A plane's instruments say it is flying north at $200\ \text{m s}^{-1}$, but a crosswind is pushing it sideways. Its real velocity over the ground is a combination of the two, a single resultant vector. This lesson brings together everything so far: resolve, add, and divide by time to find velocity in two dimensions, and solve for any missing quantity in a 2D motion problem.
A boat moves north at $3\ \text{m s}^{-1}$ while a current carries it east at $4\ \text{m s}^{-1}$.
Before reading: what do you think the boat's actual speed over the ground is, more or less than $7\ \text{m s}^{-1}$? Write your guess and one sentence on why.
A 2D velocity is the resultant of an east–west part and a north–south part
When something moves through a moving medium, its true velocity is the vector sum of its own velocity and the medium's velocity, and if those two are perpendicular, the result comes straight from Pythagoras.
The plane's airspeed (north) and the wind (east) are perpendicular velocity vectors. Add them head-to-tail and the resultant ground velocity has magnitude and direction:
$$v = \sqrt{v_x^2 + v_y^2} \qquad\qquad \theta = \tan^{-1}\!\left(\frac{v_y}{v_x}\right)$$
A 2D velocity is the resultant of its perpendicular components: magnitude $v = \sqrt{v_x^2 + v_y^2}$ and direction $\theta = \tan^{-1}(v_y/v_x)$. The components are independent, so they combine by the right-triangle rule, not by simple addition.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? A swimmer moves north at $6\ \text{m s}^{-1}$ while a current pushes east at $8\ \text{m s}^{-1}$. What is the magnitude of the resultant velocity?
Find the 2D displacement first, then divide by time
We just saw how to combine two velocities. That raises a question: what if we know a 2D journey and the time, and need the average velocity? This card answers it → find the resultant displacement, then apply $v_{av} = \Delta s / \Delta t$.
In two dimensions the velocity relationship works exactly as before, but the displacement on top is the 2D resultant found with Pythagoras.
A drone flies $120\ \text{m}$ east, then $160\ \text{m}$ north, taking $40\ \text{s}$ for the whole trip. Find its average velocity (magnitude and direction).
For 2D average velocity, find the resultant displacement $\Delta s = \sqrt{s_x^2 + s_y^2}$, then apply $v_{av} = \dfrac{\Delta s}{\Delta t}$. The direction is $\theta = \tan^{-1}(s_y/s_x)$.
Add the highlighted method to your notes before the check below.
Quick check: True or false: in two dimensions you must find the resultant displacement before dividing by time to get the average velocity.
Pattern A, Combine then divide
A kayaker paddles north at $2.0\ \text{m s}^{-1}$ while a river current flows east at $1.5\ \text{m s}^{-1}$. In your book: (1) draw the two velocity vectors head-to-tail; (2) find the magnitude of the resultant ground velocity; (3) find its direction as an angle east of north.
Two truths, one lie: which statement about velocity in two dimensions is the lie?
Dot point: Solve problems involving the calculation of velocity in 2 dimensions
Rearranging the velocity relationship for speed, displacement or time
We just learned to find velocity from displacement and time. That raises a question: what if the unknown is the displacement or the time instead? This card answers it → rearrange the same relationship to solve for whichever quantity is missing.
A single relationship links velocity, displacement and time, so if you know any two you can always rearrange to find the third.
From $v_{av} = \dfrac{\Delta s}{\Delta t}$ you can rearrange to $\Delta s = v_{av}\,\Delta t$ (displacement) or $\Delta t = \dfrac{\Delta s}{v_{av}}$ (time). Identify which quantity is unknown, choose the matching form, and substitute. In 2D, treat the displacement as the resultant magnitude.
From $v_{av} = \dfrac{\Delta s}{\Delta t}$, rearrange to find any unknown: $\Delta s = v_{av}\,\Delta t$ or $\Delta t = \dfrac{\Delta s}{v_{av}}$. The same relationship solves for speed, velocity, displacement, distance or time.
Pause, copy the highlighted rearrangements into your book before the check below.
Last check: A cyclist travels at an average velocity of $8.0\ \text{m s}^{-1}$ for $25\ \text{s}$. What displacement did they cover?
Odd one out: three of these can be found from $v_{av} = \Delta s / \Delta t$ if the other two are known. Which cannot?
Dot point: Solve problems in one and 2 dimensions using v_av = Δs/Δt
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 boat is steered due north across a river at $5.0\ \text{m s}^{-1}$, while the current flows east at $12\ \text{m s}^{-1}$. Determine the magnitude and direction of the boat's resultant velocity over the ground. Show your working.
1 mark: head-to-tail diagram of the two velocities · 1 mark: $v = \sqrt{5.0^2 + 12^2}$ · 1 mark: $v = 13\ \text{m s}^{-1}$ · 1 mark: direction $= \tan^{-1}(12/5.0) \approx 67°$ east of north
ApplyBand 5(4 marks) 2. A swimmer heads directly north across a river at $1.5\ \text{m s}^{-1}$ while the current carries them east at $2.0\ \text{m s}^{-1}$. Determine the swimmer’s resultant velocity, giving magnitude and direction.
1 mark: recognises the components are perpendicular · 1 mark: magnitude $= 2.5\ \text{m s}^{-1}$ · 1 mark: angle $= 53^\circ$ · 1 mark: direction stated relative to north
Dot point: Solve problems involving the calculation of velocity in 2 dimensions
ApplyBand 5(3 marks) 3. A drone flies $120\ \text{m}$ east and $160\ \text{m}$ north in a total of $40\ \text{s}$. Calculate its average velocity, and state how the calculation would differ in one dimension.
1 mark: displacement magnitude $= 200\ \text{m}$ · 1 mark: average velocity $= 5.0\ \text{m s}^{-1}$ with direction · 1 mark: in 1D the displacement is a signed length, so no Pythagoras is needed
Dot point: Solve problems in one and 2 dimensions using v_av = Δs/Δt
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 boat's northward velocity ($5.0\ \text{m s}^{-1}$) and the eastward current ($12\ \text{m s}^{-1}$) are perpendicular, so drawing them head-to-tail forms a right triangle. The resultant magnitude is $v = \sqrt{5.0^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\ \text{m s}^{-1}$. The direction, measured from north toward east, is $\theta = \tan^{-1}\!\left(\frac{12}{5.0}\right) \approx 67°$. So the boat travels at $13\ \text{m s}^{-1}$ at about $67°$ east of north over the ground, pushed well downstream of its intended heading.
Q2 (4 marks): The swimmer’s own velocity ($1.5\ \text{m s}^{-1}$ north) and the current ($2.0\ \text{m s}^{-1}$ east) are perpendicular, so they combine by Pythagoras: $$v = \sqrt{1.5^2 + 2.0^2} = \sqrt{2.25 + 4.00} = \sqrt{6.25} = 2.5\ \text{m s}^{-1}$$ The angle east of north is $$\theta = \tan^{-1}\!\left(\frac{2.0}{1.5}\right) = 53^\circ$$ so the resultant velocity is $2.5\ \text{m s}^{-1}$ at $53^\circ$ east of north. The swimmer does not travel straight across: the current carries them downstream the whole way.
Q3 (3 marks): First find the displacement. The legs are perpendicular, so $$s = \sqrt{120^2 + 160^2} = \sqrt{40000} = 200\ \text{m}$$ at $\tan^{-1}(120/160) = 37^\circ$ east of north. Then $$v_{av} = \frac{\Delta s}{\Delta t} = \frac{200}{40} = 5.0\ \text{m s}^{-1}$$ at $37^\circ$ east of north. The relationship $v_{av} = \Delta s / \Delta t$ is identical in one dimension; the only difference is that in 1D the displacement is a single signed length, so it is obtained by adding signed legs rather than by resolving and applying Pythagoras, and the direction is carried entirely by the sign.
You can now combine perpendicular velocities, find 2D velocity from displacement and time, and solve for any unknown. Next lesson visualises all of this as motion graphs, displacement, velocity and acceleration against time.
↩ Back to focus-area hubAt the start you had a boat moving north at $3\ \text{m s}^{-1}$ and east at $4\ \text{m s}^{-1}$. The actual ground speed is $5\ \text{m s}^{-1}$, from $\sqrt{3^2 + 4^2}$, less than the $7\ \text{m s}^{-1}$ you might get by adding directly, because the velocities are perpendicular. If you guessed $7$, revisit Card 1.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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