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IGCSE · Cambridge (CIE) · 0654

Plot distance-time graphs and calculate straight gradients

Read both axes and use a large gradient triangle to find a speed.

Section P1.2 · Motion

Motion · Cambridge IGCSE 0654 · P1.2

Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.

Before you begin

  • Plot ordered coordinate pairs.
  • Subtract two times and two distances.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which quantity belongs on the horizontal axis of a distance–time graph?

What you will learn

  • Plot a distance–time record.
  • Interpret rest and constant speed using gradient.
Cumulative distance in metres against time in seconds, with straight sections through (0,0), (4,12), (6,12) and (10,20). The gradients are 3, 0 and 2 m/s. The horizontal middle section means rest. Axis values are quantitative; straight intervening motion is the stated model.Distance gradient gives speedDistance / m0102004610Time / sGradients: 3, 0, 2 m/sHorizontal distance: at restDistance continues after the stopStraight sections are the stated model
Cumulative distance in metres against time in seconds, with straight sections through (0,0), (4,12), (6,12) and (10,20). The gradients are 3, 0 and 2 m/s. The horizontal middle section means rest. Axis values are quantitative; straight intervening motion is the stated model.

Plot the quantities rather than square counts

Put time on the horizontal axis and cumulative distance travelled on the vertical axis, labelled with units. Data (0 s, 0 m), (4 s, 12 m), (6 s, 12 m), (10 s, 20 m) give three straight sections if the motion is specified as constant within each interval. Read the values represented by each grid interval on both axes.

The first line rises at a constant rate, the middle is horizontal and the last rises less steeply. A horizontal cumulative-distance line means no further distance is travelled, so the object is at rest. A straight rising line means constant speed, not increasing speed merely because distance keeps growing.

A gradient is a rate of distance change

For a straight section, speed equals change in distance divided by change in time. From four to six seconds the numerator is zero. From six to ten seconds it is (20 − 12)/(10 − 6) = 2 m/s. Use endpoints on the same straight section; spanning a stop gives an interval average rather than that section's speed.

Choose a large triangle on a straight graph to reduce the relative effect of small reading differences. Divide distance values by time values, not vertical squares by horizontal squares unless the axis scales happen to make that ratio meaningful. Cumulative distance travelled cannot decrease; a falling line would need a different vertical quantity, such as signed position.

Sources for this lesson

Pause and explain

What does the 4–6 s horizontal distance–time section mean?

Put the idea to work

Worked example

For (0,0), (4,12), (6,12), (10,20), find each straight-section speed.

Show the worked solution
  1. 0–4 s: (12 − 0)/(4 − 0) = 3 m/s.
  2. 4–6 s: no distance increase, so 0 m/s.
  3. 6–10 s: (20 − 12)/(10 − 6) = 2 m/s.

Answer 3 m/s; 0 m/s; 2 m/s.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Plot the given four points (0,0), (4,12), (6,12), (10,20) in s and m, with straight sections. Find the speeds.

Give me a hint

Calculate the gradient of each section separately.

Compare my reasoning
  1. Label time in s and distance in m.
  2. Find changes 12/4, 0/2 and 8/4.
  3. The section speeds are 3, 0 and 2 m/s.

3, 0 and 2 m/s.

Look for these in your work

  • Label axes and plot correctly.
  • Use differences within each section.
2 · Independent

A straight distance–time section runs from (5 s, 18 m) to (9 s, 30 m). Find speed.

Give me a hint

Subtract both coordinates before dividing.

Compare my reasoning
  1. Distance change = 30 − 18 = 12 m.
  2. Time change = 9 − 5 = 4 s.
  3. Speed = 12/4 = 3 m/s.

3 m/s.

Look for these in your work

  • Subtract the initial distance.
  • Subtract the initial time.
3 · Transfer

A student says a straight rising distance–time line means acceleration. Explain the mistake.

Give me a hint

Compare gradients at different times on the straight line.

Compare my reasoning
  1. The gradient is the same everywhere on the section.
  2. Equal intervals of time add equal distances.
  3. The speed is constant; a changing speed needs a changing gradient.

A straight rising line has constant speed, not increasing speed.

Look for these in your work

  • Link slope to speed.
  • Distinguish distance increase from speed increase.

Common mistakes

  • A horizontal distance line is rest, not constant non-zero speed.
  • Use changes in axis values.
  • A cumulative-distance graph cannot fall.
Recall without your notes

What can you explain now?

Distance rises from 12 m at 6 s to 20 m at 10 s. What is speed?

Compare with the explanation

2 m/s.

The gradient is 8 m over four seconds.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Draw one journey with a stop and two different constant speeds, then label each gradient.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Sciences — Co-ordinated (Double)

Course scope and assessment

This 2028–2029 pathway maps all 152 detailed subject-content sections in the published 0654 syllabus to original teaching and staged practice. It adapts teaching from the related single-science pathways and excludes clear 0610-only topics such as classification systems, human excretion, tropic responses, food supply, population dynamics, detailed pollution and biotechnology. Some explanations may include wider single-science context. A 0654-specific objective-by-objective review, Core/Supplement audit, human subject-teacher review, supervised laboratory experience and cumulative assessment remain in progress. Written practice is self-checked and does not predict an official grade. The 2025–2027 syllabus has not been reconciled here; choose the 2028–2029 syllabus only.

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