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Cambridge IGCSE0580

Graphs in practical situations

Mathematics 0580 Chapter Notes

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Graphs in practical situations
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In this chapter

  1. 1Reading a travel graph
  2. 2Gradient as a rate of change
  3. 3Drawing graphs from data
  4. 4Gradient of a tangent to a curve
  5. 5Acceleration and deceleration
  6. 6Distance from a speed–time graph

1. Reading a travel graph

A travel graph plots distance against time. A horizontal line means the object is at rest. A steeper line means a faster speed.

The gradient of a straight section is the speed. Read times from the time axis and take care with the scale.

Times are in hours and minutes, so 1 hour 30 minutes is 1.5 hours, not 1.3.

speed = distance/time

Examiner insight

“Some candidates tried to divide by an actual time, rather than the travelling time.”

Worked example 15 marks

Do not use a calculator. A cyclist leaves home at 09 00 and cycles 12 km to a lake in 30 minutes. She rests at the lake for 30 minutes. She then cycles home, arriving at 11 00.

(a) Work out her speed, in km / h, on the way to the lake.

(b) Work out her average speed for the whole journey.

  1. 1

    (a) speed = distance ÷ time, with 30 minutes = 0.5 hours [1]

  2. 2

    12 ÷ 0.5 = 24 km/h [1]

  3. 3

    (b) total distance = 12 + 12 = 24 km [1]

  4. 4

    total time = 2 hours, as the rest counts [1]

  5. 5

    average speed = 24 ÷ 2 = 12 km/h [1]

2. Gradient as a rate of change

The gradient of a straight-line graph is a rate of change. It tells you how much the vertical quantity changes for each 1 unit of the horizontal quantity.

The unit of the gradient is the vertical unit per the horizontal unit. On a conversion graph the gradient is the conversion rate, such as dollars per euro.

A steeper line means a greater rate of change.

gradient = (change in y)/(change in x)

Worked example 13 marks

Do not use a calculator. A water tank fills at a constant rate. After 2 minutes it holds 30 litres. After 8 minutes it holds 90 litres. Find the gradient of the straight-line graph of volume against time and say what it means.

  1. 1

    gradient = (90 - 30)/(8 - 2) [1]

  2. 2

    = 60/6 = 10 [1]

  3. 3

    10 litres per minute: the rate at which the tank fills [1]

3. Drawing graphs from data

To draw a graph from data, plot each pair of values as a point, then join the points with a ruled line.

For a travel graph, a rest is a flat section: time goes on, distance stays the same.

A conversion graph is a straight line through (0, 0). Plot (0, 0) and one other point from the rate, then join them with a ruled line.

Examiner insight

“However, a significant number did not realise they only needed to plot (0, 0) and (50, 540) to create a fit for purpose conversion graph, and found many intermediate points; some of which were plotted slightly inaccurately so the line became fragmented rather than a smooth straight line.” “A small but significant number read the scale wrong and mis-plotted the first two points.”

Worked example 14 marks

Do not use a calculator. The rate of exchange is 1 euro = 1.10 dollars. A conversion graph of dollars against euros is to be drawn.

(a) Describe how to draw the graph. Give the coordinates of two points on the line.

(b) Work out the number of dollars for 40 euros.

  1. 1

    (a) one point is (0, 0) [1]

  2. 2

    another point: 50 × 1.10 = 55, so (50, 55) [1]

  3. 3

    join with a ruled line [1]

  4. 4

    (b) 40 × 1.10 = 44 dollars [1]

4. Gradient of a tangent to a curve

On a curve the gradient changes. To estimate the gradient at a point, draw a tangent: a ruled line that touches the curve at that point only.

Choose two points on the tangent and find the change in y over the change in x. Do not use points on the curve.

On a speed–time graph this gives an estimate of the acceleration at that time.

gradient of tangent = (y₂ - y₁)/(x₂ - x₁)

Examiner insight

“Some tangents were less accurate and had a gap between the ruled line and the curve, others were drawn at different points slightly off from the time of 7.5 seconds and some did not touch the curve at time 7.5 seconds.” “A common error for the gradient was to give an answer of 0.93 from 7 divided by 7.5, using the coordinate values of the point at 7.5 seconds.”

Worked example 12 marks

Do not use a calculator. A speed–time curve has speed in m / s and time in seconds. A tangent is drawn to the curve at t = 2. The tangent passes through the points (0, 1) and (4, 3). Estimate the acceleration at t = 2. Give the unit.

  1. 1

    gradient of tangent = (3 - 1)/(4 - 0) [1]

  2. 2

    = 0.5 m/s² [1]

5. Acceleration and deceleration

On a speed–time graph the gradient is the acceleration. A rising line is acceleration, a horizontal line is constant speed, and a falling line is deceleration.

A deceleration is a positive number: the acceleration is negative.

On a distance–time graph the gradient is the speed. Both are applications of rate of change.

a = (change in speed)/(change in time)

Examiner insight

“Some showed the division the wrong way round and there were many who gave the answer as the reduction of speed from 18 m/s to 6 m/s = 12 m/s.” “Strictly, as deceleration was asked for rather than acceleration, this should have been a positive value although –1.5 was condoned.”

Worked example 14 marks

Do not use a calculator. A car's speed rises steadily from 6 m / s to 21 m / s in 5 seconds. It later slows steadily from 21 m / s to rest in 6 seconds.

(a) Find the acceleration in the first 5 seconds.

(b) Find the deceleration in the last 6 seconds.

  1. 1

    (a) change in speed = 21 - 6 = 15, so a = 15 ÷ 5 [1]

  2. 2

    = 3 m/s² [1]

  3. 3

    (b) deceleration = 21 ÷ 6 [1]

  4. 4

    = 3.5 m/s² [1]

6. Distance from a speed–time graph

The distance travelled is the area under the speed–time graph. The sections are straight lines, so the area is made of rectangles, triangles and trapeziums.

Split the area into rectangles and triangles, or use one trapezium. Add the areas.

Distance is speed × time only for a constant speed.

distance = area under the graph

trapezium = 1/2(a + b)h

Examiner insight

“Some candidates attempted to use distance = speed × time instead of area under the graph.” “Some candidates correctly found the area under the graph correctly and then omitted to convert from metres to kilometres.”

Worked example 15 marks

The diagram shows the speed–time graph for 65 seconds of a journey.

(a) The acceleration during the first 48 seconds of this part of the journey is 0.25 m / s². Show that V = 32 m / s.

(b) Work out the total distance travelled during the 65 seconds. Give your answer in kilometres.

  1. 1

    (a) 0.25 = (V - 20)/48 [1]

  2. 2

    V - 20 = 0.25 × 48 = 12, so V = 32 [1]

  3. 3

    (b) trapezium: 1/2 × (20 + 32) × 48 = 1248 m [1]

  4. 4

    triangle: 1/2 × (65 - 48) × 32 = 272 m [1]

  5. 5

    total = 1520 m = 1.52 km [1]

Recap

  • Travel graph: distance against time. Gradient = speed. Horizontal = at rest.
  • Gradient of a straight line = change in y ÷ change in x = rate of change, in units per unit.
  • Conversion graph: a ruled line through (0, 0) and one more point.
  • Draw graphs from data by plotting the points and joining with a ruled line.
  • Extended: the gradient of a tangent to a curve estimates the rate of change at a point.
  • Extended: gradient of a speed–time graph = acceleration. Deceleration is positive.
  • Extended: distance travelled = area under the speed–time graph. Speed × time works only for a constant speed.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A cyclist's distance–time graph has a horizontal section lasting 20 minutes. State what this tells you about the cyclist during these 20 minutes.1 mark
  2. Do not use a calculator. On a travel graph a straight line runs from 0 km at 09 00 to 90 km at 10 30. Find the speed, in km / h.2 marks
  3. A conversion graph has pounds (£) on the horizontal axis and dollars on the vertical axis. It is a straight line through (0, 0) and (50, 65). Find the number of dollars for £80.2 marks
  4. Do not use a calculator. A runner runs 300 m in 60 s. She rests for 30 s. She then runs a further 300 m in 120 s. Starting at (0, 0), give the coordinates (time in s, distance in m) of the other points needed to draw her distance–time graph.3 marks
  5. The depth of water in a tank is measured every 5 minutes. The points are plotted on a graph of depth against time. (a) Explain how the table shows that the points lie on one straight line. (b) Find the gradient of the line. Give its unit. | time (minutes) | 0 | 5 | 10 | 15 | | --- | --- | --- | --- | --- | | depth (cm) | 4 | 9 | 14 | 19 |3 marks
  6. Do not use a calculator. A speed–time curve has speed in m / s and time in seconds. A tangent drawn to the curve at t = 3 passes through the points (1, 2) and (5, 12). Estimate the acceleration at t = 3.2 marks
  7. Do not use a calculator. A lorry slows steadily from 28 m / s to 8 m / s in 5 seconds. Find its deceleration.2 marks
  8. Do not use a calculator. The speed of a train rises steadily from 0 to 20 m / s in 10 s, stays at 20 m / s for 20 s, then falls steadily to 0 in a further 10 s. Find the total distance travelled.3 marks
  9. Do not use a calculator. A lift starts from rest at the ground floor. It accelerates steadily for 4 seconds to a speed of v m / s. It stays at v m / s until 16 seconds after the start. It then decelerates steadily and stops 20 seconds after the start. The lift rises 64 m altogether. (a) Find the value of v. (b) Find the deceleration of the lift. (c) Find the height of the lift above the ground floor 18 seconds after the start.6 marks
  10. A tram leaves a stop from rest and accelerates steadily to 15 m / s in 20 seconds. It travels at 15 m / s for T seconds. It then decelerates steadily at 0.6 m / s² and comes to rest at the next stop. The two stops are 1.2 km apart. (a) Find the acceleration in the first 20 seconds. (b) Find the value of T. (c) Find the average speed of the tram between the stops, in m / s.7 marks

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