A swimmer crosses a river at 1.2 m s^-1 perpendicular to the bank; the current is 0.5 m s^-1 downstream. Calculate the resultant velocity magnitude and bearing relative to the bank. [3 marks]

A Sunshine Coast drifter experiences a 0.6 m s^-1 northeast current and a 0.3 m s^-1 north wind-driven drift. (a) Resolve each into east-north components. (b) Sum the components.

QLDPhysicsSyllabus dot point

Topic 1: Linear motion and force

Distinguish between scalar and vector quantities, including identifying examples and applying operations of addition and subtraction in one and two dimensions

A focused answer to the QCE Physics Unit 2 dot point on scalar and vector quantities. Defines the distinction with examples, walks through vector addition (head-to-tail and component methods), subtraction as adding the opposite, and the standard QCAA component resolution students use throughout Unit 2 motion and Unit 3 fields.

Generated by Claude Opus 4.88 min answerUpdated 2026-05-22

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What this dot point is asking
Scalars and vectors
Adding vectors graphically
Resolving into components
Subtracting vectors
Examples in context
Try this

What this dot point is asking

QCAA wants you to separate the two families of physical quantity. Scalars are fully described by a magnitude and a unit. Vectors need a magnitude, a unit, and a direction. The same arithmetic rules do not apply to both. This distinction underpins every motion, force and field problem in Units 2, 3 and 4.

Scalars and vectors

Type	Definition	Examples
Scalar	Magnitude only	mass, time, distance, speed, energy, temperature, work, power
Vector	Magnitude and direction	displacement, velocity, acceleration, force, momentum, impulse

Two scalars combine by ordinary arithmetic. Two vectors combine by vector addition, which depends on their directions.

Adding vectors graphically

Place the tail of the second vector at the head of the first. The resultant runs from the tail of the first to the head of the last. The order does not matter (vector addition is commutative).

For perpendicular vectors, magnitude follows Pythagoras and direction follows the inverse tangent. For non-perpendicular vectors, resolve into components or use the cosine rule.

Resolving into components

A vector $\vec{v}$ of magnitude $v$ at angle $\theta$ above the horizontal has components:

v_x = v \cos\theta, \quad v_y = v \sin\theta

Components allow you to add and subtract vectors numerically. Sum the $x$ -components, sum the $y$ -components, then recombine with Pythagoras and arctan.

Subtracting vectors

$\vec{a} - \vec{b}$ means $\vec{a} + (-\vec{b})$ . To subtract $\vec{b}$ , reverse its direction and add. This step is critical for change-in-velocity problems (for example $\Delta \vec{v} = \vec{v}_f - \vec{v}_i$ in uniform circular motion or in a collision).

Change in velocity when a car turns

A car travels east at $20$ m s $^{-1}$ and then turns so that it travels north at $20$ m s $^{-1}$ . Although the speed has not changed, the direction has, so the velocity has changed. Find the change in velocity.

Step 1: Write the initial and final velocity vectors

Express each velocity using unit vectors: $\hat{\imath}$ for east and $\hat{\jmath}$ for north.

\vec{v}_i = 20\hat{\imath} \text{ (east)}, \qquad \vec{v}_f = 20\hat{\jmath} \text{ (north)}

Step 2: Subtract vectors to find the change

Vector subtraction means adding the negative. The change in velocity is:

\Delta\vec{v} = \vec{v}_f - \vec{v}_i = 20\hat{\jmath} - 20\hat{\imath} = -20\hat{\imath} + 20\hat{\jmath}

This vector points west ( $-20\hat{\imath}$ ) and north ( $+20\hat{\jmath}$ ) simultaneously.

Step 3: Find the magnitude and direction

The two components are perpendicular, so Pythagoras gives the magnitude:

|\Delta\vec{v}| = \sqrt{(-20)^2 + 20^2} = \sqrt{800} = 28.3 \text{ m s}^{-1}

The angle between the north component and the resultant is $\arctan(20/20) = 45°$ , so the direction is $45°$ north of west.

Final answer: change in velocity $= 28.3$ m s $^{-1}$ directed $45°$ north of west. The speed is unchanged at $20$ m s $^{-1}$ , but the velocity has changed because the direction changed. This kind of velocity change, perpendicular to the motion, is exactly what produces centripetal acceleration in circular motion.

Examples in context

Example 1. A Sunshine Coast tidal current is reported by the buoy as $0.6 \text{ m s}^{-1}$ northeast. The component form is $0.6\cos 45^\circ = 0.424 \text{ m s}^{-1}$ east and $0.424 \text{ m s}^{-1}$ north. A drifter released into a $0.3 \text{ m s}^{-1}$ northward wind-driven current experiences a total velocity vector summed component-wise: $0.424 \text{ m s}^{-1}$ east and $0.724 \text{ m s}^{-1}$ north, magnitude $0.84 \text{ m s}^{-1}$ at bearing $030^\circ$ . QCAA EA Unit 2 data-test scenarios use exactly this oceanographic addition.

Example 2. A Cairns crane lifts a $400 \text{ kg}$ load while the boom swings horizontally at $1.5 \text{ m s}^{-1}$ . The load's velocity is the vector sum of the boom's horizontal and the hoist's vertical: $\vec{v} = (1.5, 0.5) \text{ m s}^{-1}$ giving magnitude $1.58 \text{ m s}^{-1}$ at $18.4^\circ$ above horizontal. Distinguishing these vectors from the scalar quantity speed ( $1.58 \text{ m s}^{-1}$ without direction) is the kind of QCAA-style multi-mark answer markers expect.

Try this

Q1. Distinguish scalar from vector quantities and give two examples of each. [2 marks]

Cue. Scalar: magnitude only (speed, energy); vector: magnitude and direction (velocity, force).

Q2. A swimmer crosses a river at $1.2 \text{ m s}^{-1}$ perpendicular to the bank; the current is $0.5 \text{ m s}^{-1}$ downstream. Calculate the resultant velocity magnitude and bearing relative to the bank. [3 marks]

Cue. Magnitude $\sqrt{1.2^2 + 0.5^2} = 1.30 \text{ m s}^{-1}$ ; angle $\arctan(0.5/1.2) = 22.6^\circ$ downstream of perpendicular.

Q3. A Sunshine Coast drifter experiences a $0.6 \text{ m s}^{-1}$ northeast current and a $0.3 \text{ m s}^{-1}$ north wind-driven drift. (a) Resolve each into east-north components. (b) Sum the components. (c) State the magnitude and bearing of the total velocity. [3+2+2 marks; ISMG: Analysis and interpretation]

Cue. (a) NE: $(0.424, 0.424)$ , wind: $(0, 0.3)$ ; (b) total $(0.424, 0.724)$ ; (c) $0.84 \text{ m s}^{-1}$ at $030^\circ$ .

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Year 11 SAC4 marksA hiker walks

4.0

km east, then

3.0

km north. Calculate (a) the total distance travelled and (b) the magnitude and direction of the resultant displacement.

Show worked answer →

(a) Total distance is a scalar sum along the path: $4.0 + 3.0 = 7.0$ km.

(b) Resultant displacement is the vector from start to finish. With the legs at right angles, use Pythagoras and trig.

Magnitude: $|s| = \sqrt{4.0^2 + 3.0^2} = \sqrt{25} = 5.0$ km.

Direction: $\theta = \tan^{-1}(3.0 / 4.0) = 36.9°$ measured north of east.

Markers reward the explicit distinction between scalar distance and vector displacement, the right-angle Pythagoras step, and a direction given as an angle from a stated reference axis.