Reference frames — Galilean transformations: Worked examples

1.1.22 · D3 · Physics › Measurement, Vectors & Kinematics › Reference frames — Galilean transformations

Shuru karne se pehle, ek reminder ki har symbol ka kya matlab hai, seedhi baat mein, taaki kuch bhi bina samjhe use na ho:

Scenario matrix

Is topic ka har problem inhi cells mein se ek hai. Har row ek case class ka naam, uski trickiness ki wajah, aur woh worked example batata hai jo neeche us case ko cover karta hai.

#	Case class	Iska chhupa hua trap	Covered by
A	1D, same direction (dono velocities parallel, same sign)	kuch nahi — scalar add kaam karta hai	Ex 1
B	1D, opposite direction (signs differ)	signs respect karne chahiye; closing speed	Ex 2
C	2D perpendicular	speeds add nahi kar sakte; chahiye components + magnitude + angle	Ex 3 (fig)
D	2D general angle (koi bhi axis ke along nahi)	pehle dono vectors ko components mein resolve karo	Ex 4 (fig)
E	Zero / degenerate input ( $u^{'} = 0$ ya $V = 0$ )	check karo ki formula phir bhi theek behave karta hai	Ex 5
F	Limiting behaviour (kya hota hai jab speed → 0 ya badhti hai)	trends ke baare mein reasoning, ek number nahi	Ex 6
G	Acceleration invariance (forces/decel)	$a^{'} = a$ ; velocity irrelevant	Ex 7
H	Real-world word problem (straight land karne ke liye upstream aim karo)	sahi unknown choose karo	Ex 8 (fig)
I	Exam twist (frame implicitly diya gaya / relative-velocity phrasing)	words ko $u_{A / B} = u_{A} - u_{B}$ mein translate karo	Ex 9

Worked examples

Worked example Ex 2 — Cell B: 1D, opposite direction

Do cars head-on approach kar rahe hain, dono road ke relative $30 m/s$ par. Car A ki velocity car B ke frame mein dhundho — closing speed.

Forecast: Kya yeh $0$ , $30$ , ya $60 m/s$ hai? Jawab commit karo.

East $= + x$ choose karo. Maano A east move karta hai ( $u_{A} = + 30$ ), B west move karta hai ( $u_{B} = - 30$ ). Yeh step kyun? Opposite motion ka matlab opposite signs — isi cell ka poora point.
Relative-velocity form use karo $u_{A / B} = u_{A} - u_{B}$ . Yahan $S^{'}$ = car B ka frame $V = u_{B}$ par move kar raha hai, toh $V$ subtract karna hai hi $u_{B}$ subtract karna. $u_{A / B} = 30 - (- 30) = 60 m/s .$ Yeh step kyun? B ki seat se, A dono road-speeds ke sum par rush karta hai.

Verify: Do minus ek plus banate hain, toh speeds add hoti hain — yeh us gut feeling se match karta hai ki head-on approach scary-fast hoti hai. Units m/s ✓.

Worked example Ex 3 — Cell C: 2D perpendicular (river-boat)

Ek river east mein $V = 3 m/s$ par flow karti hai. Ek boat paani ke relative due north point karta hai $u^{'} = 4 m/s$ par. Boat ki ground velocity dhundho: speed aur direction.

Forecast: Ground speed $7$ , $5$ , ya $1 m/s$ hogi? Kis direction mein — pure north?

Dono ko components ke roop mein likho $(x, y)$ : $u^{'} = (0, 4)$ (north), $V = (3, 0)$ (east). Yeh step kyun? Perpendicular arrows ko numbers ki tarah add nahi kiya ja sakta; components hume axis-by-axis add karne dete hain.
Component-wise add karo: $u = u^{'} + V = (0 + 3, 4 + 0) = (3, 4) m/s$ . Yeh step kyun? Yeh head-to-tail rule hai — figure mein green resultant dekho.
Magnitude Pythagoras se: $∣ u ∣ = 3^{2} + 4^{2} = 25 = 5 m/s$ . Yeh step kyun? Do components ek right triangle ki legs hain; resultant hypotenuse hai.
Direction: $θ = tan^{- 1} (\frac{u _{y}}{u _{x}}) = tan^{- 1} (4/3) \approx 53.1 3^{\circ}$ north of east measure kiya gaya. Yeh step kyun? $tan$ us triangle par "opposite over adjacent" hai; $tan^{- 1}$ poochta hai "kis angle ka yeh ratio hai?" Dono components positive hain, toh hum safely first quadrant mein hain — koi sign fix needed nahi.

Verify: $3$ - $4$ - $5$ triangle ✓. Boat east drift karti hai bhaave usne straight north aim kiya — yeh eastward push bilkul current hai. Units m/s ✓.

Worked example Ex 4 — Cell D: 2D general angle

Ek train par jo east mein $V = 10 m/s$ par move kar rahi hai, ek passenger ek ball $u^{'} = 10 m/s$ par throw karta hai jo train ke relative $3 0^{\circ}$ north of east aim kiya gaya hai. Ground velocity?

Forecast: Kya ground speed $20 m/s$ se zyada hogi, ya $19.3$ ke paas aayegi? Guess karo.

$u^{'}$ ko components mein resolve karo. $cos 3 0^{\circ} = \frac{3}{2} \approx 0.8660$ , $sin 3 0^{\circ} = 0.5$ ke saath: $u^{'} = (10 cos 3 0^{\circ}, 10 sin 3 0^{\circ}) = (8.6603, 5.0) m/s .$ Yeh step kyun? Jab koi bhi vector axis par nahi hota toh add karne se pehle use east/north parts mein todna zaroori hai.
Frame velocity add karo $V = (10, 0)$ : $u = (8.6603 + 10, 5.0 + 0) = (18.6603, 5.0) m/s .$ Yeh step kyun? Galilean addition phir — lekin ab eastward parts pile up ho rahe hain.
Magnitude: $∣ u ∣ = 18.660 3^{2} + 5. 0^{2} \approx 348.21 + 25 = 373.21 \approx 19.32 m/s .$ Yeh step kyun? Same right-triangle idea; resultant summed legs ka hypotenuse hai.
Direction: $θ = tan^{- 1} (5.0/18.6603) \approx 15. 0^{\circ}$ north of east. Yeh step kyun? Dono summed components positive hain → first quadrant → plain $tan^{- 1}$ sahi hai, koi $\pm 18 0^{\circ}$ correction nahi.

Verify: Angle $3 0^{\circ}$ se $1 5^{\circ}$ tak shrink hua — purely eastward $V$ add karna ball ko zyada eastward tilt karta hai, toh chhota north-angle sahi hai ✓. Speed $19.32 < 20$ kyunki dono vectors parallel nahi hain; parallel hote toh exactly $20$ milta ✓.

Worked example Ex 5 — Cell E: zero / degenerate input

(a) Train seat par ek cup rakhha hai: $u^{'} = 0$ . Train $V = 20 m/s$ east par. Ground velocity? (b) Train parked hai, $V = 0$ . Passenger $u^{'} = 1.4 m/s$ par walk karta hai. Ground velocity?

Forecast: (a) mein, kya ground cup ko move karte dekhta hai? (b) mein, kya frame kuch change karta hai?

(a) $u = u^{'} + V = 0 + 20 = 20 m/s$ east. Yeh step kyun? Train mein rest par koi object train ki speed par carry ho jaata hai — formula "cup bas $V$ ke saath ride karta hai" mein reduce ho jaata hai.
(b) $u = u^{'} + V = 1.4 + 0 = 1.4 m/s$ . Yeh step kyun? $V = 0$ ke saath dono frames coincide karte hain; $u = u^{'}$ exactly, toh transformation kuch nahi karta — ek sanity check ki formula sahi se degrade hota hai.

Verify: (a) us roz ki baat se match karta hai ki moving train par coffee ground ke upar $20 m/s$ par tumhare saath move karti hai ✓. (b) mein $V = 0$ ke saath map identity hai ✓. Units m/s ✓.

Worked example Ex 6 — Cell F: limiting behaviour

Ex 3 ki river mein ek swimmer north cross karta hai $u^{'} = 4 m/s$ par. Drift angle $ϕ$ dekho (ground path kitna east of north tilt hota hai) jab current $V$ $0$ se upar range karta hai.

Forecast: Jab $V \to 0$ , kya $ϕ \to 0^{\circ}$ ya $\to 9 0^{\circ}$ ? Jab $V$ bahut bada ho jaata hai?

East-of-north drift angle likho: $ϕ = tan^{- 1} (V / u^{'}) = tan^{- 1} (V /4)$ . Yeh step kyun? East component $V$ hai, north component $u^{'} = 4$ hai; north se tilt opposite/adjacent $= V / u^{'}$ hai.
Limit $V \to 0$ : $tan^{- 1} (0) = 0^{\circ}$ . Path pure north hai. Yeh step kyun? Koi current nahi matlab koi sideways push nahi, toh koi drift nahi — formula ko exactly $0^{\circ}$ dena chahiye, aur deta hai.
Limit $V \to \infty$ : $tan^{- 1} (huge) \to 9 0^{\circ}$ . Path almost pure east ban jaata hai. Yeh step kyun? Overwhelming current swimmer ko sideways sweep kar leta hai; tiny $4 m/s$ north barely register karta hai. $tan^{- 1}$ $9 0^{\circ}$ par saturate ho jaata hai, kabhi exceed nahi karta.
Sample $V = 4$ : $ϕ = tan^{- 1} (1) = 4 5^{\circ}$ — equal east aur north pushes. Yeh step kyun? Ek concrete midpoint confirm karta hai ki trend smooth aur monotone hai.

Verify: $ϕ$ monotonically $0^{\circ}$ se $9 0^{\circ}$ ki taraf rise karta hai aur $4 5^{\circ}$ hit karta hai jab $V = u^{'}$ ✓ — intuition se match karta hai ki "faster current = zyada drift, lekin kabhi sideways se aage nahi."

Worked example Ex 7 — Cell G: acceleration invariance

Ex 4 ki ball par air drag $a^{'} = 2 m/s^{2}$ deceleration feel karti hai apni motion ke opposite, train par measure kiya gaya. Ground observer kya acceleration measure karta hai?

Forecast: Bada, chhota, ya exactly $2 m/s^{2}$ ?

$a^{'} = a$ yaad karo. $u = u^{'} + V$ differentiate karo; kyunki $V$ constant hai, $d V / d t = 0$ , toh $a = a^{'}$ . Yeh step kyun? Acceleration velocity ka time-derivative hai, aur frames ke beech ka fark ( $V$ ) ek constant hai jo differentiation mein zero ho jaata hai.
Answer read karo: ground $a = 2 m/s^{2}$ measure karta hai, same magnitude, same direction. Yeh step kyun? Invariance ka matlab koi translation needed nahi — number seedha carry ho jaata hai.

Verify: Train ki speed ( $10 m/s$ ) kahin bhi appear nahi hoti — sahi hai, kyunki constant velocity zero acceleration contribute karti hai ✓. Units m/s² ✓. Yahi wajah hai ki $F = m a$ har inertial frame mein hold karta hai (dekho Newton's laws of motion).

Worked example Ex 8 — Cell H: real-world word problem (straight land karne ke liye aim karo)

River phir east mein $V = 3 m/s$ par flow karti hai. Boat paani ke relative $u^{'} = 5 m/s$ kar sakta hai. Kis angle par aim karna chahiye taaki ground path due north (seedha across) ho? Aur resulting ground speed kya hai?

Forecast: Thoda east aim karo, thoda west, ya dead north?

Zero east-drift demand karo. Ground velocity pure north point kare iske liye, boat ka khud ka eastward component current cancel kare: $u^{'} sin α = V$ , jahan $α$ aim angle hai west of north. Yeh step kyun? Hum unknown heading solve kar rahe hain, toh hum yeh condition impose karte hain ki east components ka sum zero ho.
$α$ solve karo: $sin α = V / u^{'} = 3/5 = 0.6 \Rightarrow α = sin^{- 1} (0.6) \approx 36.8 7^{\circ}$ west of north (upstream). Yeh step kyun? $sin^{- 1}$ jawab deta hai "kis angle ka yeh sine hai?"; kyunki $V < u^{'}$ ek real angle exist karta hai — boat current ko beat kar sakti hai.
Ground speed = surviving north component: $u = u^{'} cos α = 5 cos (36.8 7^{\circ}) = 5 \times 0.8 = 4 m/s$ . Yeh step kyun? $u^{'}$ ka sirf north part bachta hai jab east part cancel ho jaata hai.

Verify: $3$ - $4$ - $5$ triangle yahan chhupa hai: $u^{'} = 5$ , cancelled east $= 3$ , net north $= 4$ ✓. Boat upstream (west of north) aim karti hai, jo intuition se match karta hai — tum current se ladte ho seedha jaane ke liye ✓. Degenerate check: agar $V > u^{'}$ toh $sin α > 1$ ka koi solution nahi → boat seedha north path hold nahi kar sakti, sahi physics.

Worked example Ex 9 — Cell I: exam twist (implicit frames)

"Train P par ek observer (east mein $15 m/s$ par move kar raha hai) train Q ko east se $25 m/s$ par approach karte dekhta hai (yani Q, P par $25 m/s$ par close karta hai). Q ki velocity ground ke relative kya hai?" Frame wording mein chhupa hua hai — ise suljhao.

Forecast: Kya Q ki ground speed $40$ , $25$ , ya $10 m/s$ hai? Kis direction mein?

Axis ke saath sab kuch naam do. East $= + x$ . P ki ground velocity $u_{P} = + 15$ . Q east se approach karta hai, toh P ke relative, Q west move karta hai: $u_{Q / P} = - 25$ . Yeh step kyun? "East se approach karna" sign fix karta hai — Q P ki taraf ja raha hai, yaani P ke view mein westward.
$u_{Q / P} = u_{Q} - u_{P}$ use karo, $u_{Q}$ solve karo: $u_{Q} = u_{Q / P} + u_{P} = - 25 + 15 = - 10 m/s .$ Yeh step kyun? Yeh Galilean velocity addition rearranged hai — dekho Relative velocity.
Sign interpret karo. $u_{Q} = - 10$ matlab Q west mein $10 m/s$ par ground ke relative move karta hai. Yeh step kyun? Answer ka sign direction decode karta hai; ise ignore karna classic exam slip hai.

Verify: Forward check karo: P se (jo $+ 15$ move kar raha hai), ek ground-westward Q ( $- 10$ ) appear hota hai $- 10 - 15 = - 25 m/s$ par, yani $25 m/s$ par closing ✓. Statement se consistent.

Recall Self-test: cell ka naam batao

Ek helicopter $u^{'} = 50 m/s$ due north fly karta hai; hawa $V = 50 m/s$ due east blow karti hai. Konsa matrix cell, aur ground speed aur heading kya hai? Answer ::: Cell C (perpendicular). $∣ u ∣ = 5 0^{2} + 5 0^{2} = 502 \approx 70.7 m/s$ at $4 5^{\circ}$ north of east.

Connections

Reference frames — Galilean transformations — woh parent machinery jo har example use karta hai.
Vectors — addition and components — resolve/Pythagoras/ $tan^{- 1}$ steps.
Relative velocity — cells B aur I $u_{A / B} = u_{A} - u_{B}$ par lean karte hain.
Inertial and non-inertial frames — kyun cell G ki invariance ko constant $V$ chahiye.
Newton's laws of motion — acceleration invariance ka payoff.
Special relativity — Lorentz transformation — jahan $t^{'} = t$ assumption finally break hoti hai.