What this covers
Linear Motion Graphs is part of the Force and Motion I chapter. We teach it the way it is tested: the concept in plain English, then a worked example, then a question the student tries while the teacher checks the method.
How it is examined
It can appear in Paper 1 (objective) and Paper 2 (structured), and where an experiment applies, in Paper 3. We do not predict which questions appear; we prepare the technique for all of them.
A common mistake
How to study it
Learn the definition precisely, practise one or two SPM-style questions with full working, and link it to the rest of Force and Motion I. If it keeps costing marks, a one-to-one lesson fixes exactly that.
Force and Motion I · Formulas · Exam Papers
What you need to know
Motion can be represented graphically using two key graphs. On a displacement-time graph, the gradient at any point gives the velocity at that instant, since velocity is the rate of change of displacement.
A straight line means constant velocity, a horizontal line means the object is at rest, and a curved line means the velocity is changing. The steeper the gradient, the greater the magnitude of the velocity, and a negative gradient means the object is moving back towards its starting point.
On a velocity-time graph, the gradient at any point gives the acceleration, since acceleration is the rate of change of velocity; a horizontal line means zero acceleration, that is, constant velocity. The area under a velocity-time graph, taken between the graph line and the time axis, gives the displacement travelled over that time interval.
For a simple case where velocity increases uniformly from 0 to 12 m s⁻¹ over 4 s, the area is a triangle: displacement = ½ × base × height = ½ × 4 s × 12 m s⁻¹ = 24 m.
Worked example
Question: A velocity-time graph shows an object accelerating uniformly from 4 m s⁻¹ to 16 m s⁻¹ over 6 s, then travelling at a constant 16 m s⁻¹ for a further 5 s. Find the acceleration during the first stage and the total displacement.
Acceleration during the first stage: a = (v − u) / t = (16 − 4) m s⁻¹ ÷ 6 s = 12 m s⁻¹ ÷ 6 s = 2 m s⁻². The area under the graph during the first stage is a trapezium: displacement₁ = ½ × (4 m s⁻¹ + 16 m s⁻¹) × 6 s = ½ × 20 m s⁻¹ × 6 s = 60 m.
The area during the second stage is a rectangle: displacement₂ = 16 m s⁻¹ × 5 s = 80 m. Total displacement = displacement₁ + displacement₂ = 60 m + 80 m = 140 m.
This shows how splitting a velocity-time graph into simple shapes makes the area, and therefore the displacement, straightforward to calculate.
How it is examined
Paper 1 objective items commonly show a displacement-time or velocity-time graph and ask candidates to read off the gradient, describe the motion in a particular section, or identify which section shows acceleration, deceleration or rest. Paper 2 structured questions frequently require candidates to calculate a gradient or an area under a graph with full working and units, or to sketch a graph representing a described motion.
Paper 3 practical work may involve plotting a displacement-time or velocity-time graph from measured data and using the gradient to find velocity or acceleration.
Common mistakes include confusing the gradient of a displacement-time graph with the gradient of a velocity-time graph, calculating the area under a displacement-time graph when it has no physical meaning, forgetting to apply a trapezium or triangle area formula correctly, and leaving out units on the gradient or the area value. Sketches must also show the correct shape, starting point and any changes in slope, since a rough sketch without these details loses marks.
Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)