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

Rule hai "subtract the seer" — observer B hai, toh ${\vec{v}}_B$ subtract karo: $v_{AB}=v_A-v_B=30-18=+12\text{ m/s}$ Matlab: B ke passenger ko train A $12$ m/s par aage badhti lagti hai. $+$ sign kehta hai "$+$ (rightward) direction mein." ✓

$4$ m/s engine ki koshish hai paani ke through → woh hai $==v_{bw}==$. Bank-measured speed $v_{bg}$ hai (generally alag), aur $v_w$ current hai — ek alag number. ✓

Seedha across point karne ka matlab hai ki saari engine speed $x$-axis (across) par baith jaati hai, toh across-speed poori $6$ m/s hai. (a) $t_{\min }=\frac{d}{v_{bw}}=\frac{120}{6}=20\text{ s}$ (b) Drift downstream hai, yaani $y$-axis ke along, toh hum ise $y_{\text{drift}}$ likhte hain. Un $20$ s mein current boat ko $2$ m/s par downstream slide karta hai: $y_{\text{drift}}=v_w{t}_{\min }=2\times 20=40\text{ m}$ Kyun current ne crossing slow nahi kiya: current purely $y$ ke along hai, $x$ ke along crossing motion ke perpendicular. Perpendicular velocity $x$-distance ko touch nahi kar sakti. ✓

Figure mein words mein: cyan engine vector upstream-and-across aim kiya hua, ek white current vector uski tip par tip-to-tail chain se laga downstream point karta hua, aur amber resultant page par seedha upar khada (pure across) — proof ki sideways parts cancel ho gaye. Ek chhota white arc steering angle $\theta$ mark karta hai. Saari downstream drift khatam karne ke liye, boat ka upstream component exactly current ko cancel karna chahiye: $v_{bw}\sin \theta =v_w\Rightarrow \sin \theta =\frac{v_w}{v_{bw}}=\frac{2}{6}=\frac{1}{3}$ (a) $\theta =\arcsin (1/3)\approx 19.47^{\circ }$ upstream from straight-across (figure mein "theta" marked white arc). (b) Bacha hua component across point karta hai. Kyun hum likh sakte hain $v_{bw}\cos \theta =\sqrt{v_{bw}^2-v_w^2}$: engine vector $v_{bw}$ ek right triangle ka hypotenuse hai jiske do legs hain upstream part $v_{bw}\sin \theta$ aur across part $v_{bw}\cos \theta$. Humne pehle se upstream leg ko $v_w$ ke barabar force kar diya. Pythagoras se across leg hai $\sqrt{v_{bw}^2-(v_{bw}\sin \theta {)}^2}=\sqrt{v_{bw}^2-v_w^2}$ — wahi identity jaise $\cos \theta =\sqrt{1-{\sin }^2\theta }$, bas $v_{bw}$ se scale kiya gaya. Isliye: $v_{\text{across}}=v_{bw}\cos \theta =\sqrt{v_{bw}^2-v_w^2}=\sqrt{36-4}=\sqrt{32}\approx 5.657\text{ m/s}$ (c) $t=\frac{d}{v_{\text{across}}}=\frac{120}{\sqrt{32}}\approx 21.21\text{ s}$ Notice karo: $21.21>20$ — zero drift time khaata hai, exactly jaisa parent note ne warn kiya tha. ✓

Pehle zero-drift condition test karo: $\sin \theta =\frac{v_w}{v_{bw}}=\frac{2.0}{1.5}\approx 1.333>1$ Kisi angle ka sine $1$ se bada nahi ho sakta, toh seedha opposite land karna impossible hai — current hamesha jeet jaata hai.

Ab least-drift heading. Figure dekho. Figure mein words mein: ek white current vector downstream axis ke along; $v_{bw}$ radius ka ek cyan circle current ki tip par centred (boat ki engine tip is circle par kahin land hoti hai jaise heading swing karti hai); aur origin se us circle par ek point tak amber ground-velocity arrow. Least drift angle $\delta$ tab hota hai jab amber arrow circle par tangent ho.

Kyun "tangent" answer hai: jaise hum heading swing karte hain, amber tip circle ke around chalti hai. $\delta$ angle jo ${\vec{v}}_{bg}$ across-direction ke saath banata hai, tab sabse chhota hota hai jab amber line circle ko bass graze kare (tangent ho) — us se aage push karo aur arrow wapas swing karta hai, $\delta$ badhaata hai. Tangency point par radius tangent line ke perpendicular hoti hai. Yahan radius ${\vec{v}}_{bw}$ hai aur tangent line ${\vec{v}}_{bg}$ ke along hai, toh optimum par ${\vec{v}}_{bg}\perp {\vec{v}}_{bw}$.

Optimum par right triangle. ${\vec{v}}_{bg}\perp {\vec{v}}_{bw}$ ke saath, teen vectors velocity space mein ek right triangle banate hain: hypotenuse $v_w$ (current), ek leg $v_{bw}$ (engine, $v_{bg}$ ke perpendicular), aur doosra leg $v_{bg}$ khud. Toh: $v_{bg}=\sqrt{v_w^2-v_{bw}^2}=\sqrt{4-2.25}=\sqrt{1.75}\approx 1.3229\text{ m/s.}$ Heading. Us triangle mein current aur engine ke beech angle satisfy karta hai $\cos {\theta }_{\text{best}}=v_{bw}/v_w$. Straight-across se upstream measure kiya gaya (puri note mein $\theta$ ka same reference), heading ${\theta }_{\text{best}}=90^{\circ }-\arccos (v_{bw}/v_w)$ nikalta hai. Numerically $\arccos (1.5/2.0)=41.41^{\circ }$, toh ${\theta }_{\text{best}}=90^{\circ }-41.41^{\circ }=48.59^{\circ }$ upstream from straight-across.

Ab ${\vec{v}}_{bg}$ ko uske across aur downstream parts mein split karo — ye drift decide karte hain. Drift angle $\delta$ across-axis se ${\vec{v}}_{bg}$ ka tilt hai, aur same right triangle se $\sin \delta =v_{bw}/v_w$, $\cos \delta =\sqrt{v_w^2-v_{bw}^2}/v_w$. Isliye:

• across-speed $v_x=v_{bg}\cos \delta =\frac{v_{bw}}{v_w}\sqrt{v_w^2-v_{bw}^2}=\frac{1.5}{2.0}\sqrt{1.75}\approx 0.9922$ m/s,
• downstream-speed $v_y=v_{bg}\sin \delta =\frac{(v_w^2-v_{bw}^2)}{v_w}...$ — lekin humein sirf drift ratio chahiye, jo hai $\tan \delta$: $\frac{v_y}{v_x}=\tan \delta =\frac{\sqrt{v_w^2-v_{bw}^2}}{v_{bw}}=\frac{\sqrt{1.75}}{1.5}\approx \mathrm{0.8819.}$

Crossing time across-speed use karta hai: $t=\frac{d}{v_x}=\frac{60}{0.9922}\approx 60.47\text{ s.}$ Least drift = (drift ratio) $\times$ width: $y_{\min }=\frac{v_y}{v_x}d=\tan \delta \cdot d=0.8819\times 60\approx 52.92\text{ m downstream.}$ Toh swimmer drift se bach nahi sakta; best wo kar sakta hai woh hai lagbhag $52.9$ m downstream land karna. ✓

Observer (Q) subtract karo, slot by slot (tuple order hai across, downstream): ${\vec{v}}_{PQ}={\vec{v}}_{Pg}-{\vec{v}}_{Qg}=(4-4,3-(-1))=(0,4)\text{ m/s}$ Q ko, boat P purely downstream $4$ m/s par slide karti lagti hai aur koi across-motion nahi dikhata. Speed $=\sqrt{0^2+4^2}=4$ m/s. Component subtraction ke liye Vectors — addition, components, unit vectors dekho. ✓

(a) Min-time landing drift $=v_{w}d/v_{bw}=3\cdot 90/5=54$ m downstream. Jetty sirf $30$ m downstream hai, yaani min-time point ke upstream, toh humein thoda upstream aim karna hoga — lekin fully nahi (woh $0$ m par land karega).

(b) Maano boat heading straight-across se $\theta$ angle upstream banaye. Across-velocity: $v_x=v_{bw}\cos \theta$. Net downstream velocity: $v_y=v_w-v_{bw}\sin \theta$. Cross time $t=d/v_x$. Required drift: $v_{y}t=30$. $\frac{(v_w-v_{bw}\sin \theta )d}{v_{bw}\cos \theta }=30$ Numbers plug karo ($d=90,v_{bw}=5,v_w=3$): $\frac{(3-5\sin \theta )\cdot 90}{5\cos \theta }=30\Rightarrow 18(3-5\sin \theta )=30\cos \theta$ $54-90\sin \theta =30\cos \theta \Rightarrow 9-15\sin \theta =5\cos \theta .$ Solve karo: $\theta \approx 16.26^{\circ }$ upstream test karo, jahan $\sin \theta =0.28$ ($5\sin \theta =1.4$, toh $3-1.4=1.6$) aur $5\cos \theta =4.80$; phir $\frac{1.6\cdot 90}{4.80}=30$ ✓.

(c) $v_x=5\cos 16.26^{\circ }=4.80$ m/s, toh $t=\frac{90}{4.80}=18.75\text{ s}.$ ✓

Yeh Rain-man umbrella problem hai river problem ke upar ride karta hua — frames ko layer karo. Tuple order hai (across, downstream, vertical) jaisa page ke top par fix kiya gaya. Nayi notation (pehli use par define karo): subscript $r$ ka matlab hai rain. Toh ${\vec{v}}_{rg}$ = rain ki ground ke relative velocity (jo ek still bystander dekhta hai), aur ${\vec{v}}_{rb}$ = rain ki boat ke relative velocity (jo moving passenger dekhta hai). Same "subtract the observer" rule se, ${\vec{v}}_{rb}={\vec{v}}_{rg}-{\vec{v}}_{bg}$.

Step 1 — boat ki ground velocity. Min-time: seedha across heading, toh ${\vec{v}}_{bg}=(v_{bw},v_w,0)=(5,4,0)\text{ m/s (across, downstream, vertical)}.$ Step 2 — rain relative to boat. Vertical $z$ hai, upar positive, toh girne wali baarish negative hai: ${\vec{v}}_{rg}=(0,0,-8)$. Observer (boat) subtract karo, slot by slot: ${\vec{v}}_{rb}={\vec{v}}_{rg}-{\vec{v}}_{bg}=(0-5,0-4,-8-0)=(-5,-4,-8)\text{ m/s}.$ Step 3 — speed aur direction. Horizontal part magnitude $=\sqrt{5^2+4^2}=\sqrt{41}\approx 6.40$ m/s. Total speed $=\sqrt{5^2+4^2+8^2}=\sqrt{105}\approx 10.25$ m/s. Vertical se angle: $\tan \alpha =\frac{\text{horizontal}}{\text{vertical}}=\frac{\sqrt{41}}{8}\Rightarrow \alpha =\arctan \frac{\sqrt{41}}{8}\approx 38.66^{\circ }.$ Passenger umbrella $\approx 38.7^{\circ }$ vertical se tilt karta hai, apni motion ki direction ki taraf jhukate hue (apparent rain mein). ✓

L1 :: velocities padho/naam do aur seer subtract karo. L2 :: min-time ($t=d/v_{bw}$) aur zero-drift ($\sin \theta =v_w/v_{bw}$) formulas apply karo. L3 :: feasibility judge karo ($v_{bw}>v_w$) aur least-drift tangent-circle se dhundho. L4 :: target landing point ke liye coupled drift$=v_{y}t$ equation solve karo. L5 :: boat frame par ek doosri relative-motion problem (rain) stack karo.