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Cambridge IGCSE Mathematics — 0580 Extended

Topic 2.9: Algebra and graphs — Graphs in Practical Situations

Conversion graphs

A conversion graph is a straight line linking two units (currency, \(\mathrm{^\circ C}\) to \(\mathrm{^\circ F}\), and so on). Read a value on one axis, go across to the line, then down or across to the other axis. If the value sits between grid lines, interpolate.

Method

  1. Find the given number on its axis.
  2. Draw a line to the graph, then to the other axis.
  3. Read the scale carefully — check what each small division is worth.

Use the \(\mathrm{^\circ C}\)–\(\mathrm{^\circ F}\) graph to convert \(25\mathrm{^\circ C}\).

Conversion graph of Celsius to Fahrenheit with 25 degrees Celsius reading 77 degrees Fahrenheit
\(25\mathrm{^\circ C}\) sits halfway from \(20\) to \(30\). Read off \(77\mathrm{^\circ F}\).

Distance–time graphs

On a distance–time graph the gradient is speed. A horizontal section means the object is stationary (gradient \(0\)). A return journey slopes down if distance is measured from the start.

A cyclist rides \(80\,\mathrm{km}\) in \(2\,\mathrm{h}\), rests for \(1\,\mathrm{h}\), then returns the \(80\,\mathrm{km}\) in \(2\,\mathrm{h}\). Sketch the graph and find each speed.

Distance-time graph: 80 km in 2 hours, 1 hour rest, return 80 km in 2 hours
Outward speed \(80/2 = 40\,\mathrm{km/h}\). Rest from \(2\,\mathrm{h}\) to \(3\,\mathrm{h}\). Return \(80/2 = 40\,\mathrm{km/h}\).

Speed–time graphs

On a speed–time graph the gradient is acceleration. For linear sections only, the area under the graph is distance.

Method

  1. Split the graph into triangles and rectangles (linear pieces only).
  2. Distance \(=\) sum of those areas, using the units on the axes.
  3. Acceleration \(=\) change in speed \(\div\) time for that section. Deceleration is negative acceleration.

A particle accelerates from \(0\) to \(10\,\mathrm{m/s}\) in \(2\,\mathrm{s}\), stays at \(10\,\mathrm{m/s}\) for \(3\,\mathrm{s}\), then decelerates to rest in \(2\,\mathrm{s}\). Find the distance and the accelerations.

Speed-time graph from 0 to 10 metres per second, constant, then back to 0, area 50 metres
Distance \(\tfrac{1}{2}\times 2\times 10 + 3\times 10 + \tfrac{1}{2}\times 2\times 10 = 10 + 30 + 10 = 50\,\mathrm{m}\). First \(2\,\mathrm{s}\): acceleration \(5\,\mathrm{m/s}^2\). Last \(2\,\mathrm{s}\): \(-5\,\mathrm{m/s}^2\).

Tangents and instantaneous rate

If the graph is a curve, a chord gives average rate. A tangent at a point estimates the instantaneous rate of change (the gradient of that tangent).

The distance–time curve is \(s = t^2\). Estimate the speed at \(t = 3\,\mathrm{s}\) from the tangent.

Tangent to s equals t squared at t equals 3 seconds with gradient 6 metres per second
Rise \(6\,\mathrm{m}\) over run \(1\,\mathrm{s}\). Instantaneous speed \(\approx 6\,\mathrm{m/s}\).

Paper 2 (non-calculator)

Keep units consistent: \(\mathrm{km/h}\) is not \(\mathrm{m/s}\). Convert mixed times before using a gradient — \(2\,\mathrm{h}\,30\,\mathrm{min} = 2.5\,\mathrm{h}\), not \(2.3\,\mathrm{h}\).

Try this

From the \(50\,\mathrm{m}\) speed–time example, what distance is covered in the first \(2\,\mathrm{s}\)?

Show answer
Answer

  1. The first \(2\,\mathrm{s}\) is a triangle of base \(2\,\mathrm{s}\) and height \(10\,\mathrm{m/s}\)

    \[ \text{distance} = \tfrac{1}{2} \times 2 \times 10 \]

  2. Final answer

    \[ 10\,\mathrm{m} \]

Exam Traps

  • Area under a speed–time graph is distance, not speed.
  • Gradient of a distance–time graph is speed, not acceleration. Acceleration is the gradient of a speed–time graph.

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