4.5 — Relative motion — inertial frames

Simple but deep question: "velocity" — with respect to whom? Every physical measurement happens in a "reference frame" — and the same event has different values in different frames. Relative velocity is the translation tool.

The Galilean relativity principle

If you observe A relative to frame O, and A observes B relative to itself:

\[ \vec v_{B/O} = \vec v_{B/A} + \vec v_{A/O} \]

Subscript rule: B/O = "B's velocity as seen by O". In addition, the middle subscripts drop out.

Example 1 — train and passenger

A train moves at 30 m/s east relative to ground. A passenger inside walks 2 m/s forward (east) relative to the train.

Passenger's velocity relative to ground?

\[ \vec v_{pass/ground} = 2\hat i + 30\hat i = 32\hat i\ m/s \]

Walking backward: -2 + 30 = 28 m/s.

Example 2 — boat and river

A boat wants to cross a river flowing east at 3 m/s, heading due north. Boat can do 5 m/s in still water.

Question: what heading should the pilot aim?

Analysis: \vec v_{boat/ground} = \vec v_{boat/water} + \vec v_{water/ground}

Pilot heads at angle \theta west of north:

\[ \vec v_{boat/water} = -5\sin\theta\hat i + 5\cos\theta\hat j \]

\[ \vec v_{water/ground} = 3\hat i \]

For net velocity to be pure \hat j, x must vanish:

\[ -5\sin\theta + 3 = 0 \Rightarrow \sin\theta = 0.6 \Rightarrow \theta = 36.9° \]

Actual northward speed: v = 5\cos 36.9° = 4\ m/s

Result: aim 36.9° west of north but actually move at 4 m/s due north.

Example 3 — wind effect on a plane

A plane wants to fly east from A to B. Wind is 50 km/h toward northwest. Plane does 300 km/h in still air. What heading?

Analysis:

• NW wind: \vec v_{air/ground} = -35.4\hat i + 35.4\hat j
• Goal: \vec v_{plane/ground} should be positive \hat i only

So:

• Plane's y component relative to air: -35.4 (cancel the wind)
• \sin\phi = 35.4/300 = 0.118 \Rightarrow \phi \approx 6.8° south of east

Final ground speed: \cos 6.8° \cdot 300 - 35.4 \approx 263\ km/h due east.

Example 4 — two cars on different roads

Car A at 20 m/s east, car B at 15 m/s north. B's velocity relative to A?

\[ \vec v_{B/A} = \vec v_{B/ground} - \vec v_{A/ground} = -20\hat i + 15\hat j \]

Magnitude: \sqrt{400 + 225} = 25\ m/s. Angle: \arctan(15/(-20)) in quadrant II — \approx 143° from +x.

Interpretation: from A's driver's view, B moves at 25 m/s northwest.

What is an inertial frame?

Inertial frame: a frame where Newton's first law holds — a free object stays at constant velocity.

• Earth (approximately): a good inertial frame for most everyday problems
• Constant-velocity train: another inertial frame — closed relative to ground
• Accelerating or turning train: non-inertial — a free object appears to accelerate to an observer inside (with no visible force)

Galilean relativity principle: the laws of physics are the same in all inertial frames. Being inside a smoothly cruising airplane feels the same as standing on the ground.

Historical note: this idea generalized in Einstein's special relativity (1905) — extending to electromagnetism, resulting in the space-time symmetry.

A few notes and common mistakes

1. Direction matters. \vec v_{A/B} means "A relative to B". The reverse: \vec v_{B/A} = -\vec v_{A/B}.

2. Vector addition, not scalar. The plane example showed you can't just add velocities — directions differ.

3. Relative velocity may exceed or fall below the components. Two cars at 50 km/h head-on: relative velocity 100 km/h. Same-direction: 0.

What you should be able to do

• Apply the relative-velocity addition rule
• Solve boat-in-river and plane-in-wind problems
• Recognize inertial frames and the Galilean relativity principle
• Compute one object's velocity relative to another by subtracting frame velocities

Preview of §4.6

Worked problems — applying all Chapter 4 tools to real scenarios.

📚 See also: Halliday Vol 1, Ch 4, §4.6 — Relative Motion.

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