Rain falling straight down looks slanted from a moving car. Another car overtaking you at almost your speed seems to barely creep forward. Velocity is not absolute, it depends on who is watching. This lesson shows how to switch between observers using a single subtraction of vectors, the foundation of frames of reference you will meet again in Year 12 relativity.
Two trains move along parallel tracks in the same direction: yours at $60\ \text{km h}^{-1}$ and the other at $60\ \text{km h}^{-1}$.
Before reading: how fast does the other train appear to move as seen from your window? Write your answer and one sentence on why.
The same motion has different velocities for different watchers
There is no single "true" velocity, what you measure depends on how fast you yourself are moving, which is why a parallel train at the same speed appears motionless.
The velocity of $A$ as measured by an observer on $B$ is the relative velocity $\vec{v}_{AB}$. It is found by subtracting $B$'s velocity from $A$'s:
$$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$$
Both $\vec{v}_A$ and $\vec{v}_B$ are usually measured relative to the ground. If two trains move at $60\ \text{km h}^{-1}$ in the same direction, $\vec{v}_{AB} = 60 - 60 = 0$, the other train appears stationary.
Velocity depends on the observer's frame of reference. The velocity of $A$ relative to $B$ is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$, where both velocities are measured relative to the ground. Same-direction, equal-speed motion gives zero relative velocity.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? Car A travels east at $30\ \text{m s}^{-1}$ and car B travels east at $20\ \text{m s}^{-1}$. What is the velocity of A relative to B?
Subtracting velocities that point in different directions
We just saw relative velocity along a line. That raises a question: what if the two velocities point in different directions, like a car and falling rain? This card answers it → subtract the vectors, which means adding the reverse and using Pythagoras.
To subtract $\vec{v}_B$ you reverse it and add it head-to-tail to $\vec{v}_A$, exactly the vector subtraction from earlier, now giving the velocity one observer sees of the other.
Rain falls straight down at $80\ \text{km h}^{-1}$ (object A). A car drives north at $80\ \text{km h}^{-1}$ (object B). Find the velocity of the rain relative to the car.
In 2D, $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ is found by reversing $\vec{v}_B$ and adding head-to-tail to $\vec{v}_A$. For perpendicular parts the magnitude is $\sqrt{v_A^2 + v_B^2}$ and the direction comes from $\tan^{-1}$.
Add the highlighted method to your notes before the check below.
Quick check: True or false: to find the velocity of A relative to B in 2D, you reverse $\vec{v}_B$ and add it head-to-tail to $\vec{v}_A$.
Pattern A, Subtract then combine
Boat A moves north at $6.0\ \text{m s}^{-1}$ and boat B moves east at $8.0\ \text{m s}^{-1}$. In your book: (1) write $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ and draw the subtraction; (2) find the magnitude of the velocity of A relative to B; (3) state its direction.
Fill the gaps. The velocity of A relative to B is $v_{AB} = v_A - $ . Two cars driving side by side at the same velocity have a relative velocity of . Reversing the order of the subtraction reverses the .
Dot point: Analyse problems of relative velocity involving motion in two dimensions using v_AB = v_A − v_B
Why overtaking is slow but a head-on pass is fast
We just learned to subtract velocity vectors. That raises a question: why does an overtaking car crawl past while an oncoming one flashes by? This card answers it → the subtraction gives a small relative velocity for same-direction motion and a large one for opposite directions.
When two objects move the same way, subtracting their velocities leaves only the small difference; when they move opposite ways, the subtraction adds their speeds.
For same-direction motion, $\vec{v}_{AB} = v_A - v_B$ is small (e.g. $62 - 60 = 2\ \text{km h}^{-1}$, so overtaking feels slow). For opposite directions, $\vec{v}_B$ is negative, so $\vec{v}_{AB} = v_A - (-v_B) = v_A + v_B$ is large (e.g. $60 - (-60) = 120\ \text{km h}^{-1}$, so an oncoming car rushes past). The subtraction relationship handles both cases automatically once signs are correct.
Same-direction motion subtracts to a small relative velocity ($v_A - v_B$); opposite-direction motion, with $\vec{v}_B$ negative, gives a large one ($v_A + v_B$). The single relationship $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ covers both once directions are signed correctly.
Pause, copy the highlighted comparison into your book before the check below.
Last check: Two cars approach head-on, each at $50\ \text{km h}^{-1}$. What is the speed of one relative to the other?
Match each pair of motions with the relative velocity of A with respect to B:
Dot point: Analyse problems of relative velocity involving motion in two dimensions using v_AB = v_A − v_B
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A plane flies north at $200\ \text{m s}^{-1}$ (relative to the ground). A second plane flies east at $150\ \text{m s}^{-1}$. Determine the velocity of the first plane relative to the second. Show your working.
1 mark: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ set up (200 N and 150 W) · 1 mark: $|\vec{v}_{AB}| = \sqrt{200^2 + 150^2}$ · 1 mark: $= 250\ \text{m s}^{-1}$ · 1 mark: direction $= \tan^{-1}(150/200) \approx 37°$ west of north
ApplyBand 6(4 marks) 2. A car travels north at $25\ \text{m s}^{-1}$. A second car travels east at $25\ \text{m s}^{-1}$. Determine the velocity of the first car relative to the second, giving magnitude and direction.
1 mark: uses $v_{AB} = v_A - v_B$ · 1 mark: reverses $v_B$ to west · 1 mark: magnitude $= 35.4\ \text{m s}^{-1}$ · 1 mark: direction $45^\circ$ west of north
Dot point: Analyse problems of relative velocity involving motion in two dimensions using v_AB = v_A − v_B
UnderstandBand 5(3 marks) 3. Explain why two aircraft flying side by side at $250\ \text{m s}^{-1}$ appear stationary relative to each other, and what this shows about velocity measurements.
1 mark: $v_{AB} = v_A - v_B = 0$ · 1 mark: each pilot sees no change in the other’s position · 1 mark: velocity is only defined relative to a chosen frame of reference
Dot point: Analyse problems of relative velocity involving motion in two dimensions using v_AB = v_A − v_B
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): The velocity of plane A relative to plane B is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$. Subtracting $\vec{v}_B$ (150 east) means adding its reverse (150 west) to $\vec{v}_A$ (200 north). These two parts are perpendicular, so the magnitude is $|\vec{v}_{AB}| = \sqrt{200^2 + 150^2} = \sqrt{40000 + 22500} = \sqrt{62500} = 250\ \text{m s}^{-1}$. The direction, measured from north toward west, is $\theta = \tan^{-1}\!\left(\frac{150}{200}\right) \approx 37°$. So plane A appears to move at $250\ \text{m s}^{-1}$ at about $37°$ west of north as seen from plane B.
Q2 (4 marks): Relative velocity is a vector subtraction: $$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$$ Here $\vec{v}_A = 25\ \text{m s}^{-1}$ north and $\vec{v}_B = 25\ \text{m s}^{-1}$ east, so $-\vec{v}_B$ is $25\ \text{m s}^{-1}$ west. Adding $25$ north and $25$ west head-to-tail gives perpendicular components, so $$v_{AB} = \sqrt{25^2 + 25^2} = \sqrt{1250} = 35.4\ \text{m s}^{-1}$$ and since the two components are equal the direction is $45^\circ$ west of north. To an observer in the second car, the first car appears to move north-west at $35.4\ \text{m s}^{-1}$.
Q3 (3 marks): The velocity of one aircraft relative to the other is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$. The two velocities are identical in both magnitude and direction, so the subtraction gives zero. Each pilot therefore sees the other holding a fixed position beside them, even though both are moving at $250\ \text{m s}^{-1}$ relative to the ground. This shows that a velocity is never an absolute property of an object: it is always measured with respect to a chosen frame of reference, and the same motion has different velocities in different frames. Stating a velocity without stating the frame is incomplete.
You can now find relative velocity in one and two dimensions using $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$. That closes the kinematics half of the focus area. Next lesson turns to the causes of motion, forces, free-body diagrams and net force.
↩ Back to focus-area hubAt the start, two trains both moved at $60\ \text{km h}^{-1}$ in the same direction. The other train appears stationary ($\vec{v}_{AB} = 60 - 60 = 0$), even though both move at $60$ relative to the ground. Velocity always depends on the frame of reference. Revisit Card 1 if the zero result surprised you.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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