PHYSICS · LESSON 03 OF 9
Relative motion and reference frames
Every velocity is measured relative to an observer. Name the frame, then add velocities as vectors: vA/C = vA/B + vB/C.
What this lesson explains
A speed is incomplete until you say what it is measured relative to. A person can walk towards the back of a moving train while still moving forwards relative to the platform. Naming the observer removes the apparent contradiction.
Relative motion is used whenever a vehicle moves through a moving medium — boats in currents, aircraft in wind, conveyors and moving walkways — and whenever two moving objects approach or overtake each other.
Before you begin
Vector components
A velocity at angle θ has components v cos θ and v sin θ along perpendicular axes; its magnitude is √vx2 + vy2 (see Vectors and motion).
Constant velocity
With constant velocity, distance along a direction = (velocity component in that direction) × time.
The idea, made visible
Write vA/B for the velocity of object A measured by observer B. Keep one set of axis directions when combining velocities. In one dimension the signs carry direction; in two dimensions add the x-components and y-components separately. Speeds, which are magnitudes, cannot generally be added as scalars.
Position vectors satisfy rA/C = rA/B + rB/C: go from C to B, then from B to A. Differentiating with respect to the same time gives vA/C = vA/B + vB/C. The middle observer B connects the two terms. This is the classical, nonrelativistic rule, valid for speeds far below the speed of light and axes that do not rotate relative to each other.
For two objects observed from the ground G, vA/B = vA/G − vB/G. Reversing the observer reverses the vector: vB/A = −vA/B. If the frames move at constant velocity relative to one another, their measured accelerations agree. An accelerating or rotating frame needs further care when applying Newton’s laws.
For a river or wind problem, distinguish the vehicle’s velocity relative to the fluid from the fluid’s velocity relative to the ground. A boat’s heading is the direction it points through the water; its track is the path seen from the bank. They differ when a current has a sideways component.
| Heading through water | Across speed | Time | Downstream drift |
|---|---|---|---|
| Straight across | 2.5 m/s | 40 s | 60 m |
| 36.9° upstream from across | 2.0 m/s | 50 s | 0 m |
| Current as fast as boat; aim to cancel drift | 0 m/s | No finite crossing time | No crossing |
The formulas and what they mean
| Symbol | Meaning | Unit |
|---|---|---|
| vA/B | velocity of A relative to B | m/s |
| u | boat speed relative to water | m/s |
| c | current speed relative to bank | m/s |
| w | river width, measured perpendicular to banks | m |
| α | upstream heading angle measured from straight across | ° or rad |
Add velocities through a named frame
vA/C = vA/B + vB/C
Conditions and limits: Vectors expressed using consistent axes and the same time; classical speeds.
Observe A from B
vA/B = vA/G − vB/G
Conditions and limits: Both given velocities are measured from the same frame G.
Point straight across a uniform river
t = w/u; downstream drift = ct
Conditions and limits: The boat points perpendicular to the banks. Current is parallel to the banks; u and c are constant.
Land directly opposite
sin α = c/u; vacross = √u2 − c2; t = w / √u2 − c2
Conditions and limits: Aim upstream by α from straight across. A positive crossing speed requires u > c. At u = c, cancellation uses the whole boat speed and the boat cannot cross.
A first worked example
How to solve a relative-motion problem
- Name every velocity with two labels: the moving object and the observer (vboat/water, vwater/bank).
- Choose axes once and write every velocity in components with signs.
- Chain the frames so the middle label cancels: vA/C = vA/B + vB/C.
- Use the component along a required direction to find times; use the magnitude only for a speed along the actual path.
Walking backwards on a moving train
Problem. A train moves east at 12 m/s. A passenger walks west along it at 1.5 m/s relative to the train. Find the passenger’s velocity relative to the platform.
Choose east positive. vP/T = −1.5 m/s and vT/G = +12 m/s.
The westward walking velocity must have the opposite sign.
vP/G = −1.5 + 12 = +10.5 m/s.
The train frame cancels between the two terms.
Result: The passenger moves east at 10.5 m/s relative to the platform.
What it means: Walking towards the back of the train does not necessarily mean moving west over the ground.
A different case
A faster vehicle approaches from behind
Problem. Car A is 180 m behind car B. Both move east at constant speeds: A at 24 m/s and B at 18 m/s. Find when A reaches B.
vA/B = 24 − 18 = 6 m/s east.
The gap closes at the relative speed, not at either ground speed.
t = 180/6 = 30 s.
Initial separation divided by constant closing speed.
A travels 24 × 30 = 720 m; B travels 18 × 30 = 540 m.
A covers 180 m more, exactly the initial gap.
Result: A reaches B after 30 s.
What it means: Adding 24 and 18 would apply to head-on motion, not to vehicles travelling in the same direction.
More worked cases
Each case below uses a different skill. Every step and result is shown.
Point straight across and accept the drift
Problem. A boat travels at 2.5 m/s relative to water, pointing straight across a 100 m river. The current is 1.5 m/s downstream. Find crossing time, drift and speed relative to the bank.
Across component = 2.5 m/s; downstream component = 1.5 m/s.
The current changes only the downstream component in this model.
t = 100/2.5 = 40 s; drift = 1.5 × 40 = 60 m.
Use the across component to cross the width.
Ground speed = √2.52 + 1.52 = 2.92 m/s.
The perpendicular components form a right triangle.
Result: 40 s, landing 60 m downstream; speed relative to the bank ≈ 2.92 m/s.
What it means: Dividing 100 m by 2.92 m/s would mix a perpendicular width with speed along a diagonal path.
Aim upstream to land directly opposite
Problem. Use the same river and boat. Find the upstream heading and crossing time for zero downstream drift.
Choose the boat’s upstream component to be −1.5 m/s.
It must cancel the water’s +1.5 m/s downstream velocity.
sin α = 1.5/2.5 = 0.60, so α = 36.9° upstream from straight across.
The total velocity through water remains 2.5 m/s.
Across component = √2.52 − 1.52 = 2.0 m/s; t = 100/2.0 = 50 s.
Part of the boat’s velocity is now used to cancel the current.
Result: Aim 36.9° upstream from straight across; crossing takes 50 s.
What it means: Zero drift takes longer than simply pointing straight across. If the current is at least the boat’s water speed, a direct opposite-bank crossing is impossible in this model.
Common misunderstandings
Misunderstanding: Adding speeds without directions.
Correct idea: Add signed components of velocities. Then calculate a magnitude if needed.
Misunderstanding: Using diagonal ground speed to cross a perpendicular river width.
Correct idea: Time uses the velocity component in the direction of that width.
Misunderstanding: Using the same velocity for every observer.
Correct idea: Write both the moving object and the reference frame on each velocity.
Keep in mind
- vA/C = vA/B + vB/CAdd velocities through a named frame
- vA/B = vA/G − vB/GObserve A from B
- t = w/u; downstream drift = ctPoint straight across a uniform river
- sin α = c/u; vacross = √u2 − c2; t = w / √u2 − c2Land directly opposite
Scope of this lesson
- Classical (Galilean) velocity addition only: speeds far below the speed of light, frames that do not rotate relative to each other.
- Accelerating frames are not treated.
Next: Newton laws and force diagrams. Forces change motion. Isolate one object, draw every force acting on it, and apply ΣF = ma along each axis.
Further reading: OpenStax University Physics 1 — Relative motion. Sources and credits.