.

Step 2 — $θ$ ke upar minimize karo. Drift expression differentiate karke zero set karo (ya standard result use karo) toh drift exactly smallest hoti hai jab: $cos θ = \frac{v _{b w}}{v _{w}} = \frac{2}{3} = 0.6667 \Rightarrow θ \approx 48.1 9^{\circ} upstream .$ Yeh step kyun? Yeh heading woh sweet spot hai jo do opposing effects balance karta hai: zyada upstream aim karna current se zyada ladta hai lekin kam across-speed bacha hai (toh leftover current ke liye zyada time). Minimum wahan hai jahan woh trade off karte hain. Note hai $cos$ , $sin$ nahi — Ex 4 ke zero-drift rule se alag hai.

Step 3 — Minimum drift compute karo. $cos θ = 2/3$ ke saath $sin θ = 1 - (2/3)^{2} = 5 /3 \approx 0.7454$ milta hai. Back substitute karo: $drift_{m i n} = \frac{d v _{w}^{2} - v _{b w}^{2}}{v _{b w}} = \frac{60 9 - 4}{2} = \frac{60 5}{2} = 305 \approx 67.08 m .$ Yeh step kyun? Optimal angle ko $drift (θ)$ mein plug karne se messy trig is tidy closed form mein collapse ho jaati hai, jo physically achieve hone wala sabse chhota downstream distance hai — figure mein curve ke bottom pe violet point.

Step 4 — Heading pe sanity check. Optimal $θ \approx 48. 2^{\circ}$ dead upstream ( $9 0^{\circ}$ ) se kam upstream hai: fully upstream point karne se across-speed khatam ho jaati aur swimmer strand ho jaata. Toh best plan ek partial upstream aim hai. Yeh step kyun? Yeh tempting wrong answer "jitna ho sake upstream aim karo" se bachata hai — forecast trap — yeh dikhake ki extreme heading actually kyun buri hai. Figure mein drift curve steeply oopar jaata hai jab $θ \to 9 0^{\circ}$ .

Verify: Clean formula $d v_{w}^{2} - v_{b w}^{2} / v_{b w}$ require karta hai $v_{w} > v_{b w}$ (yahan $3 > 2$ ) — exactly woh regime jahan opposite-landing fail hua tha. $305 \approx 67.08$ m downstream ✓. Yeh woh "consolation prize" heading hai jo strong current tumhare liye chhod deta hai.

Recall Tumhare liye kaun sa cell sabse mushkil tha?

Har example ko apni matrix row se trace karo. Agar koi bhi cell abhi bhi shaky lagti hai, sabse jaldi fix yeh hai ki uska heading rule ( $sin θ$ zero drift ke liye, $cos θ$ min drift ke liye) component picture se re-derive karo. Zero-drift heading rule ::: $sin θ = v_{w} / v_{b w}$ (upstream aim karo) Min-drift heading jab $v_{w} > v_{b w}$ ::: $cos θ = v_{b w} / v_{w}$ Min-drift value jab opposite impossible ho ::: $d v_{w}^{2} - v_{b w}^{2} / v_{b w}$ Boundary case $v_{w} = v_{b w}$ ::: $sin θ = 1$ , $θ = 9 0^{\circ}$ , across-speed $0$ , crossing time infinite — $v_{b w} > v_{w}$ strictly chahiye

Connections

Vectors — addition, components, unit vectors — har 2D example ek component split + add hai.
Kinematics in 2D — projectile motion — "independent perpendicular axes" trick reused.
Frames of reference & Galilean transformation — subtraction rule ka formal basis.
Newton's laws — inertial frames — kyun constant relative velocity physics same rakhti hai.
Rain-man umbrella problem — Ex 8 uska twin hai.
Relative motion — 1D and 2D; river-boat problems — parent topic.