Reference frames — Galilean transformations: Worked examples

Before anything, one reminder of what each symbol means, in plain words, so nothing is used unearned:

The scenario matrix

Every problem in this topic is one of these cells. Each row names a case class, why it is tricky, and which worked example below covers it.

#	Case class	The trap it hides	Covered by
A	1D, same direction (both velocities parallel, same sign)	none — scalar add works	Ex 1
B	1D, opposite direction (signs differ)	must respect signs; closing speed	Ex 2
C	2D perpendicular	can't add speeds; need components + magnitude + angle	Ex 3 (fig)
D	2D general angle (neither along an axis)	resolve both vectors into components first	Ex 4 (fig)
E	Zero / degenerate input ( $u^{'} = 0$ or $V = 0$ )	check the formula still behaves	Ex 5
F	Limiting behaviour (what happens as a speed → 0 or grows)	reasoning about trends, not one number	Ex 6
G	Acceleration invariance (forces/decel)	$a^{'} = a$ ; velocity irrelevant	Ex 7
H	Real-world word problem (aim upstream to land straight)	choose the unknown correctly	Ex 8 (fig)
I	Exam twist (frame given implicitly / relative-velocity phrasing)	translate words to $u_{A / B} = u_{A} - u_{B}$	Ex 9

The worked examples

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

A river flows east at $V = 3 m/s$ . A boat points due north relative to the water at $u^{'} = 4 m/s$ . Find the boat's ground velocity: speed and direction.

Forecast: Will the ground speed be $7$ , $5$ , or $1 m/s$ ? Which direction — pure north?

Write both as components $(x, y)$ : $u^{'} = (0, 4)$ (north), $V = (3, 0)$ (east). Why this step? Perpendicular arrows can't be added as numbers; components let us add them axis-by-axis.
Add component-wise: $u = u^{'} + V = (0 + 3, 4 + 0) = (3, 4) m/s$ . Why this step? This is the head-to-tail rule — see the green resultant in the figure.
Magnitude by Pythagoras: $∣ u ∣ = 3^{2} + 4^{2} = 25 = 5 m/s$ . Why this step? The two components are the legs of a right triangle; the resultant is the hypotenuse.
Direction: $θ = tan^{- 1} (\frac{u _{y}}{u _{x}}) = tan^{- 1} (4/3) \approx 53.1 3^{\circ}$ measured north of east. Why this step? $tan$ is "opposite over adjacent" on that triangle; $tan^{- 1}$ asks "which angle has this ratio?" Both components are positive, so we sit safely in the first quadrant — no sign fix needed.

Verify: $3$ - $4$ - $5$ triangle ✓. The boat drifts east even though it aimed straight north — that eastward push is exactly the current. Units m/s ✓.

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

On a train moving east at $V = 10 m/s$ , a passenger throws a ball at $u^{'} = 10 m/s$ aimed $3 0^{\circ}$ north of east relative to the train. Ground velocity?

Forecast: Will the ground speed exceed $20 m/s$ , or land near $19.3$ ? Guess.

Resolve $u^{'}$ into components. With $cos 3 0^{\circ} = \frac{3}{2} \approx 0.8660$ , $sin 3 0^{\circ} = 0.5$ : $u^{'} = (10 cos 3 0^{\circ}, 10 sin 3 0^{\circ}) = (8.6603, 5.0) m/s .$ Why this step? When neither vector lies on an axis you must break it into east/north parts before adding.
Add the frame velocity $V = (10, 0)$ : $u = (8.6603 + 10, 5.0 + 0) = (18.6603, 5.0) m/s .$ Why this step? Galilean addition again — but now the eastward parts pile up.
Magnitude: $∣ u ∣ = 18.660 3^{2} + 5. 0^{2} \approx 348.21 + 25 = 373.21 \approx 19.32 m/s .$ Why this step? Same right-triangle idea; the resultant is the hypotenuse of the summed legs.
Direction: $θ = tan^{- 1} (5.0/18.6603) \approx 15. 0^{\circ}$ north of east. Why this step? Both summed components are positive → first quadrant → the plain $tan^{- 1}$ is correct, no $\pm 18 0^{\circ}$ correction.

Verify: The angle shrank from $3 0^{\circ}$ to $1 5^{\circ}$ — adding a purely eastward $V$ should tilt the ball more eastward, so a smaller north-angle is right ✓. Speed $19.32 < 20$ because the two vectors aren't parallel; parallel would give exactly $20$ ✓.

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

(a) A cup rests on the train seat: $u^{'} = 0$ . Train at $V = 20 m/s$ east. Ground velocity? (b) The train is parked, $V = 0$ . Passenger walks at $u^{'} = 1.4 m/s$ . Ground velocity?

Forecast: In (a), does the ground see the cup moving? In (b), does the frame change anything?

(a) $u = u^{'} + V = 0 + 20 = 20 m/s$ east. Why this step? An object at rest in the train is carried along at the train's speed — the formula reduces to "the cup just rides with $V$ ."
(b) $u = u^{'} + V = 1.4 + 0 = 1.4 m/s$ . Why this step? With $V = 0$ the two frames coincide; $u = u^{'}$ exactly, so the transformation does nothing — a sanity check that the formula degrades correctly.

Verify: (a) matches the everyday fact that coffee on a moving train moves with you at $20 m/s$ over the ground ✓. (b) with $V = 0$ the map is the identity ✓. Units m/s ✓.

Worked example Ex 6 — Cell F: limiting behaviour

A swimmer swims north across the river of Ex 3 at $u^{'} = 4 m/s$ . Watch the drift angle $ϕ$ (how far east of north the ground path tilts) as the current $V$ ranges from $0$ up.

Forecast: As $V \to 0$ , does $ϕ \to 0^{\circ}$ or $\to 9 0^{\circ}$ ? As $V$ grows very large?

Write the drift angle east-of-north: $ϕ = tan^{- 1} (V / u^{'}) = tan^{- 1} (V /4)$ . Why this step? East component is $V$ , north component is $u^{'} = 4$ ; the tilt off north is opposite/adjacent $= V / u^{'}$ .
Limit $V \to 0$ : $tan^{- 1} (0) = 0^{\circ}$ . The path is pure north. Why this step? No current means no sideways push, so no drift — the formula must give exactly $0^{\circ}$ , and it does.
Limit $V \to \infty$ : $tan^{- 1} (huge) \to 9 0^{\circ}$ . The path becomes almost pure east. Why this step? An overwhelming current sweeps the swimmer sideways; the tiny $4 m/s$ north barely registers. $tan^{- 1}$ saturates at $9 0^{\circ}$ , never exceeding it.
Sample $V = 4$ : $ϕ = tan^{- 1} (1) = 4 5^{\circ}$ — equal east and north pushes. Why this step? A concrete midpoint confirms the trend is smooth and monotone.

Verify: $ϕ$ rises monotonically from $0^{\circ}$ toward $9 0^{\circ}$ and hits $4 5^{\circ}$ when $V = u^{'}$ ✓ — matches intuition that "faster current = more drift, but never past sideways."

Worked example Ex 7 — Cell G: acceleration invariance

The ball from Ex 4 feels air drag decelerating it at $a^{'} = 2 m/s^{2}$ opposite its motion, as measured on the train. What acceleration does the ground observer measure?

Forecast: Bigger, smaller, or exactly $2 m/s^{2}$ ?

Recall $a^{'} = a$ . Differentiate $u = u^{'} + V$ ; since $V$ is constant, $d V / d t = 0$ , so $a = a^{'}$ . Why this step? Acceleration is the time-derivative of velocity, and the only difference between frames ( $V$ ) is a constant that dies under differentiation.
Read off the answer: ground measures $a = 2 m/s^{2}$ , same magnitude, same direction. Why this step? Invariance means no translation is needed — the number just carries across.

Verify: The train's speed ( $10 m/s$ ) does not appear anywhere — correct, since constant velocity contributes zero acceleration ✓. Units m/s² ✓. This is exactly why $F = m a$ holds in every inertial frame (see Newton's laws of motion).

Worked example Ex 8 — Cell H: real-world word problem (aim to land straight)

The river again flows east at $V = 3 m/s$ . The boat can do $u^{'} = 5 m/s$ relative to the water. At what angle must it aim so its ground path is due north (straight across)? And what is the resulting ground speed?

Forecast: Aim slightly east, slightly west, or dead north?

Demand zero east-drift. For the ground velocity to point pure north, the boat's own eastward component must cancel the current: $u^{'} sin α = V$ , where $α$ is the aim angle west of north. Why this step? We're solving for the unknown heading, so we impose the condition that the east components sum to zero.
Solve for $α$ : $sin α = V / u^{'} = 3/5 = 0.6 \Rightarrow α = sin^{- 1} (0.6) \approx 36.8 7^{\circ}$ west of north (upstream). Why this step? $sin^{- 1}$ answers "which angle has this sine?"; since $V < u^{'}$ a real angle exists — the boat can beat the current.
Ground speed = the surviving north component: $u = u^{'} cos α = 5 cos (36.8 7^{\circ}) = 5 \times 0.8 = 4 m/s$ . Why this step? Only the north part of $u^{'}$ remains after the east part is cancelled.

Verify: $3$ - $4$ - $5$ triangle hides here: $u^{'} = 5$ , cancelled east $= 3$ , net north $= 4$ ✓. The boat aims upstream (west of north), which matches intuition — you fight the current to go straight ✓. Degenerate check: if $V > u^{'}$ then $sin α > 1$ has no solution → the boat cannot hold a straight north path, correct physics.

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

"An observer on train P (moving east at $15 m/s$ ) sees train Q approaching from the east at $25 m/s$ (i.e. Q closes on P at $25 m/s$ ). What is Q's velocity relative to the ground?" The frame is hidden in the wording — untangle it.

Forecast: Is Q's ground speed $40$ , $25$ , or $10 m/s$ ? Which direction?

Name everything with an axis. East $= + x$ . P's ground velocity $u_{P} = + 15$ . Q approaches from the east, so relative to P, Q moves west: $u_{Q / P} = - 25$ . Why this step? "Approaching from the east" fixes the sign — Q heads toward P, i.e. westward in P's view.
Use $u_{Q / P} = u_{Q} - u_{P}$ , solve for $u_{Q}$ : $u_{Q} = u_{Q / P} + u_{P} = - 25 + 15 = - 10 m/s .$ Why this step? This is Galilean velocity addition rearranged — see Relative velocity.
Interpret the sign. $u_{Q} = - 10$ means Q moves west at $10 m/s$ relative to the ground. Why this step? The sign of the answer decodes the direction; ignoring it is the classic exam slip.

Verify: Check by going forward: from P (moving $+ 15$ ), a ground-westward Q ( $- 10$ ) appears at $- 10 - 15 = - 25 m/s$ , i.e. closing at $25 m/s$ ✓. Consistent with the statement.

Recall Self-test: name the cell

A helicopter flies at $u^{'} = 50 m/s$ due north; wind blows $V = 50 m/s$ due east. Which matrix cell, and what's the ground speed and heading? 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 — the parent machinery every example uses.
Vectors — addition and components — the resolve/Pythagoras/ $tan^{- 1}$ steps.
Relative velocity — cells B and I lean on $u_{A / B} = u_{A} - u_{B}$ .
Inertial and non-inertial frames — why cell G's invariance needs constant $V$ .
Newton's laws of motion — the payoff of acceleration invariance.
Special relativity — Lorentz transformation — where the $t^{'} = t$ assumption finally breaks.