AcademyMotion in Space
Academy
Relative Motion
Level 1 - Physics topic page in Motion in Space.
Principle
Relative motion relates measurements made from different frames by adding displacement, velocity, or acceleration vectors.
The subscript order matters: the first label is the object being described, and the second label is the reference frame.
Notation
\(\vec r_{PA}\)
position of P relative to A
\(\mathrm{m}\)
\(\vec v_{PA}\)
velocity of P relative to A
\(\mathrm{m\,s^{-1}}\)
\(\vec a_{PA}\)
acceleration of P relative to A
\(\mathrm{m\,s^{-2}}\)
\(A,B\)
two reference frames or observers
Method
Derivation 1: Build the position relation
To locate P from A, you can go from A to B and then from B to P. That path is vector addition.
Go A to B
\[\vec r_{BA}\]
Then B to P
\[\vec r_{PB}\]
Add the path
\[\vec r_{PA}=\vec r_{PB}+\vec r_{BA}\]
Read this as: P relative to A equals P relative to B plus B relative to A.
Derivation 2: Differentiate the relation
Velocity and acceleration relations come from differentiating the position relation.
Differentiate position
\[\frac{d\vec r_{PA}}{dt}=\frac{d\vec r_{PB}}{dt}+\frac{d\vec r_{BA}}{dt}\]
Velocity transform
\[\vec v_{PA}=\vec v_{PB}+\vec v_{BA}\]
Acceleration transform
\[\vec a_{PA}=\vec a_{PB}+\vec a_{BA}\]
Derivation 3: Use components when directions differ
Relative motion is vector addition, so crossed directions must be handled by components.
Write each velocity
\[\vec v=v_x\hat{\imath}+v_y\hat{\jmath}\]
Add matching components
\[v_{PA,x}=v_{PB,x}+v_{BA,x},\qquad v_{PA,y}=v_{PB,y}+v_{BA,y}\]
Find speed if needed
\[v_{PA}=\sqrt{v_{PA,x}^2+v_{PA,y}^2}\]
The vector sketch shows the common boat-current pattern: ground velocity is boat-relative-to-water plus water-relative-to-ground.
Rules
These are the compact results from the derivations above.
Position transform
\[\vec r_{PA}=\vec r_{PB}+\vec r_{BA}\]
Velocity transform
\[\vec v_{PA}=\vec v_{PB}+\vec v_{BA}\]
Acceleration transform
\[\vec a_{PA}=\vec a_{PB}+\vec a_{BA}\]
Examples
Question
A boat moves east at
\[4\,\mathrm{m\,s^{-1}}\]
relative to water. The river flows north at \[3\,\mathrm{m\,s^{-1}}\]
Find boat velocity relative to ground.Answer
Use
\[\vec v_{BG}=\vec v_{BW}+\vec v_{WG}\]
\[\vec v_{BG}=4\hat{\imath}+3\hat{\jmath}\,\mathrm{m\,s^{-1}}\]
Its speed is \[v_{BG}=\sqrt{4^2+3^2}=5\,\mathrm{m\,s^{-1}}\]
Checks
- Order of subscripts matters.
- Add velocities as vectors, not just speeds.
- Use components for crossed directions.
- Accelerations are equal only when the frames have zero relative acceleration.