A sloping line does not have one universal meaning. On a distance-time graph, its gradient can describe speed. On a speed-time graph, its height describes speed, while the area beneath it describes distance travelled. Reading the wrong feature can turn an otherwise accurate calculation into an answer to a different question.
This Secondary 3 Mathematics Learning Guide follows complete journeys rather than isolated graph rules. You will distinguish distance from a starting point from total distance travelled, include stops correctly, find distances from areas, recover missing speeds and durations, and decide whether a calculated meeting time belongs to the correct journey stage. Each worked example makes the axis labels and model assumptions explicit.
Scope and source: the 2027 SEC G3 Mathematics syllabus includes interpreting distance-time and speed-time graphs in real-world problems. This is a G3-oriented learning guide, not a claim that every school teaches every application in the same Secondary 3 term. All journeys below are original, idealised teaching models.
Learning route: diagnostic · whole journey · speed-time area · catch-up problems · independent practice · explained answers.
Return to the Secondary Mathematics Hub. For the underlying line skills, use Coordinate Geometry and Transformations.
Read the Vertical Axis Before Reading the Line
Three graphs can describe the same traveller while looking different. A graph of total distance travelled records accumulated movement and cannot decrease. A graph of distance from home can decrease when the traveller returns towards home. A speed-time graph records how fast the traveller moves, regardless of whether the direction is outward or homeward.
The distinction is especially important because school questions sometimes use the broad phrase “distance-time graph” for a vertical axis labelled distance from a reference point. Read the complete label. Do not assume that every downward segment represents negative speed or that the final vertical reading is the total length of the journey.
For the straight-road examples here, moving away from home makes the distance-from-home graph rise, and moving towards home makes it fall. In a more general two-dimensional journey, change in distance from home alone does not describe every part of the motion. A person could move around a circle centred on home while remaining the same distance away. Our one-dimensional road assumption prevents that ambiguity.
A Diagnostic About Meaning, Not Just Calculation
Answer these before the main lesson. A straight-road traveller’s distance from home is unchanged for ten minutes: what does that horizontal section represent? A speed-time graph is horizontal at 12 m/s: is the traveller stationary? How far is travelled at 12 m/s for 20 seconds? A traveller goes 5 km from home and returns: what are the final distance from home and the total distance travelled?
The answers are a stop under the stated straight-road model; no, the speed is constant at 12 m/s; 240 m; and 0 km from home but 10 km travelled. The two horizontal graphs do different jobs. One fixes position relative to home; the other fixes speed. The final pair distinguishes where the traveller finishes from how much movement has accumulated.
When one of these answers is wrong, repair the label-to-meaning connection first. More gradient arithmetic will not fix a student who is reading a speed graph as a position graph. Before any numerical work, write a sentence such as “The vertical number tells me how far from home the traveller is.” That sentence sets the interpretation for everything that follows.
The Graph Is Not a Picture of the Road
A rising distance-time line does not mean the road goes uphill. The horizontal axis is time, not east-west position, and the vertical axis is a distance measurement, not elevation. A sharp corner in the graph describes a change in the modelled rate, not a corner on a map.
A graph can also simplify a journey by joining recorded points with straight segments. Doing so assumes a constant rate between those points. The endpoints alone do not prove that the actual traveller maintained that rate at every instant. Whenever this guide calculates from straight segments, the straight-segment model is part of the given information.
Gradient: Use Changes, Not Coordinates in Isolation
A line segment through (3 min, 600 m) and (9 min, 1800 m) has gradient (1800 − 600)/(9 − 3) = 200 m/min. In metres per second, this is 200/60 = 10/3 m/s. In kilometres per hour, it is 12 km/h. The three values describe the same rate in different units.
The calculation uses changes between two points. Dividing a single vertical coordinate by its horizontal coordinate works only when the relevant line also passes through the origin of those axes. A delayed start, an initial distance from home or an offset clock can make that shortcut wrong. The gradient formula remains valid because it subtracts the two readings consistently.
Worked Journey 1: Travel, Stop, Travel and Return
A cyclist moves on one straight road. The table gives time since departure and distance from home. Join consecutive points by straight lines to represent constant speed during each moving stage. At the horizontal section, the cyclist is stopped.
| Time since departure | Distance from home |
|---|---|
| 0 min | 0 km |
| 12 min | 3 km |
| 20 min | 3 km |
| 35 min | 6 km |
| 50 min | 0 km |
During the first twelve minutes, the cyclist travels 3 km. The speed is 3/(12/60) = 15 km/h. From minute 12 to minute 20, the distance remains 3 km, giving an eight-minute stop. From minute 20 to minute 35, the cyclist travels another 3 km in fifteen minutes, so the speed is 12 km/h.
The final segment falls from 6 km to 0 km over fifteen minutes. Its signed gradient is −24 km/h, describing movement towards home. The cyclist’s speed is the magnitude, 24 km/h. The negative gradient does not mean a negative amount of distance was travelled. It tells us the distance from home was decreasing.
The total distance is 3 + 0 + 3 + 6 = 12 km. The final distance-from-home reading is zero because the cyclist returns home. Adding signed changes, 3 + 0 + 3 − 6, also gives zero, but that is the net change in position along this road, not the accumulated journey length.
The Same Journey Has a Different Accumulated-Distance Graph
To plot total distance travelled for that journey, use the same times but vertical readings 0, 3, 3, 6 and 12 km. The return stage now rises from 6 to 12 because the cyclist continues accumulating distance. The stop remains horizontal because no new distance is added during it.
This comparison is a useful transfer task. The events have not changed; only the quantity on the vertical axis has changed. A student who can reconstruct both graphs understands more than the appearance of one familiar diagram. The speed during the return stage can be recovered from either graph, but one has a negative position gradient and the other a positive accumulated-distance gradient.
Worked Journey 2: Average Speed Including and Excluding a Stop
For the complete fifty-minute journey, average speed is total distance divided by total elapsed time: 12/(50/60) = 14.4 km/h. The stop belongs in the denominator because it is part of the time from departure to return.
A different question could explicitly ask for average speed while moving. Moving time is 50 − 8 = 42 minutes, so that average is 12/(42/60) = 120/7, approximately 17.1 km/h. Neither value is an alternative rounding of the other. They answer different questions because their time intervals are different.
Do not average the three moving speeds as (15 + 12 + 24)/3. They apply for unequal durations. More generally, travelling 9 km at 6 km/h and another 9 km at 12 km/h takes 1.5 + 0.75 = 2.25 hours. The whole-journey average is 18/2.25 = 8 km/h, not 9 km/h. The reliable definition always returns to total distance and total time.
Why Area Under a Speed-Time Graph Gives Distance
At constant speed v for duration t, distance is vt. On a speed-time graph, those same quantities form the height and width of a rectangle, whose area is vt. The units confirm the connection: (m/s) × s = m.
For a straight sloping speed segment, the speed changes uniformly. Its average over that interval is the average of its endpoint speeds, so distance is [(initial speed + final speed)/2] × duration. This is exactly the area of the corresponding trapezium. When one endpoint speed is zero, the trapezium becomes a triangle.
For a curved speed-time graph, geometric pieces may provide an estimate rather than an exact area. Do not apply the endpoint-average shortcut automatically to an arbitrary curve. The detailed shape between the endpoints can change the total distance. In the worked calculations below, straight lines or constant-speed sections are explicitly stated.
Worked Journey 3: Acceleration, Cruising and Deceleration
A trolley starts from rest, increases speed uniformly to 12 m/s in eight seconds, maintains 12 m/s until time 26 seconds, and then decreases uniformly to rest at time 38 seconds. Its speed-time vertices are (0, 0), (8, 12), (26, 12) and (38, 0), with time in seconds and speed in m/s.
The initial triangle has area 1/2 × 8 × 12 = 48 m. The cruising rectangle lasts 26 − 8 = 18 seconds, so its area is 18 × 12 = 216 m. The final triangle lasts 38 − 26 = 12 seconds, so its area is 1/2 × 12 × 12 = 72 m. Total distance is 336 m.
Average speed over the full journey is 336/38, approximately 8.84 m/s. The final speed is zero, but the travelled distance is not zero. The journey keeps accumulating distance during deceleration because the trolley is still moving until the endpoint.
The initial speed-time gradient is 12/8 = 1.5 m/s². The final gradient is (0 − 12)/12 = −1 m/s². In this straight-line, fixed-direction model, these describe acceleration and deceleration. They are rates of change of speed, not distances. The unit m/s² is a useful warning that a gradient calculation cannot substitute for the area calculation.
Worked Journey 4: A Trapezium That Does Not Start at Rest
A vehicle’s speed rises uniformly from 6 m/s to 18 m/s over ten seconds. The distance travelled is [(6 + 18)/2] × 10 = 120 m. Using only the triangle above the initial speed would give 60 m and omit the rectangle underneath it.
The same area can be separated into a rectangle 6 × 10 = 60 m and a triangle 1/2 × 10 × (18 − 6) = 60 m. Their sum checks the trapezium calculation. The speed-time gradient is (18 − 6)/10 = 1.2 m/s², a different quantity with different units.
The important subtraction is in the triangle’s height, not in the total trapezium height formula. The triangle represents only the increase above 6 m/s. Drawing or mentally identifying that baseline prevents confusion between the total area and the additional area caused by speeding up.
Worked Journey 5: Convert the Time Axis Before Finding Area
A vehicle travels at 72 km/h for 2.5 minutes. Multiplying 72 by 2.5 gives a number with units km·min/h, not kilometres. Convert time to hours: distance = 72 × (2.5/60) = 3 km. Alternatively, 72 km/h is 20 m/s and 2.5 minutes is 150 seconds, giving 3000 m.
Now suppose it accelerates uniformly from rest to 72 km/h over thirty seconds. Convert 72 km/h to 20 m/s before calculating the triangular area: 1/2 × 30 × 20 = 300 m. The small number of operations does not make the unit choice optional.
A dependable habit is to write the units beside the graph coordinates before decomposing its area. A graph can legitimately mix minutes with kilometres per hour, but the calculation must reconcile them. The same dimensional control is developed in Accuracy and Calculator Discipline.
Worked Journey 6: Recover an Unknown Maximum Speed
A speed-time graph rises uniformly from zero to V m/s in ten seconds, remains at V for twenty seconds, and falls uniformly to zero over fifteen seconds. The total distance is 520 m. Find V.
The three areas are 5V, 20V and 7.5V. Their sum gives 32.5V = 520, so V = 16 m/s. The graph height has been found from an area condition. This is an algebra problem created by a graphical representation, not a different formula to memorise.
Check the recovered distance: 80 + 320 + 120 = 520 m. The total duration is 45 seconds, so average speed is 520/45, approximately 11.6 m/s. This is below the maximum 16 m/s, as it should be. A mean above the graph’s maximum height would expose an error in the area, duration or units.
Worked Journey 7: Recover an Unknown Cruising Duration
A traveller accelerates uniformly from rest to 10 m/s in six seconds, travels at 10 m/s for T seconds, then decelerates uniformly to rest in four seconds. The total distance is 250 m. Find the cruising duration and complete journey time.
The two triangular distances are 30 m and 20 m. The remaining 200 m belongs to the cruising rectangle. Therefore 10T = 200 and T = 20 seconds. Complete journey time is 6 + 20 + 4 = 30 seconds.
The answer T is a duration, not the clock coordinate where cruising ends. Cruising ends at t = 6 + 20 = 26 seconds. Confusing duration with endpoint time is a common source of errors in multi-stage graphs, particularly when the horizontal axis already contains nonzero start times.
Worked Journey 8: A Delayed Start and a Catch-Up Point
Walker A leaves a meeting point at 4 km/h. Walker B leaves the same point fifteen minutes later at 6 km/h. Both continue along the same straight route without stopping. Let t be time in hours measured from A’s departure.
A’s distance is 4t. After B has departed, B’s distance is 6(t − 0.25). At a meeting, the distances from their common start are equal: 4t = 6(t − 0.25). This gives 2t = 1.5 and t = 0.75 hours.
They meet forty-five minutes after A leaves, or thirty minutes after B leaves, at distance 3 km. All three pieces belong in the interpretation because a question can ask for elapsed time from either departure. The solution is inside the assumed interval t ≥ 0.25, so B’s formula is valid there.
Writing B’s distance as 6t would incorrectly allow B to move during the fifteen-minute delay. A shared clock must be used, but a shared clock does not mean equal travel durations. The brackets in t − 0.25 preserve that distinction.
Worked Journey 9: Check Which Stage Contains the Meeting
Cyclist A rides at 12 km/h for thirty minutes, reaches a point 6 km away, and waits there until one hour after departure. Cyclist B starts from A’s original starting point fifteen minutes after A and rides at 16 km/h. Both use the same route. When does B reach A?
During A’s moving stage, an equation would be 12t = 16(t − 0.25). It gives t = 1 hour, but that is outside the stage 0.25 ≤ t ≤ 0.5 in which both moving formulas apply. The algebra is not wrong; its result does not belong to its assumed stage.
During the waiting stage, A’s distance is fixed at 6 km. Solve 6 = 16(t − 0.25), giving t = 0.625 hours = 37.5 minutes after A starts. This lies between thirty and sixty minutes, so the waiting-stage model accepts it. B has travelled for 22.5 minutes, covering 6 km.
This is a general lesson in piecewise modelling: solve the equation, then check the interval over which its formula was defined. A neat answer from the wrong interval is not a valid journey solution. Keep the stage boundaries next to each equation so they cannot disappear during calculation.
Worked Journey 10: Equal Speeds Do Not Mean Equal Positions
Two travellers start together on the same straight path. For 0 ≤ t ≤ 8 seconds, A’s speed is v = 2t m/s and B’s speed is a constant 8 m/s. Their speed-time graphs intersect when 2t = 8, at t = 4 seconds.
By that time, A has travelled the triangular area 1/2 × 4 × 8 = 16 m. B has travelled the rectangular area 4 × 8 = 32 m. They have equal speeds at that instant but are 16 m apart.
For a later catch-up, compare accumulated areas. At time t, A’s area is 1/2 × t × 2t = t², while B’s is 8t. Solve t² = 8t to obtain t = 0 or t = 8. The first is their shared departure; the later meeting is at 8 seconds and 64 m, within the given interval.
The distinction is exact: an intersection on a speed-time graph means equal speeds. An intersection on a shared position-time graph means equal positions. In a straight-road model, equal positions at the same time establish a meeting; equal speeds do not.
Curved Distance Graphs and Tangents
When a distance graph is curved, the rate changes. A chord between two curve points gives an average gradient over an interval. A tangent gives a local gradient estimate at one point. The point of contact determines which instant is being investigated.
For example, suppose a tangent to an accumulated-distance graph at a chosen instant passes through (1 s, 0 m) and (3 s, 16 m). Its gradient is 16/2 = 8 m/s. These are points on the tangent line; they need not be actual positions occupied by the traveller at times 1 and 3 seconds. Extending a tangent is a mathematical reading method, not an additional journey prediction.
The more detailed graph-reading method is in Graphical Solutions, Intersections and Tangent Gradients. For motion questions, attach the correct units and distinguish a local rate from a whole-journey average.
Independent Practice: Identify the Graph Quantity First
All moving stages below use the stated straight-line graph models. Show the distance, duration or area relationship before calculating. No official examination marks or time limits are implied by this original practice set.
1. A straight distance-time segment runs from (0 min, 0 km) to (15 min, 4.5 km). Find the speed in km/h.
2. Along one straight road, a traveller’s distance from home falls from 8 km to 2 km in twenty minutes at constant speed. Find the distance travelled and the speed.
3. Speed rises uniformly from zero to 15 m/s in six seconds. Find distance travelled and the speed-time gradient.
4. A speed-time graph is horizontal at 18 m/s for twenty-five seconds. Find the distance. Explain why a horizontal line does not represent a stop here.
5. Speed changes uniformly from 4 m/s to 12 m/s over nine seconds. Find the distance.
6. A speed-time graph joins (0, 0), (5, 10), (17, 10) and (25, 0), using seconds and m/s. Find total distance, average speed and the magnitude of the final deceleration.
7. Speed rises from zero to V in four seconds, remains at V for eight seconds and falls to zero in six seconds. All sloping sections are straight. Total distance is 156 m. Find V.
8. A traveller accelerates uniformly from rest to 12 m/s in ten seconds, cruises for T seconds and decelerates uniformly to rest in ten seconds. Total distance is 360 m. Find T and complete journey time.
9. A leaves a point at 5 km/h. B follows twelve minutes later at 7.5 km/h on the same route. Find the catch-up time from A’s departure and the distance from the starting point.
10. A journey covers 12 km in thirty minutes of moving time and includes an additional six-minute stop. Find the average speed including the stop and the average while moving.
11. Two speed-time graphs intersect. State exactly what follows, and one conclusion that does not follow without further information.
12. A learner draws a falling segment on a graph labelled total distance travelled. Explain the difficulty and give a different vertical-axis label that could make a falling segment meaningful on a straight-road return trip.
Explained Answers
1. Fifteen minutes is 0.25 hours. Speed = 4.5/0.25 = 18 km/h. The gradient must use hours to produce kilometres per hour.
2. The traveller covers 8 − 2 = 6 km. Speed is 6/(20/60) = 18 km/h. The position gradient is negative, but speed is nonnegative.
3. Distance is the triangular area 1/2 × 6 × 15 = 45 m. Gradient is 15/6 = 2.5 m/s². Different units confirm that these are different quantities.
4. Distance = 18 × 25 = 450 m. The vertical reading is a nonzero constant speed. A stop would lie at speed zero, on the time axis.
5. Trapezium area = 1/2 × (4 + 12) × 9 = 72 m. Using a triangle alone would omit the distance associated with the initial 4 m/s baseline.
6. The areas are 25, 120 and 40 m, giving 185 m. Average speed = 185/25 = 7.4 m/s. Final deceleration magnitude = 10/(25 − 17) = 1.25 m/s².
7. Total area is 2V + 8V + 3V = 13V. Thus V = 156/13 = 12 m/s. Check the separate distances: 24 + 96 + 36 = 156 m.
8. The two triangles contribute 60 m each. Therefore 120 + 12T = 360, giving T = 20 seconds. Complete journey time is 40 seconds, not twenty seconds.
9. Twelve minutes is 0.2 hours. Solve 5t = 7.5(t − 0.2) to get t = 0.6 hours, or 36 minutes. The distance is 5 × 0.6 = 3 km. B travels for twenty-four minutes.
10. Including the stop, elapsed time is 0.6 hours and average speed is 20 km/h. While moving, it is 12/0.5 = 24 km/h. State which denominator answers the requested question.
11. The travellers have the same speed at the intersection time. They need not have the same position or total distance travelled. Those depend on starting positions and accumulated motion.
12. Total distance travelled cannot decrease. The label “distance from home” could allow a falling segment during movement back along the same road towards home.
How to Correct a Motion-Graph Error
Locate the first disagreement between the graph and its interpretation. If the student uses height for distance on a speed graph, return to the axis meaning and rectangle area. If the area is correct but the answer is sixty times too large, inspect minutes versus hours. If a meeting time falls after a stage has ended, inspect the formula’s interval rather than changing the algebra blindly.
A useful corrected solution contains the graph quantity, the selected interval, the mathematical operation and the contextual answer. Then change one condition: include a stop, delay one traveller, replace a triangular acceleration stage with a nonzero starting speed, or ask for moving-time average instead of whole-journey average. The learner must notice what changes before recalculating.
Questions Students and Parents Often Ask
Does every downward line mean slowing down? No. A downward distance-from-home line can represent constant-speed travel towards home. A downward speed-time line represents decreasing speed. The vertical label decides.
Should stops be included in average speed? Include them when the question asks for the entire journey from departure to arrival. Exclude them only when the question explicitly asks for a moving-time average or another narrower interval.
Why not keep trying one equation until it gives a reasonable meeting time? Different stages can have different formulas. A valid answer must satisfy both the equation and its time interval. Changing a formula without reference to the journey can hide the original modelling error.
What should a parent ask before supplying a method? Ask what the vertical axis measures, which interval is being used and whether the answer requires height, gradient or area. These three questions expose the decision without doing the calculation for the learner.
Continue the Secondary 3 Learning Route
Use Probability with Unknown Quantities and Changing Sample Spaces when changing conditions alter a denominator; Missing Values, Combined Means and Statistical Data Corrections when totals and group sizes control an average; and Geometric Algebra, Constraints and Maximum-Area Problems when a valid formula must also obey a geometric domain.
A motion graph is secure when the axis meaning, chosen interval, units and final interpretation all describe the same journey. Return to the Secondary Mathematics Hub for the connected learning map.
