What is the difference between a scalar and a vector?

A scalar has a size only, such as 20 m s^−1 (a speed). A vector has a size and a direction, such as 20 m s^−1 north (a velocity). Distance, speed and time are scalars; displacement, velocity, acceleration and force are vectors.

How do you add two vectors at right angles?

Draw them head to tail. The resultant's size is √(a^2 + b^2) and its direction is tan^−1(opposite ÷ adjacent), measured from a line you name. For vectors that aren't at right angles, resolve each into components first, add the components, then recombine.

What does vAB = vA − vB mean?

It is the velocity of A as seen by someone moving with B. Subtracting B's velocity removes the observer's own motion. Rearranged, vA = vAB + vB, which is how you add a boat's velocity through the water to the water's velocity to get the boat's velocity relative to the bank.

Which way should a boat point to cross a river straight across?

Partly upstream, so that the upstream part of its velocity through the water cancels the current. If the boat's speed in the water is v and the current is c, the angle upstream from straight across is sin^−1(c ÷ v), and the crossing takes longer than pointing straight across.

A ball hits a wall at 10 m s^−1 and bounces straight back at 8.0 m s^−1. It is in contact with the wall for 0.050 s. What is its average acceleration?

360 m s^−2 away from the wall

Year 11 Physics · Fundamentals of mechanics · Vectors and motion in two dimensions · 12 min read

Vectors and relative velocity

A vector has a size and a direction; a scalar has only a size. That one difference changes how you add things up: 4 m east and 3 m north is not 7 m of displacement, it is 5 m. Year 11 Physics needs vectors from the first week, and every later mechanics topic leans on them.

This guide covers adding and subtracting vectors in two dimensions, resolving a vector into components, change in velocity, and relative velocity, such as a boat crossing a river or a plane in a crosswind. Then the common mistakes, a quick quiz and a full worked example.

What you need to know

Classify displacement, distance, speed, velocity, time and acceleration as scalars or vectors.
Represent an object's acceleration, velocity and displacement with vector diagrams.
Find vector sums and differences in two dimensions.
Resolve a vector into components, using F_x = F cos θ and F_y = F sin θ with θ measured from the x-axis.
Calculate velocities in two dimensions to solve problems.
Apply a = Δv ÷ Δt to one- and two-dimensional problems.
Use v_AB = v_A − v_B to analyse relative velocity problems in two dimensions.

The equations you will use:

F_x = F cos θ F_y = F sin θComponents of any vector, with θ measured from the x-axis.

R = √(R_x² + R_y²) θ = tan⁻¹(R_y ÷ R_x)Put the components back together: size by Pythagoras, direction by inverse tan.

Δv = v − u = v + (−u)Change in velocity is a vector subtraction: reverse u, then add.

a = Δv ÷ ΔtThe average acceleration points the same way as Δv.

v_AB = v_A − v_BThe velocity of A as seen by an observer moving with B.

Dot points: PY11-FM-01, PY11-FM-02, PY11-FM-03, PY11-FM-04, PY11-FM-08, PY11-FM-10, PY11-FM-16, PY11-FM-17 (2025 Physics syllabus, Year 11), in our own words.

Mini lesson

Scalars and vectors

A scalar is fully described by a number and a unit. A vector also needs a direction, given as a compass direction ("30° north of east"), as up, down, left or right, or in one dimension as a + or − sign.

Scalars: distance, speed, time, mass, energy.
Vectors: displacement, velocity, acceleration, force.
Pairs to keep apart: distance (scalar) and displacement (vector); speed (scalar) and velocity (vector).
In a diagram, a vector is an arrow: its length shows the size (to scale) and it points in the vector's direction.

Adding vectors: head to tail

To add vectors, draw them head to tail: start the second arrow where the first one ends. The resultant runs from the tail of the first arrow to the head of the last. The order doesn't matter; the resultant is the same.

When the two vectors are at right angles, the resultant is the hypotenuse of a right-angled triangle. Find its size with Pythagoras and its direction with inverse tan, and always say which line the angle is measured from.

4.0 m east plus 3.0 m north: R = √(4.0² + 3.0²) = 5.0 m, at tan⁻¹(3.0 ÷ 4.0) = 37° north of east. Not 7.0 m.

Resolving a vector into components

Any vector can be split into two perpendicular parts that add back to it. With the angle θ measured from the x-axis, the x-component is F cos θ and the y-component is F sin θ. The side next to the angle uses cos; the side opposite uses sin.

Components make any addition easy, even when the vectors are not at right angles. A force of 80 N at 30° above the horizontal has a horizontal part of 80 cos 30° = 69 N and a vertical part of 80 sin 30° = 40 N.

Step 1: resolve every vector into x- and y-components, with signs (left and down are negative).
Step 2: add all the x-components, then all the y-components.
Step 3: recombine: size by Pythagoras, direction by inverse tan, with the reference line stated.

Subtracting vectors: change in velocity

Acceleration depends on the change in velocity, Δv = v − u. To subtract a vector, reverse it and add: v − u = v + (−u).

In one dimension, signs do the work. A ball hits a wall at 10 m s⁻¹ and rebounds at 8 m s⁻¹: with towards the wall as positive, Δv = (−8) − (+10) = −18 m s⁻¹, so 18 m s⁻¹ away from the wall. That is far more than the 2 m s⁻¹ you get by subtracting speeds.

In two dimensions, draw it. A car turning a corner from 15 m s⁻¹ north to 15 m s⁻¹ east has the same speed, but its velocity has changed by 21 m s⁻¹ towards the south-east, so it has accelerated.

Δv = v + (−u): reverse the initial velocity (north becomes south) and add it head to tail. √(15² + 15²) = 21 m s⁻¹, towards the south-east. The car's acceleration points the same way.

Relative velocity

Every velocity is measured relative to something, usually the ground. The velocity of A relative to B is v_AB = v_A − v_B: what an observer moving with B would see. In one dimension: a car at 25 m s⁻¹ east passing a truck at 18 m s⁻¹ east moves forward at only 7 m s⁻¹ relative to the truck.

In two dimensions the same rule needs a vector diagram. A boat heads straight across a 50 m river at 2.5 m s⁻¹ relative to the water, and the current flows at 1.5 m s⁻¹. Relative to the bank, the boat's velocity is the vector sum: √(2.5² + 1.5²) = 2.9 m s⁻¹, at 31° downstream from straight across.

The crossing time depends only on the across-river part: 50 ÷ 2.5 = 20 s. In that time the current carries the boat 1.5 × 20 = 30 m downstream. To land directly opposite, the driver must point the boat partly upstream.

v_boat,bank = v_boat,water + v_water,bank. The boat heads straight across but moves at 2.9 m s⁻¹ at 31° downstream of straight across.

Common mistakes

Adding perpendicular vectors as plain numbers: 4.0 m east + 3.0 m north = 7.0 m.Vectors at right angles combine by Pythagoras: √(4.0² + 3.0²) = 5.0 m. The resultant is always shorter than the two sizes added together.
Using sin for the component next to the angle.With θ measured from the x-axis, F_x = F cos θ (adjacent) and F_y = F sin θ (opposite). Sketch the triangle and mark which side is which.
Giving a direction as just '37°'.Name the reference line: '37° north of east' or '37° above the horizontal'. An angle without a reference could point in several different directions.
Finding a change in velocity by subtracting speeds, so a car turning a corner at constant speed has Δv = 0.Δv = v − u is a vector subtraction. Reverse u, add it to v head to tail, and you get a real change (21 m s⁻¹ south-east for 15 m s⁻¹ north to 15 m s⁻¹ east).
Swapping the order in v_AB = v_A − v_B, so the relative velocity comes out pointing the wrong way.v_AB is A as seen from B: the observer's own velocity, v_B, is the one subtracted.
Thinking a boat pointed straight across a river lands straight across.The current carries it downstream the whole time. To land directly opposite, aim upstream so that the boat's upstream component cancels the current.

Worked example

A light plane flies at 180 km h⁻¹ relative to the air. The pilot needs to travel due north to a town 240 km away. A steady wind blows towards the east at 50 km h⁻¹.

(a) Determine the direction in which the pilot must point the plane. (b) Calculate the plane's speed relative to the ground. (c) Calculate the flight time in minutes.

(5 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. The wind is much weaker than the plane (50 is just over a quarter of 180), so the pilot only needs to aim a little into the wind. Since sin 15° ≈ 0.26, I predict about 15° west of north. Some of the airspeed is spent cancelling the wind, so the ground speed should be a little under 180 km h⁻¹, about 170, and the flight a bit longer than 240 ÷ 180 h = 80 min.

K · Knowns

I sketch the vector triangle with every value on it, and name each velocity with two letters: P for the plane, A for the air, G for the ground. The ground velocity has to point due north; that is the condition the pilot wants.

v_PA = 180 km h⁻¹ (direction unknown)
v_AG = 50 km h⁻¹ east
v_PG must point due north; d = 240 km

Knowns on the sketch: v_PA = 180 km h⁻¹ at an unknown angle θ west of north, v_AG = 50 km h⁻¹ east. The resultant v_PG must point due north; its size is unknown.

N · Not known

Three answers: an angle with a reference line, a speed, and a time in minutes.

(a) θ = ? (west of north) (b) v_PG = ? km h⁻¹ (c) t = ? min

O · Organise

The relative velocity rule v_PA = v_PG − v_AG rearranges to v_PG = v_PA + v_AG, which is my head-to-tail triangle. The triangle has a right angle between north and east, and v_PA is the hypotenuse because it is the longest side. The wind is opposite θ, so I use sin.

v_PG = v_PA + v_AG
sin θ = v_AG ÷ v_PA
v_PG = √(v_PA² − v_AG²), t = d ÷ v_PG

W · Work it out

I keep full calculator values all the way through and round only the final answers. Pythagoras here subtracts, because I'm finding a shorter side, not the hypotenuse.

(a) sin θ = 50 ÷ 180 = 0.2778, θ = 16.13°
(b) v_PG = √(180² − 50²) = √(32 400 − 2500) = √29 900 = 172.9 km h⁻¹
(c) t = 240 ÷ 172.9 = 1.388 h = 1.388 × 60 = 83.28 min

U · Units

The speeds are all in km h⁻¹ and the distance in km, so the time comes out in hours and I convert to minutes as asked. The data have two or three significant figures, so I'll give two or three.

θ ≈ 16°, v_PG ≈ 173 km h⁻¹, t ≈ 83 min

D · Direction

The heading must say what the angle is measured from: 16° west of north, which is into the wind. The ground velocity is due north, as required. As a sense check, the ground speed is less than the airspeed, which it must be when part of the airspeed goes into cancelling the wind.

Point the plane 16° west of north (N 16° W); it then travels due north at 173 km h⁻¹

(a) Point the plane 16° west of north. (b) 173 km h⁻¹ due north relative to the ground. (c) About 83 minutes.

Check against the prediction. I predicted about 15° west of north, a ground speed of about 170 km h⁻¹ and a bit over 80 min. 16°, 173 km h⁻¹ and 83 min all match.

Exam advice

What markers look for

When two velocities are at right angles, combine them as a vector sum: Pythagoras for the size and inverse tan for the direction. The stronger answers added the velocities as vectors and backed this up with a vector diagram.NESA marking feedback, 2022, summarised
A vector answer needs a direction as well as a size. Give it against a clear reference, such as '16° west of north' or 'away from the wall'.NESA marking feedback, 2023, summarised
Draw and label the vector diagram, even when the question doesn't ask for one. A labelled triangle shows the marker your method, and it is where most sin and cos mix-ups get caught.NESA marking feedback, 2022, summarised
Check the angle you calculated against your sketch before you state it. tan⁻¹(40 ÷ 30) gives the angle from east, not from north.CortexHSC teaching notes
Show the equation, the substitution and the answer with a unit, every time. Correct working earns marks even when the final number slips.General advice from NESA marking feedback, 2021 to 2025, summarised

Summaries in our own words. Read NESA's own feedback with the exam packs: 2022 · 2023.

Where this leads in Year 12

Year 12 mechanics starts with vectors. In projectile motion the first step is always to resolve the launch velocity into horizontal and vertical components (our projectile motion guide is written for the current HSC course, but this physics carries straight into your Year 12). In uniform circular motion you draw the change in velocity, v − u, as a vector diagram and find that it points to the centre of the circle. Frames of reference, the idea behind relative velocity, are also the starting point for special relativity.

FAQ

What is the difference between a scalar and a vector?: A scalar has a size only, such as 20 m s⁻¹ (a speed). A vector has a size and a direction, such as 20 m s⁻¹ north (a velocity). Distance, speed and time are scalars; displacement, velocity, acceleration and force are vectors.
How do you add two vectors at right angles?: Draw them head to tail. The resultant's size is √(a² + b²) and its direction is tan⁻¹(opposite ÷ adjacent), measured from a line you name. For vectors that aren't at right angles, resolve each into components first, add the components, then recombine.
What does v_AB = v_A − v_B mean?: It is the velocity of A as seen by someone moving with B. Subtracting B's velocity removes the observer's own motion. Rearranged, v_A = v_AB + v_B, which is how you add a boat's velocity through the water to the water's velocity to get the boat's velocity relative to the bank.
Which way should a boat point to cross a river straight across?: Partly upstream, so that the upstream part of its velocity through the water cancels the current. If the boat's speed in the water is v and the current is c, the angle upstream from straight across is sin⁻¹(c ÷ v), and the crossing takes longer than pointing straight across.

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