The location of an object relative to a reference point, given as a coordinate (often x or r). Vector.

The total length of the path travelled. Scalar.

What is displacement?

The change in position from start to finish, r = r_f - r_i. Vector. Independent of path.

Distance per unit time. Scalar.

QLDPhysicsSyllabus dot point

Topic 1: Linear motion and force

Recall, describe and apply the concepts of position, displacement, distance, speed, velocity and acceleration, distinguishing between scalar and vector quantities and between average and instantaneous values

A focused answer to the QCE Physics Unit 2 dot point on the basic kinematic quantities. Defines position, displacement, distance, speed, velocity and acceleration; distinguishes average and instantaneous values; and works the QCAA short-answer style problem on average versus instantaneous velocity that recurs in IA1 and the EA.

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

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
Definitions
Average vs instantaneous
Sign conventions
Examples in context
Try this

What this dot point is asking

QCAA expects you to use the standard kinematic vocabulary with precision. The same SI units appear in pairs (distance and displacement in metres; speed and velocity in m s $^{-1}$ ) but the pair members are not interchangeable. The dot point also requires the distinction between average values (over an interval) and instantaneous values (at one moment).

Definitions

Position: The location of an object relative to a reference point, given as a coordinate (often $x$ or $\vec{r}$ ). Vector.
Distance: The total length of the path travelled. Scalar.
Displacement: The change in position from start to finish, $\Delta \vec{r} = \vec{r}_f - \vec{r}_i$ . Vector. Independent of path.
Speed: Distance per unit time. Scalar.

\text{average speed} = \frac{\text{total distance}}{\text{total time}}

Velocity. Rate of change of displacement. Vector.

\vec{v}_{\text{avg}} = \frac{\Delta \vec{r}}{\Delta t}

Acceleration. Rate of change of velocity. Vector.

\vec{a}_{\text{avg}} = \frac{\Delta \vec{v}}{\Delta t}

Acceleration has SI unit m s $^{-2}$ and points in the direction of the change in velocity, not the direction of motion. A car slowing down has acceleration opposite its velocity.

Average vs instantaneous

Average quantities use the endpoints of an interval. Instantaneous quantities use the limit as $\Delta t \to 0$ , equivalent to the slope of the position-time graph (instantaneous velocity) or the velocity-time graph (instantaneous acceleration) at that point.

A speedometer reads instantaneous speed. A police speed trap measuring time over a known distance reads average speed.

Sign conventions

Pick a positive direction at the start of a problem and apply it consistently to position, velocity and acceleration. A negative velocity means motion in the negative direction; a negative acceleration means the velocity is becoming more negative (which can mean speeding up if velocity is already negative).

Worked example

A car has position $x(t) = 2.0t^2 - 4.0t + 1.0$ metres, with $t$ in seconds. We find the instantaneous velocity and acceleration at $t = 3.0$ s by differentiating the position function.

Step 1: Differentiate position to find the velocity function

Instantaneous velocity is the rate of change of position, so we differentiate $x(t)$ with respect to $t$ . Each power of $t$ follows the power rule.

v(t) = \frac{dx}{dt} = 4.0t - 4.0

Step 2: Evaluate velocity at $t = 3.0$ s

Substituting the specific time gives the instantaneous velocity at that moment.

v(3.0) = 4.0(3.0) - 4.0 = 12.0 - 4.0 = 8.0\ \text{m s}^{-1}

Step 3: Differentiate velocity to find acceleration

Acceleration is the rate of change of velocity. Differentiating the linear function $v(t) = 4.0t - 4.0$ gives a constant.

a(t) = \frac{dv}{dt} = 4.0\ \text{m s}^{-2}

Final answer: at $t = 3.0$ s the velocity is $8.0$ m s $^{-1}$ and the acceleration is $4.0$ m s $^{-2}$ . Because the acceleration is constant, the instantaneous and average accelerations are the same throughout the motion.

Examples in context

Example 1. A Cairns light-rail train accelerates from rest over $400 \text{ m}$ between Smithfield stops, reaching $14 \text{ m s}^{-1}$ in $60 \text{ s}$ . Average velocity is $\bar{v} = \Delta x / \Delta t = 400/60 = 6.67 \text{ m s}^{-1}$ , distinct from the instantaneous velocity at the end of the leg. Average acceleration is $\bar{a} = \Delta v / \Delta t = 14/60 = 0.23 \text{ m s}^{-2}$ . The QCAA Unit 2 distinction between average and instantaneous quantities is exactly the data-test analysis required when track sensors stream tachograph data into a stimulus stem.

Example 2. A Sunshine Coast tidal-study buoy logs displacement of a drifter every $30 \text{ s}$ . Over one hour the drifter moves $720 \text{ m}$ along a curved track but its net displacement is only $480 \text{ m}$ east. The distance (scalar) is therefore $720 \text{ m}$ , the displacement (vector) is $480 \text{ m}$ east, and average speed ( $0.20 \text{ m s}^{-1}$ ) exceeds the magnitude of average velocity ( $0.133 \text{ m s}^{-1}$ ). QCAA Unit 2 IA1 stimulus often uses such an oceanographic dataset to test the scalar versus vector distinction.

Try this

Q1. Define displacement, velocity and acceleration and identify which are vectors. [3 marks]

Cue. Displacement (vector) = change in position; velocity (vector) = rate of change of displacement; acceleration (vector) = rate of change of velocity.

Q2. A car covers $1200 \text{ m}$ around a loop in $80 \text{ s}$ , returning to its start. State the distance, the displacement, the average speed and the average velocity. [3 marks]

Cue. Distance $1200 \text{ m}$ ; displacement $0$ ; average speed $15 \text{ m s}^{-1}$ ; average velocity $0$ .

Q3. A Cairns light-rail train accelerates from rest to $14 \text{ m s}^{-1}$ in $20 \text{ s}$ . (a) Calculate the average acceleration. (b) Sketch a velocity-time graph and identify the area-under-graph as displacement. (c) Explain why instantaneous velocity at $t = 10 \text{ s}$ may differ from the average if acceleration is non-uniform. [2+3+2 marks; ISMG: Analysis and interpretation]

Cue. (a) $0.70 \text{ m s}^{-2}$ ; (b) area = displacement, $140 \text{ m}$ if uniform; (c) instantaneous = tangent slope, may be greater or less than chord.

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 SAC3 marksA runner jogs

400

m east in

80

s, then

200

m west in

50

s. Calculate (a) the average speed and (b) the average velocity over the entire trip.

Show worked answer →

Total time: $80 + 50 = 130$ s.

(a) Average speed uses total distance.

Total distance: $400 + 200 = 600$ m.

Average speed: $600 / 130 = 4.6$ m s $^{-1}$ .

(b) Average velocity uses displacement (final position minus initial).

Net displacement: $400 - 200 = 200$ m east.

Average velocity: $200 / 130 = 1.5$ m s $^{-1}$ east.

Markers reward the explicit use of distance versus displacement, the inclusion of a direction on velocity, and units throughout.