Real-Life and Travel Graphs

1. Quantities, scales and contextual models

Put the input (often time) horizontally and label both axes with quantities and units. Choose equal steps on each axis that fit all the data and use a useful amount of the grid; the two axes can use different scales. Plot ordered pairs, then use straight sections only where the rate is constant. Use a smooth curve where the model changes smoothly. Do not extend a model beyond a sensible range without justification.

Worked example — time units

A time reading of 1.5 hours is 1 hour 30 minutes: 0.5 × 60 = 30. It is not 1 hour 50 minutes. Similarly, 20 minutes = 20 ÷ 60 = ⅓ hour. If a graph shows 4 km travelled in those 20 minutes, speed = 4 ÷ ⅓ = 12 km/h.

Conversion graph

Using the approximate conversion 5 miles ≈ 8 km, plot (0, 0), (5, 8) and (10, 16) and join them with a straight line. Read 7.5 miles ≈ 12 km, or work backwards from 12 km to 7.5 miles. This model passes through the origin: zero distance in one unit is zero in the other.

Fixed-charge tariff — C = 3 + 2d

Here d is distance in kilometres and C is cost in pounds. The vertical intercept £3 is the starting charge; the gradient £2/km adds £2 for each kilometre. For 4 km, C = 3 + 2 × 4 = £11. A £13 fare corresponds to d = (13 − 3) ÷ 2 = 5 km. The cost is not directly proportional to distance because there is a fixed charge.

2. A journey with a stop and return

From 0–20 min the traveller moves 4 km away from home; from 20–30 min the horizontal section is a 10-minute stop; from 30–60 min the traveller covers 4 km back along the same road.

Outward speed: 4 ÷ (20/60) = 12 km/h. Return speed: 4 ÷ (30/60) = 8 km/h. The return gradient is −8 km/h, but speed is the non-negative magnitude, 8 km/h.

Overall average speed: total distance ÷ total elapsed time = (4 + 4) ÷ 1 = 8 km/h, including the stop. The average while moving is 8 ÷ (50/60) = 9.6 km/h. Neither is the simple mean (12 + 8) ÷ 2 = 10: the two moving speeds lasted for different times.

Choose the right meaning: distance from home can decrease on a return. Cumulative distance travelled only stays level or increases; it finishes at 8 km. Displacement is signed position relative to a chosen origin and direction. On this straight road, taking away from home as positive, displacement finishes at 0 km. Distance from home is never negative; displacement can be negative if the traveller goes beyond the origin in the opposite direction.

The graph is not a drawing of the road. A falling line means the plotted position is decreasing, not that the road goes downhill. A flat position graph means stopped in these straight-line journeys; it does not generally prove rest if all you know is distance from a point and someone could move around that point.

3. When do two travellers meet?

A leaves town at 09:00 at 6 km/h. B starts 12 km from town at 09:00 and travels towards it at 12 km/h on the same road. Their lines intersect at about (40 min, 4 km): they meet at 09:40, 4 km from town.

Check the graph: in 40/60 = ⅔ hour, A covers 6 × ⅔ = 4 km and B covers 12 × ⅔ = 8 km, so 4 + 8 = 12 km. Both coordinates matter. A crossing means a meeting only when the vertical axis is position measured from the same place and the travellers are on the same route. Crossings of cumulative-distance graphs need not be meetings.

4. Speed–time and velocity–time graphs

To construct this speed graph, plot (0, 0), (4, 6), (10, 6), (14, 0) and join successive points by straight lines because each stage has constant acceleration or deceleration.

At t = 8 s, the height gives 6 m/s. The flat section from 4–10 s means constant speed, not a stop. At 12 s the speed is 3 m/s; the object stops at 14 s.

Acceleration, 0–4 s: (6 − 0) ÷ 4 = 1.5 m/s². From 10–14 s, the speed changes at (0 − 6) ÷ 4 = −1.5 m/s², so the deceleration has magnitude 1.5 m/s². One m/s² means the velocity changes by one m/s each second.

Velocity includes direction. At 5 s, velocity is −2 m/s: speed is 2 m/s in the negative direction. From 2–6 s the velocity gradient is (−4 − 4) ÷ (6 − 2) = −2 m/s². The object slows while velocity is positive, reaches zero at 4 s, reverses, then speeds up in the negative direction. A negative acceleration does not always mean slowing down.

To construct a velocity graph, include a scale below zero when motion reverses. A flat velocity line means constant velocity; a line on zero means rest. Height gives speed or velocity on these graphs, while gradient gives acceleration. Area applications come next in Areas Under Graphs and Accumulated Change (Higher).

5. A non-standard context: water height

Read this graph from its labels rather than expecting a familiar shape. At 1 minute the height is about 6 cm; at 2 minutes it is about 8 cm. The water keeps getting higher, but each extra centimetre takes longer.

If equal volumes enter each minute and there is no leak, the increasingly shallow gradient indicates a wider horizontal cross-section higher up: the same water must spread over more area. A straight-sided container with constant cross-sectional area would instead give a straight height–time line. The height graph alone does not tell you the volume added unless you know the container’s shape.

6. Higher — apply chords and tangents to motion

Use the chord/tangent construction from Rates of Change. Here the task is to interpret the resulting rate and its units: a distance–time gradient gives speed; a displacement–time gradient gives signed velocity; a velocity–time gradient gives acceleration.

Estimate a speed at one instant

For this forward-moving object, the tangent at 4 s passes approximately through A(2, 0) and B(6, 32). Speed ≈ (32 − 0) ÷ (6 − 2) = 8 m/s. These are points on the tangent, not two arbitrary points on the curve. The distance at 4 s is 16 m; that height is not its speed.

Over 0–4 s, the chord instead gives average speed = (16 − 0) ÷ 4 = 4 m/s. The instantaneous speed at the end is greater because the curve is getting steeper.

Estimate an acceleration at one instant

At 4 s the velocity is 8 m/s. Using the tangent points A(2, 0) and B(6, 16), acceleration ≈ (16 − 0) ÷ (6 − 2) = 4 m/s². A curve on a velocity graph means changing acceleration; it does not give acceleration directly by its height.

Graph readings and drawn tangents are estimates. For these diagrams, sensible estimates near 8 m/s and 4 m/s² are acceptable; keep units and show the triangle used.

Think Like a Mathematician

Two cars have the same distance-from-start reading. Must they have the same speed?

No. They have the same position reading, but their graph gradients may differ. In the meeting graph, A and B meet with speeds 6 km/h and 12 km/h. To compare speeds, compare the magnitudes of their gradients.

Common Mistakes

Reading height as speed: height on a distance graph is a distance; speed comes from its gradient.

Treating every flat section as a stop: a flat distance/position graph is a stop for straight-line travel; a flat speed graph above zero is steady motion.

Calling −2 m/s a negative speed: it is velocity; the speed is 2 m/s.

Making cumulative distance fall: returning adds to total distance travelled.

Drawing the physical route: graph shape describes the relationship between the labelled quantities.

Averaging speeds without time weights: use total distance and the time requested, including stops for an overall journey average.

Practice — read, calculate and explain

For readings from the supplied graphs, accept the stated reasonable ranges. Exact calculations from given coordinates should use those coordinates.

1. A graph records a 90 km journey lasting 1 hour 15 minutes. Convert the time to hours and calculate average speed.

2. Draw a distance-from-start graph for 3 km travelled in 15 min, a 10 min stop, then a 15 min return along the same road. Label both axes and choose useful scales.

3. Use the section 2 journey. Find the distance from home at 45 min and the average speed for the whole journey.

4. Use the meeting graph. State when and where the travellers meet. Does a crossing of cumulative-distance graphs always mean a meeting?

5. Use C = 3 + 2d. Find the fare for 7 km and the journey distance for a £19 fare. Explain the 3.

6. Read the section 4 speed graph at 12 s. Calculate its acceleration from 0–4 s. What does the flat section mean?

7. Use the signed velocity graph at 5 s. Give velocity, speed and acceleration. Is the object slowing down then?

8. Water enters at a constant volume rate. The height graph becomes less steep while still rising. Explain what this suggests about the container.

9. A cyclist rides at 12 km/h for 30 min, then 6 km/h for 60 min. Is the average speed 9 km/h? Explain.

10. Higher: on the distance curve in section 6, draw a tangent at 3 s and estimate the speed. Compare with the average speed from 0–3 s.

11. Higher: use the velocity tangent in section 6 to estimate acceleration at 4 s. Which two quantities determine its units?

12. Higher: on the distance curve d = t², find the chord gradient from 2 s to 4 s and interpret it.

Frequently Asked Questions

Can a distance–time graph slope downwards?

A distance-from-start graph can; a cumulative-distance graph cannot. Always check the exact vertical label. A displacement graph can also go below zero because it records direction.

Which parts are Higher?

The curved-motion chord and tangent applications in section 6 and practice 10–12 are Higher. Reading travel graphs and using constant straight-line gradients are shared Foundation and Higher skills.

Where do graph families and intersections fit?

See Quadratic Graphs and Graphical Solutions and Cubic, Reciprocal and Exponential Graphs for general curve behaviour and graphical solutions. Here we interpret those ideas in context.

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