Real graphs tell stories: how many kilometres in a mile, when a cyclist stopped for a rest, how hard a car braked. Read the gradient and the area and the story falls out.
The gradient of a graph is the rate of change of the vertical quantity per unit of the horizontal one. Distance-time: gradient = speed. Speed-time (Extended): gradient = acceleration, and area under the graph = distance travelled.
Fig. 1: a cyclist rides 12 km in 40 minutes, rests for 20 minutes, then rides home in 30 minutes
Section 01
Conversion graphs.
In the lab's conversion tab, choose a pair of units (miles and kilometres, pounds and dollars, Celsius and Fahrenheit). Slide along one axis and read across to the other. Notice which lines go through the origin and which do not.
Worked example Miles and kilometres
A conversion graph is a straight line through \((0, 0)\) and (10 miles, 16 km). (a) Convert 25 miles to km. (b) Convert 56 km to miles. (c) What does the gradient represent?
Gradient \(= \dfrac{16}{10} = 1.6\) km per mile, so km \(= 1.6 \times\) miles.
(a) \(25 \times 1.6 = 40\) km.
(b) \(56 \div 1.6 = 35\) miles.
(c) The gradient is the number of kilometres in one mile.
Answer: (a) 40 km (b) 35 miles (c) 1.6 km per mile
Worked example A line that misses the origin
Fahrenheit and Celsius are linked by \(F = 1.8C + 32\). Convert 20 °C to °F and explain why the graph does not pass through the origin.
\(F = 1.8 \times 20 + 32 = 36 + 32 = 68\).
At \(C = 0\), \(F = 32\), so the line crosses the \(F\)-axis at 32, not at 0.
Answer: 68 °F; 0 °C is 32 °F, so the quantities are not in direct proportion.
Examiner trap
Read the scale on each axis before you read off a value. On a graph where one small square is 2 units, a point three squares along is 6, not 3.
Section 02
Distance-time graphs.
In the distance-time tab, build a journey from stages and watch the graph draw itself. Make one stage steeper and the speed goes up; make it flat and the traveller has stopped. The cyclist in Fig. 1 is a typical exam journey.
Line going down: travelling back towards the start. The speed is still positive.
Average speed \(= \dfrac{\text{total distance}}{\text{total time}}\), including any stops.
Practise reading the scale: the graph scales trainer has you work out one small square, plot values and read points back, on printed scales like the exam grids.
Worked example Speed of one stage
Use Fig. 1 to find the cyclist's speed, in km/h, for the first 40 minutes.
Distance 12 km, time 40 min \(= \dfrac{40}{60} = \dfrac{2}{3}\) h.
Speed \(= 12 \div \dfrac{2}{3} = 18\) km/h.
Answer: 18 km/h
Worked example Average speed
Find the average speed for the whole journey in Fig. 1.
Total distance: 12 km out and 12 km back, so 24 km.
Total time: 90 min \(= 1.5\) h, including the 20-minute stop.
Average speed \(= 24 \div 1.5 = 16\) km/h.
Answer: 16 km/h
Your turn
A train travels 45 km in 30 minutes, stops for 15 minutes, then travels a further 30 km in 20 minutes. Find the speed on each moving stage and the average speed for the whole journey.
Stage 1: \(45 \div 0.5 = 90\) km/h. Stage 3: \(30 \div \dfrac{1}{3} = 90\) km/h. Average: \(75\) km in 65 min, \(75 \div \dfrac{65}{60} = 69.2\) km/h (3 s.f.).
Examiner trap
Time read in minutes must be changed to hours for km/h: 40 minutes is \(\dfrac{2}{3}\) h, not 0.4 h. And the average speed is NOT the mean of the stage speeds: \(\dfrac{18 + 24}{2} = 21\) is wrong because the stages last different times and the stop counts.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
10 miles = 16 km. Use this conversion to change 45 miles to kilometres.
Quick check+10 XP
On a distance-time graph, what does a horizontal line show?
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On a distance-time graph, what does the gradient represent?
Quick check+10 XP
(Extended) What does the area under a speed-time graph represent?
Section 03
Speed-time graphs.
(Extended) Open the speed-time tab and drag the corners of the graph. Watch the acceleration (the gradient) change, and the shaded area, which is the distance travelled, grow and shrink.
Gradient = acceleration in m/s2. A negative gradient is a deceleration.
Horizontal line: constant speed (zero acceleration), NOT stopped.
Area under the graph = distance travelled. Split it into triangles, rectangles and trapezia.
Worked example A car journey (Extended)
A car starts from rest and accelerates uniformly to 20 m/s in 8 s. It travels at 20 m/s for 12 s, then decelerates uniformly to rest in 5 s. Find (a) the acceleration, (b) the deceleration, (c) the total distance, (d) the average speed.
(a) Gradient \(= \dfrac{20}{8} = 2.5\) m/s2.
(b) Gradient of the last part \(= \dfrac{20}{5} = 4\) m/s2 (deceleration).
Check which graph you have. On a distance-time graph the gradient is speed; on a speed-time graph the gradient is acceleration and a flat line means steady speed. Using the wrong meaning is the commonest lost mark here.
Section 04
Curves and tangents.
(Extended) When the graph is curved, the rate of change is different at every point. Draw a tangent (a straight line that just touches the curve) at the point you want and find the gradient of the tangent.
Worked example Speed at an instant (Extended)
A ball rolls so that its distance is \(d = t^2\) metres after \(t\) seconds. A tangent drawn to the curve at \(t = 4\) passes through \((2, 0)\) and \((6, 32)\). Estimate the speed of the ball at \(t = 4\).
Use two points on the TANGENT, as far apart as possible: \((2, 0)\) and \((6, 32)\).
Do not use two points on the curve: that gives the AVERAGE speed between them, not the speed at one instant. Here the average speed from 0 to 4 s is \(16 \div 4 = 4\) m/s, only half the speed at \(t = 4\).
Stage 2 · Exam
Exam-style questions
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A cyclist rides 12 km in 40 minutes, stops for 20 minutes, then rides 12 km home in 30 minutes. Find the average speed for the whole journey.
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(Extended) A car accelerates uniformly from rest to 20 m/s in 8 s, travels at 20 m/s for 12 s, then decelerates uniformly to rest in 5 s. Find the total distance travelled.
Exam style+20 XP
(Extended) A train slows uniformly from 24 m/s to 6 m/s in 12 s. Find its deceleration.
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(Extended) A bus travels at 15 m/s for \(T\) seconds, then decelerates uniformly to rest in 6 s. The total distance is 345 m. Find \(T\).
Skill unlocked
Travel graphs, read and drawn.
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