Motion
Study the concepts, diagrams and graphs. Practise every calculation with a formula, a unit conversion and a clearly explained final answer. Use the interactive labs, then check your understanding with test questions.
Work through the five chapter sections in order. After each topic, predict an answer before you reveal the worked solution. Run the interactive graphs and practise rearranging equations. Finally, use the learning-goal checklist and question bank. All calculations use the Year 2 TTO symbols s, v, t and a.
2.1 · Motion and distance
Start with motion, distance, time, quantities and the distinction between distance travelled and displacement.
An object is moving when its position changes over time.Physics describes motion with measurable quantities and units.You need to know which symbol describes each quantity.
1Quantities and units
| Quantity | Symbol | SI unit | Symbol for unit |
|---|---|---|---|
| Distance travelled | s | metre | m |
| Time | t | second | s |
| Speed | v (average: vavg) | metre per second | m/s |
| Acceleration | a | metre per second squared | m/s² |
2Distance versus displacement
The blue path is the distance travelled.The displacement connects the start to the finish; here that straight-line displacement is zero.
Distance travelled
The whole length of the route: 4 + 2 + 4 + 2 = 12 m.
Distance cannot be negative.
Displacement
The change in position from start to end, including direction: 0 m after a complete lap.
2.2 · Speed and average speed
Based on the source lesson on speed, average speed, converting units, displacement and rearranging s = vt.
Speed tells us how much distance is travelled per unit of time.The speedometer shows instantaneous speed; a journey calculation usually gives average speed.
Find speed
v = s / t
Divide distance by time.
Find distance
s = v × t
Multiply speed by time.
Find time
t = s / v
Divide distance by speed.
1Converting speed
km/h → m/s
Divide by 3.6
72 km/h ÷ 3.6 = 20 m/s
m/s → km/h
Multiply by 3.6
12 m/s × 3.6 = 43.2 km/h
2Scientific notation
Very large or small results are easier to read with powers of ten.
For example, 72,000 m = 7.2 × 10⁴ m.
Always retain the unit; changing the notation does not change the quantity.
2.3 · Acceleration and deceleration
Source lesson: changes in speed over time, the unit m/s², and areas under speed–time graphs.
An object accelerates when its velocity changes.When it slows down in the same direction, acceleration is negative (deceleration).
Constant speed
A horizontal v–t graph means a = 0 m/s².
Every second, the object covers the same distance.
Constant acceleration
A straight upward or downward v–t line means speed changes by the same amount every second.
1Find the distance from a v–t graph
For the 0 → 20 m/s example, the triangular area is ½ × 20 m/s × 4 s = 40 m. You will explore this with adjustable starting and final speeds in the Speed–Time Graph Lab.
2.4 · Reading and drawing motion graphs
Source lesson: constant speed, increasing speed, average speed, slope and areas under velocity–time graphs.
A distance–time (s–t) graph shows how much distance has been travelled. A speed–time (v–t) graph shows how fast the object moves.Do not confuse their slopes.
On a distance–time graph
The gradient (slope) gives speed: v = Δs / Δt.
A steeper line means greater speed.Curved lines mean changing speed.
On a speed–time graph
The gradient gives acceleration: a = Δv / Δt.
Here Δv = vfinal − vinitial.
The area underneath gives distance travelled when velocity remains non-negative.
1Draw a distance–time graph from calculated points
For constant acceleration from rest to 20 m/s in 4 s, the distance after each second is s = ½ at², with a = 5 m/s².
This expression is derived from the triangular area of the v–t graph; it is not a separate rule you must memorise for every motion.
| Elapsed time | Distance travelled | How to calculate |
|---|---|---|
| 0 s | 0 m | ½ × 5 × 0² |
| 1 s | 2.5 m | ½ × 5 × 1² |
| 2 s | 10 m | ½ × 5 × 2² |
| 3 s | 22.5 m | ½ × 5 × 3² |
| 4 s | 40 m | ½ × 5 × 4² |
2.5 · Stopping distance
Source lesson: thinking distance, braking distance, speed dependence, traffic hazards and the two-second rule.
A driver continues moving during the reaction time.Only after that does braking reduce the speed.These two distances add up to the stopping distance.
Thinking distance
The driver sees, recognises and responds to a danger, while speed remains approximately constant.
On a v–t graph: rectangle.
Braking distance
The brakes reduce the car's speed from the starting speed to zero.
For constant deceleration: triangle.
Reaction factors
Distraction, fatigue, poor visibility, alcohol and some drugs can delay a response.
Braking factors
Speed, road grip, tyre condition, brakes and slippery surfaces affect braking distance.
Double the speed
For the same braking deceleration, braking distance becomes four times as large, while thinking distance doubles if reaction time is unchanged.
One motion, three connected graphs: v–t, a–t and s–t
Do not memorise the graphs as separate pictures.They all describe the same moving object.
The relationship in three steps
Move the time marker: watch the shaded area become distance
Construct the s–t graph one second at a time
Each row is another coloured strip or trapezium under the v–t graph.Select a row to jump to that second.
| Time (s) | v at start → end (m/s) | Area for this second Δs (m) | Total s (m) |
|---|
Calculate a point yourself
Choose a whole second on the time slider.Calculate the total area from t = 0 to that moment, then enter the distance in metres.
What must your graph show?
Speed–Time Graph Lab
Change the starting speed, final speed and time.Read acceleration from the slope and distance from the coloured area.
Orange rectangle: initial speed × time.Blue triangle: half the change in speed × time.If speed falls, the blue triangle is subtracted from the initial rectangle.
Distance–Time Graph Lab
Follow how a straight v–t line creates a curved s–t graph when speed increases.
| Time (s) | Speed (m/s) | Distance from start (m) |
|---|
Stopping Distance Lab
Adjust speed, reaction time and braking deceleration.Compare thinking and braking distances without hiding the formula steps.
Thinking distance is proportional to speed; braking distance is proportional to speed squared when deceleration stays constant.
Rearrange Formulas – Then Calculate
Make the unknown the subject.Use the same mathematical operation on both sides of an equation.
To find v, divide both sides by t.
To find t, divide both sides by v.
Choose a question and type the formula, e.g. t=s/v.
Use * for multiplication and / for division.
Reaction Time Lab
Test your response to a visual signal.The measured time is only this simple screen response, not a prediction of how long you would need to respond while driving.
This practical can help you understand reaction time.Real traffic reactions depend on attention, visibility and several other factors.
Formula Cards and Units
Choose the correct formula before substituting the numbers.Keep a clear distinction between a quantity symbol and a unit.
Average speed
vavg = s / t
Speed in m/s; s in m; t in s.
Distance
s = v × t
m = (m/s) × s.
Time
t = s / v
s = m ÷ (m/s).
Acceleration
a = Δv / Δt
m/s² = (m/s) ÷ s.
Average speed, uniform acceleration
vavg = (vi + vf)/2
Use when acceleration is constant.
Braking distance
sb = ½vtb
Area of the braking triangle.
Learning Goals Checklist
Tick each skill after you can explain it and solve an example without looking at the solution.
Physics concepts
Calculations and graphs
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R questions and I questions
R = Recall / Reproduction. I = Insight / Interpretation.Practise both remembering a rule and applying it to a new situation.
Select a category or start the quiz.
Physics Flashcards
Look at the front, say your answer out loud, then flip the card to check it.Repeat the difficult cards.
Practice Exam and Step-by-Step Answers
Choose topics and difficulty, enter numerical answers and reveal complete worked solutions. A new set does not change the explanations in the guide.
Applied investigations
These longer tasks combine several skills.Use the six-step calculation method for every part.