Secondary Mathematics · Worked Repair Guide 31
A motion graph compresses a story into geometry. The axes decide what the graph means. On a distance–time graph, gradient describes speed. On a speed–time graph, gradient describes acceleration and area describes distance travelled. The same triangular-looking region therefore means different things depending on the axes.
This guide develops one central habit: read the axis quantities and units before reading the shape. A horizontal segment can mean rest on a distance–time graph but constant speed on a speed–time graph. A downward segment is possible for speed during slowing, but a cumulative distance travelled graph should not decrease.
All journeys below are invented mathematical models. They are not transport schedules, safety advice or measurements of real vehicles. Use Graphs, Tables and Relationships for foundational graph reading and Real-World Mathematics: Transport, Speed, Travel Time and Fuel for applied modelling.
1. Distance, displacement, speed and velocity are not interchangeable
Distance travelled is a non-negative accumulated path length. Displacement includes direction relative to a reference position.
Speed is rate of change of distance travelled. Velocity includes direction and is rate of change of displacement.
Entry check: travel 5 km east then 5 km west. Total distance=10 km; displacement=0. Average speed over 2 h is5 km/h; average velocity is0 km/h.
This guide uses distance–time and speed–time graphs unless explicitly stated otherwise.
2. Gradient on a distance–time graph is speed
If distance increases from20 m to80 m while time increases from5 s to15 s, gradient=(80−20)/(15−5)=60/10=6 m/s.
The units confirm the interpretation: metres divided by seconds gives metres per second.
A steeper straight segment means greater constant speed when both graphs use the same axis scales.
Do not compare visual steepness across graphs with different scales without calculating gradients.
3. A horizontal distance–time segment means no additional distance
If distance remains at60 m from t=10 s to t=18 s, the gradient is0.
Under a distance-travelled interpretation, the object is stationary during that interval.
The graph has not “stopped” existing; time continues while distance does not increase.
The resting duration is18−10=8 s.
4. A cumulative distance–time graph should not slope downward
Total distance travelled cannot decrease as time passes. Once60 m has been travelled, returning toward the start adds further distance rather than removing earlier distance.
A downward-sloping position-time or displacement-time graph can be meaningful because position can decrease relative to the chosen direction.
Therefore always inspect whether the vertical axis says distance from start, total distance travelled, position or displacement.
Similar-looking axes can encode different mathematics.
5. Piecewise straight distance–time graphs describe changing constant speeds
An invented journey travels from (0 s,0 m) to(10 s,50 m), rests until15 s, then reaches(25 s,130 m).
First speed=50/10=5 m/s.
Rest speed=0.
Final segment speed=(130−50)/(25−15)=80/10=8 m/s.
The last segment is steeper and represents faster motion.
6. Average speed uses total distance divided by total time
For the journey above, total distance=130 m and total elapsed time=25 s.
Average speed=130/25=5.2 m/s.
Do not average the segment speeds (5,0,8) to get13/3. The time spent at each speed differs.
Average speed is a weighted result through total distance and total time.
7. Curved distance–time graphs represent changing speed
If a distance–time curve becomes progressively steeper, speed is increasing.
If it becomes flatter, speed is decreasing.
Over an interval, the secant gradient gives average speed across that interval.
Where tangent gradients are in scope, the tangent gradient at one time estimates instantaneous speed. This is a bridge to later calculus rather than a requirement for every Secondary route.
8. On a speed–time graph, height itself is speed
If the graph is horizontal at6 m/s from t=0 to10 s, speed remains6 m/s throughout.
This is completely different from a horizontal distance–time graph, where speed is zero.
The vertical coordinate now carries speed directly; the gradient carries something else.
Reading the axis label first prevents the two graph types from being confused.
9. Gradient on a speed–time graph is acceleration
If speed rises uniformly from2 m/s to10 m/s in4 s, acceleration=(10−2)/4=2 m/s².
The unit m/s² means the speed changes by2 m/s each second.
A negative speed–time gradient means speed is decreasing: deceleration in everyday school language, or negative acceleration relative to the speed variable’s change.
The speed itself remains non-negative on an ordinary speed–time graph.
10. Area under a speed–time graph is distance travelled
At constant speed6 m/s for10 s, the rectangular area is6×10=60 m.
The units multiply: (m/s)×s=m.
This area is not geometric decoration. It represents accumulated distance over time.
Always use the numerical axis units rather than the physical centimetres of the printed graph.
11. Uniform acceleration creates triangular or trapezoidal areas
Suppose speed increases uniformly from0 to12 m/s over6 s.
Acceleration=12/6=2 m/s².
Distance=area of triangle=1/2×6×12=36 m.
The same distance can be found using average speed under uniform acceleration: (0+12)/2=6 m/s, then6×6=36 m.
12. A non-zero initial speed gives a trapezium
Speed increases uniformly from4 m/s to10 m/s over8 s.
Acceleration=(10−4)/8=0.75 m/s².
Distance=trapezium area=1/2(4+10)×8=56 m.
Equivalently, rectangle4×8=32 plus triangle1/2×8×6=24 gives56.
13. Slowing to rest creates positive area but negative gradient
Speed falls uniformly from15 m/s to0 over5 s.
Acceleration=(0−15)/5=−3 m/s².
Distance travelled during slowing=1/2×5×15=37.5 m.
The negative gradient does not make the distance negative. The area remains above the time axis and therefore positive.
14. Stopping time and stopping distance answer different questions
If a speed–time model falls linearly from20 m/s to0 at constant acceleration−4 m/s², stopping time solves0=20−4t, giving t=5 s.
Stopping distance is the triangular area 1/2×5×20=50 m.
One quantity is measured in seconds, the other in metres.
Unit checking distinguishes them immediately.
15. Total distance from a piecewise speed graph is the sum of areas
An invented speed profile accelerates from0 to8 m/s in4 s, remains at8 m/s for6 s, then slows to0 in2 s.
First triangle area=1/2×4×8=16 m.
Rectangle area=6×8=48 m.
Final triangle area=1/2×2×8=8 m.
Total distance=72 m over12 s, so average speed=6 m/s.
16. Convert units before combining rates and times
A speed of72 km/h equals20 m/s because72×1000/3600=20.
If the time axis is in seconds and speed is left in km/h, graph area has mixed units and cannot immediately be interpreted as kilometres or metres.
Convert to a consistent unit system before using gradient or area.
Use Units, Scale and Measurement when this is the underlying weakness.
17. Average speed cannot usually be found by averaging endpoint speeds
If speed changes uniformly from4 to10 m/s, the time-average speed over that interval is7 m/s because the speed–time graph is linear.
If speed changes nonlinearly, simply averaging endpoint speeds need not give the correct average.
Average speed is always total distance/total time.
Use graph area to obtain total distance when the speed varies.
18. Capstone: read one journey through both graph ideas
An invented object accelerates uniformly from0 to10 m/s in5 s, continues at10 m/s for8 s, then slows uniformly to0 in5 s.
Acceleration in the first stage=2 m/s²; final acceleration=−2 m/s².
Distance=first triangle25 m + rectangle80 m + final triangle25 m=130 m.
Total time=18 s, so average speed=130/18≈7.22 m/s.
A corresponding cumulative distance–time graph would rise throughout, with its gradient increasing during acceleration, constant during the middle stage, and decreasing to zero during the final stage.
19. Independent practice
- Distance rises from20 m at5 s to80 m at15 s. Find average speed over the interval.
- A distance–time graph is horizontal from10 s to18 s. Find rest duration.
- A journey covers130 m in25 s. Find average speed.
- Explain why total distance travelled should not decrease with time.
- A speed–time graph is horizontal at6 m/s for10 s. Find distance travelled.
- Speed rises from2 to10 m/s in4 s. Find acceleration.
- Speed rises uniformly from0 to12 m/s in6 s. Find distance.
- For Question7, find average speed over the6 s.
- Speed rises from4 to10 m/s in8 s. Find acceleration.
- Find distance for Question9.
- Speed falls from15 m/s to0 in5 s. Find acceleration.
- Find distance travelled while slowing in Question11.
- A speed profile rises0→8 m/s in4 s, stays at8 for6 s, then falls to0 in2 s. Find total distance.
- Find average speed for Question13.
- Convert72 km/h to m/s.
- An object travels5 km east then5 km west in2 h. Find total distance and average speed.
- For Question16, find displacement and average velocity.
- Explain the meaning of gradient on a speed–time graph.
- Explain the meaning of area under a speed–time graph.
- Why does a horizontal segment mean different things on distance–time and speed–time graphs?
20. Worked answers
1. 6 m/s. 60/10.
2. 8 s.
3. 5.2 m/s.
4. Distance travelled is accumulated path length; returning toward the start adds further distance rather than deleting earlier travel.
5. 60 m. Rectangle area6×10.
6. 2 m/s².
7. 36 m. Triangle area.
8. 6 m/s. 36/6.
9. 0.75 m/s².
10. 56 m. 1/2(4+10)×8.
11. −3 m/s².
12. 37.5 m.
13. 72 m. 16+48+8.
14. 6 m/s. 72/12.
15. 20 m/s.
16. 10 km and5 km/h.
17. 0 km and0 km/h.
18. Acceleration: change in speed per unit time.
19. Distance travelled, when the axes are speed and time in compatible units.
20. On a distance–time graph horizontal means zero gradient and therefore zero speed; on a speed–time graph the vertical height itself is a constant non-zero speed unless the line lies on the time axis.
21. Diagnose motion-graph errors by the axis meaning
Common failures include treating speed–time height as distance, using speed–time gradient as speed, averaging segment speeds without time weighting, allowing cumulative distance to fall, or calculating graph area with incompatible units.
A useful repair note says “distance–time gradient=speed”, “speed–time gradient=acceleration”, “speed–time area=distance”, or “average speed=total distance÷total time”.
Then show the same horizontal line on two differently labelled axes. A secure learner should explain why the physical meanings differ before doing any arithmetic.
22. Continue through the BTT learning routes
Return to the BTT Mathematics Hub or BTT Mathematical Lab. Use Graphs, Tables and Relationships for graph foundations, Direct and Inverse Proportion, Variation and Rates for rate structure, and Trigonometry, Pythagoras, Bearings and Elevation for spatial journeys.
Within Batch08, continue to Probability Trees, Conditional Events and Without-Replacement Reasoning, Histograms, Frequency Density and Grouped Data, or Approximation, Significant Figures, Bounds and Error Intervals.
23. Sources and scope
The journeys, speeds, graph segments and practice questions are original teaching material. Gradient and area interpretations follow from rate and accumulation definitions with consistent units.
For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match tangent-gradient, acceleration and velocity extensions to the learner’s actual subject level and school programme.
Complete Secondary Mathematics repair route: browse all 48 worked repair guides or return to the Secondary Mathematics Learning Hub.