Read and sketch straight-line motion graphs
Sketch and interpret displacement-time, distance-time, velocity-time, speed-time and acceleration-time graphs from a particle's motion.
Section 14 · Calculus
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does the gradient of a velocity-time graph represent?
What you will learn
- Sketch each standard straight-line motion graph from a stated motion
- Interpret gradients and areas on motion graphs
- Distinguish signed displacement and velocity from total distance and speed
Translate each motion quantity into its own graph
Put time on the horizontal axis. Displacement and velocity can be positive or negative; distance travelled and speed cannot decrease below zero. Acceleration can have either sign.
The gradient of a displacement-time graph is velocity; the gradient of a distance-time graph is speed; the gradient of a velocity-time graph is acceleration. A flat segment means the graph's quantity is constant.
Use rate and accumulated change to check a sketch
The signed area under a velocity-time graph is displacement change; the area under a speed-time graph is total distance; the signed area under an acceleration-time graph is velocity change. These relationships help check endpoints on displacement, distance and velocity graphs.
For piecewise-constant velocity, velocity-time and speed-time graphs are horizontal segments, while displacement-time and distance-time graphs are straight segments. When velocity changes sign, displacement can turn back but distance still increases.
Pause and explain
A particle has velocity −4 m/s for 2 seconds. What are its displacement change and distance?
Worked example
A particle starts at displacement 0, travels at 3 m/s for 2 seconds, then at −1 m/s for 3 seconds. Sketch its displacement-time, distance-time, velocity-time and speed-time graphs.
Show the worked solution
- At t = 2, displacement and distance are both 3 × 2 = 6 m.
- From t = 2 to 5, displacement changes by −1 × 3 = −3 m, ending at 3 m; distance increases by 1 × 3 = 3 m, ending at 9 m.
- Plot and join the key points: displacement (0,0) → (2,6) → (5,3); distance (0,0) → (2,6) → (5,9). Draw velocity as 3 then −1, and speed as 3 then 1, on the same time intervals.
Answer The four sketches use those points and horizontal rate segments. Displacement change is 3 m; total distance is 9 m.
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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.
A particle starts at displacement 1 m and moves at 2 m/s from t = 0 to 3 s, then stays at rest until t = 5 s. Sketch its displacement-time and distance-time graphs.
Give me a hint
Use velocity as the displacement graph's gradient; distance follows speed and cannot decrease.
Compare my reasoning
- At t = 3, displacement is 1 + 2 × 3 = 7 m and distance travelled is 6 m.
- At t = 5, both graphs remain at their t = 3 values because the particle is at rest.
- Plot displacement (0,1) → (3,7) → (5,7) and distance (0,0) → (3,6) → (5,6), joining with straight segments.
The displacement graph rises from 1 to 7 m then stays flat; distance rises from 0 to 6 m then stays flat.
Look for these in your work
- Starting displacement kept separate from distance
- Gradient matches velocity
- Rest shown by a horizontal segment
A particle moves at 4 m/s for 2 seconds, then at −2 m/s for 3 seconds. Sketch its velocity-time and speed-time graphs, then find displacement change and distance.
Give me a hint
Velocity keeps its sign; speed is its non-negative magnitude.
Compare my reasoning
- Draw velocity 4 from t = 0 to 2, then −2 from t = 2 to 5.
- Draw speed 4 from t = 0 to 2, then 2 from t = 2 to 5.
- Signed displacement is 4 × 2 − 2 × 3 = 2 m; distance is 4 × 2 + 2 × 3 = 14 m.
Velocity steps from 4 to −2 m/s; speed steps from 4 to 2 m/s. Displacement change 2 m; distance 14 m.
Look for these in your work
- Both time intervals graphed
- Velocity sign retained
- Signed and unsigned areas distinguished
A particle starts at s = 0 with velocity 4 m/s and constant acceleration −2 m/s² for 3 seconds. Sketch acceleration-time, velocity-time and displacement-time graphs; find the final velocity and displacement.
Give me a hint
Use a(t) to find v(t), then use v(t) for the displacement gradient or signed graph area.
Compare my reasoning
- Acceleration is the horizontal line a = −2; velocity is v = 4 − 2t, a straight line from 4 to −2 m/s.
- Integrating velocity gives s = 4t − t², a concave-down curve through (0,0), (2,4) and (3,3).
- The final velocity is −2 m/s and displacement is 3 m; the velocity graph's positive and negative areas are 4 m and −1 m.
Sketch a = −2, v = 4 − 2t and s = 4t − t² on 0 ≤ t ≤ 3. Final velocity −2 m/s; displacement change 3 m.
Look for these in your work
- Acceleration sign and duration shown
- Velocity gradient matches acceleration
- Displacement curve matches velocity and signed area
Common mistakes
- Taking absolute values when the question asks for displacement
- Allowing negative speed
- Using a graph gradient when an area is required
- Forgetting that acceleration-time area is velocity change
What can you explain now?
What does the area under a velocity-time graph give?
Compare with the explanation
Displacement change
Signed positive and negative regions contribute to net displacement.
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Make it stick
Which motion graphs use gradients for rates, and which use areas for changes or totals?
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