Two graphs tell the whole story of motion in a straight line. On a displacement-time graph the gradient is velocity; on a velocity-time graph the gradient is acceleration and the area is displacement.
Gradient of a displacement-time graph \(=\) velocity. Gradient of a velocity-time graph \(=\) acceleration. Area between a velocity-time graph and the time axis \(=\) displacement, with area below the axis counting as negative displacement.
Fig. 1: velocity-time graph of a journey (to scale). Gradients give accelerations of 3, 0 and −2 m s⁻²; the shaded area is the 180 m travelled
Section 01
Scalars and vectors.
In the lab, give the particle a positive initial velocity and a negative acceleration, and run it for long enough that it turns round. Watch the velocity-time line cross the axis at the moment the particle stops, and see the displacement-time curve reach its peak at that same moment.
Scalar (size only)
Vector (size and direction)
distance: total length of path
displacement: position relative to the start
speed: rate of covering distance
velocity: rate of change of displacement
acceleration: rate of change of velocity
Motion in Paper 4 is in one dimension, so direction is just a sign: choose a positive direction and stick to it.
"Deceleration" may be used to mean the speed is decreasing; a deceleration of \(2\ \text{m s}^{-2}\) is an acceleration of \(-2\ \text{m s}^{-2}\).
Worked example Distance versus displacement
A particle moves 5 m east and then 8 m west, taking 6.5 s in total. Find its distance travelled, displacement, average speed and average velocity.
Distance \(= 5 + 8 = 13\) m.
Taking east as positive, displacement \(= 5 - 8 = -3\) m, i.e. 3 m west.
Average speed \(= \dfrac{13}{6.5} = 2\ \text{m s}^{-1}\).
Average velocity \(= \dfrac{-3}{6.5} = -0.462\ \text{m s}^{-1}\).
Answer: 13 m; 3 m west; \(2\ \text{m s}^{-1}\); \(0.462\ \text{m s}^{-1}\) west.
Examiner trap
Average speed is total DISTANCE divided by time, not the average of the start and end speeds (that only works for constant acceleration).
Section 02
Displacement-time graphs.
The gradient of a displacement-time graph is the velocity. A straight line means constant velocity, a horizontal line means at rest, and a line sloping down means moving back towards the start (negative velocity).
Worked example Reading a displacement-time graph
A displacement-time graph is made of straight lines from \((0, 0)\) to \((4, 20)\), then to \((6, 20)\), then to \((11, 0)\), with \(t\) in s and \(s\) in m. Describe the motion.
\(0 \le t \le 4\): gradient \(= \dfrac{20}{4} = 5\), so velocity \(5\ \text{m s}^{-1}\).
\(4 \le t \le 6\): gradient 0, so at rest 20 m from the start.
\(6 \le t \le 11\): gradient \(= \dfrac{0 - 20}{5} = -4\), so velocity \(-4\ \text{m s}^{-1}\): back to the start at \(4\ \text{m s}^{-1}\).
Distance travelled \(= 20 + 20 = 40\) m; final displacement 0.
Answer: Velocities 5, 0 and \(-4\ \text{m s}^{-1}\); distance 40 m, displacement 0.
Examiner trap
On a displacement-time graph the HEIGHT is position, not speed. A particle can be far from the start (high on the graph) and still be at rest (flat line).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Which of these is a vector quantity?
Quick check+10 XP
What does the gradient of a displacement-time graph represent?
Quick check+10 XP
A particle moves at a constant \(4\ \text{m s}^{-1}\) for 10 s. What is the area under its velocity-time graph?
Quick check+10 XP
A velocity-time graph is a straight line from \(2\ \text{m s}^{-1}\) at \(t = 0\) to \(14\ \text{m s}^{-1}\) at \(t = 4\). Find the acceleration.
Section 03
Velocity-time graphs.
The gradient of a velocity-time graph is the acceleration, and the area under it is the displacement. Most graph questions reduce to areas of triangles, rectangles and trapezia.
Worked example Reading Fig. 1
Use Fig. 1 to find the acceleration in each stage and the total distance.
Area of the trapezium \(= \tfrac{1}{2}(10 + 20) \times 12 = 180\) m.
Answer: \(3,\ 0,\ -2\ \text{m s}^{-2}\); 180 m.
Worked example Finding an unknown time
A car starts from rest, accelerates uniformly to \(20\ \text{m s}^{-1}\) in 8 s, travels at \(20\ \text{m s}^{-1}\) for \(T\) s, then decelerates uniformly to rest in 5 s. It travels 690 m altogether. Find \(T\).
Area of first triangle \(= \tfrac{1}{2} \times 8 \times 20 = 80\) m.
Rectangle \(= 20T\) m; last triangle \(= \tfrac{1}{2} \times 5 \times 20 = 50\) m.
\(80 + 20T + 50 = 690\), so \(20T = 560\).
Answer: \(T = 28\)
Worked example When the velocity goes negative
A particle's velocity decreases uniformly from \(6\ \text{m s}^{-1}\) at \(t = 0\) to \(-4\ \text{m s}^{-1}\) at \(t = 5\). Find the distance travelled and the displacement in the 5 s.
Acceleration \(= \dfrac{-4 - 6}{5} = -2\ \text{m s}^{-2}\), so \(v = 0\) when \(t = 3\).
Area above the axis \(= \tfrac{1}{2} \times 3 \times 6 = 9\) m.
Area below the axis \(= \tfrac{1}{2} \times 2 \times 4 = 4\) m.
Distance \(= 9 + 4 = 13\) m; displacement \(= 9 - 4 = 5\) m.
Answer: Distance 13 m, displacement 5 m.
Your turn
A bus starts from rest and accelerates at \(1.5\ \text{m s}^{-2}\) for 10 s, travels at constant speed for 40 s, then decelerates at \(2.5\ \text{m s}^{-2}\) to rest. Find the total distance.
Top speed \(= 15\ \text{m s}^{-1}\). Deceleration time \(= 15 \div 2.5 = 6\) s. Distance \(= \tfrac{1}{2}(10)(15) + 40(15) + \tfrac{1}{2}(6)(15) = 75 + 600 + 45 = 720\) m.
Examiner trap
Area below the time axis is negative DISPLACEMENT but positive DISTANCE. Read the question: "distance travelled" adds the areas; "displacement" subtracts.
Stage 2 · Exam
Exam-style questions
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A car starts from rest, accelerates uniformly to \(20\ \text{m s}^{-1}\) in 8 s, moves at constant speed for \(T\) s, then decelerates uniformly to rest in 5 s. The total distance is 690 m. Find \(T\).
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A particle's velocity decreases uniformly from \(6\ \text{m s}^{-1}\) to \(-4\ \text{m s}^{-1}\) in 5 s. Find the distance it travels in this time.
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Particles P and Q leave O together. P starts from rest with acceleration \(2\ \text{m s}^{-2}\) for 6 s and then moves at constant velocity; Q moves at a constant \(9\ \text{m s}^{-1}\). When does P catch up with Q?
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A lift starts from rest, accelerates uniformly to \(3\ \text{m s}^{-1}\) in 2 s, moves at constant speed, then decelerates uniformly to rest in 3 s. It rises 30 m. Find the total time.
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