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Reference Frames &
Relative Motion

Understand how different observers describe the same motion differently, and master the velocity addition equation for AP Physics 1.

AP Physics 1

What Is a Reference Frame?

A reference frame (or frame of reference) is a coordinate system attached to a particular observer. It defines where "at rest" is and how positions, velocities, and accelerations are measured.

Origin
A chosen point from which all positions are measured (often the observer themselves).
Axes
Directions (\(x, y, z\)) that define how to describe position in space relative to the origin.
Clock
A timekeeping device that lets the observer record when events occur.

Why Reference Frames Matter

train Passenger (Frame S') Observer (Frame S) ball "The ball moves slowly" "The ball moves fast!"

Inertial vs Non-Inertial Frames

Not all reference frames behave the same way. The distinction matters because Newton's laws only work directly in inertial frames.

Inertial Frame
Moves at constant velocity (including at rest). Newton's laws apply without modification. Example: a train moving at steady speed on a straight track.
Non-Inertial Frame
Is accelerating (speeding up, slowing, or turning). Objects appear to experience "fictitious" forces. Example: a car rounding a curve.

What Is Relative Motion?

Relative motion is the calculation of the motion of an object as observed from a particular reference frame. The velocity of an object depends on which frame you measure it from.

1
All motion is relative. There is no absolute "at rest" in classical mechanics: you can only say an object is at rest relative to something.
2
Velocities transform. The velocity measured in one frame can be converted to another frame using vector addition.
3
The laws of physics are the same in all inertial reference frames (Galilean relativity).

The Velocity Addition Equation

Read the subscripts as: the velocity of A relative to C equals the velocity of A relative to B plus the velocity of B relative to C.

\(\vec{v}_{AC}\)
What you want: velocity of A as seen from C.
\(\vec{v}_{AB}\)
Velocity of A as seen from B (often given).
\(\vec{v}_{BC}\)
Velocity of B as seen from C (the "frame velocity").

Walking on a train

This separate illustration applies the velocity chain to both walking directions. Use the sign of the walking velocity relative to the train before adding the train velocity relative to the ground.

The train moves at +10 m/s relative to the ground. Walking at +2 or −2 m/s relative to the train gives ground velocities +12 or +8 m/s.
The train moves at +10 m/s relative to the ground. Walking at +2 or −2 m/s relative to the train gives ground velocities +12 or +8 m/s.

The Subscript Chain Rule

A B C \(v_{AB}\) \(v_{BC}\) \(v_{AC} = v_{AB} + v_{BC}\)

1D vs 2D Relative Velocity

1D (Same/Opposite Line) 2D (Perpendicular/Angled)
Equation \(v_{AC} = v_{AB} + v_{BC}\) \(\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}\)
Operation Scalar addition (with signs) Vector addition (components or tip-to-tail)
Magnitude Simple sum or difference Use Pythagorean theorem if perpendicular
Example Two cars on a highway Boat crossing a river with current

Flipping Subscripts

If you know the velocity of A relative to B, the velocity of B relative to A is the same magnitude but opposite direction. This lets you rearrange any problem to get the subscript chain you need.

1
Identify what you need: which object relative to which frame?
2
Check subscripts: do the inner letters match? If not, flip one velocity using \(\vec{v}_{XY} = -\vec{v}_{YX}\).
3
Add the vectors: once the chain is correct, add them.

How to Solve Relative Motion Problems

1
Label every object and frame with a letter (e.g., P = person, T = train, G = ground).
2
Write what you want using subscript notation: "I need \(\vec{v}_{PG}\)" (person relative to ground).
3
Build the chain: \(\vec{v}_{PG} = \vec{v}_{PT} + \vec{v}_{TG}\). The inner T's cancel.
4
Assign signs or components: choose a positive direction. Plug in values.
5
Compute and interpret: a positive result means motion in the positive direction; negative means the opposite.

Example 1: Same Direction

A person walks at \(2\;\text{m/s}\) toward the front of a train that moves at \(30\;\text{m/s}\) east relative to the ground. What is the person's velocity relative to the ground?

Train (T) Person (P) \(2\;\text{m/s}\) \(30\;\text{m/s}\) Ground (G) + East

Example 2: Opposite Direction

Car A travels east at \(25\;\text{m/s}\) and car B travels west at \(20\;\text{m/s}\), both relative to the ground. What is the velocity of A relative to B?

A \(+25\) B \(-20\) + East →

Example 3: River Crossing (2D)

A boat aims straight across a river at \(4\;\text{m/s}\) (relative to the water). The current flows downstream at \(3\;\text{m/s}\) relative to the ground. Find the boat's speed relative to the ground.

North Bank South Bank Boat (B) \(4\;\text{m/s}\) \(3\;\text{m/s}\) \(5\;\text{m/s}\) Current →

One boat, two velocity frames

One boat crosses a river with a visible current context. Wake and perspective are not velocity measurements; both course diagrams remain authoritative.Boat velocity relative to water and current velocity relative to the ground add as vectors to give the boat’s ground velocity.

Shortest-time river crossing

Shortest-time river crossing. Across an 80 m river, a boat aimed north at 5 m/s relative to the water takes 16 s. A 3 m/s eastward current produces 48 m downstream drift; a zero-drift heading is 36.87° west of north.Across an 80 m river, a boat aimed north at 5 m/s relative to the water takes 16 s. A 3 m/s eastward current produces 48 m downstream drift; a zero-drift heading is 36.87° west of north.

Finding the Direction in 2D

From the river example, the boat's resultant velocity is \(5\;\text{m/s}\). To find the angle the boat drifts from its intended path, use trigonometry:

The boat actually travels at \(36.9^\circ\) downstream from perpendicular. The faster the current relative to the boat's speed, the larger the drift angle.

Common Mistakes

✕ Wrong Sign Convention
Forgetting to assign negative signs to objects moving in the opposite direction. Always define a positive direction first.
✕ Subscript Mismatch
Writing \(\vec{v}_{AB} + \vec{v}_{CB}\) instead of \(\vec{v}_{AB} + \vec{v}_{BC}\). The inner subscripts must match (B with B).
✕ Scalar Addition in 2D
Adding magnitudes directly when velocities are perpendicular. Use the Pythagorean theorem: \(\sqrt{a^2 + b^2}\), not \(a + b\).

Summary of Key Ideas

Concept Key Formula / Rule Remember
Velocity addition \(\vec{v}_{AC} = \vec{v}_{AB} + \vec{v}_{BC}\) Inner subscripts cancel
Flipping subscripts \(\vec{v}_{AB} = -\vec{v}_{BA}\) Same magnitude, opposite direction
1D problems Use + / − for directions Define positive direction first
2D problems \(|\vec{v}| = \sqrt{v_x^2 + v_y^2}\) Use components or tip-to-tail
Direction (2D) \(\theta = \tan^{-1}(v_y / v_x)\) Measure from reference axis

Question 1 of 5

A passenger walks at \(1.5\;\text{m/s}\) toward the back of a train moving at \(20\;\text{m/s}\) east relative to the ground. What is the passenger's velocity relative to the ground?

Question 2 of 5

Car X moves at \(30\;\text{m/s}\) north and Car Y moves at \(20\;\text{m/s}\) south, both relative to the ground. What is the velocity of X relative to Y?

Question 3 of 5

A swimmer crosses a river at \(2\;\text{m/s}\) relative to the water, aiming straight across. The current is \(1.5\;\text{m/s}\). What is the swimmer's speed relative to the ground?

Question 4 of 5

If \(\vec{v}_{AB} = +12\;\text{m/s}\), what is \(\vec{v}_{BA}\)?

Question 5 of 5

An airplane flies at \(200\;\text{km/h}\) north relative to the air. A crosswind blows at \(50\;\text{km/h}\) east. What is the approximate direction of the airplane's velocity relative to the ground?