Read and sketch displacement time graphs using their gradients
Connect position, instantaneous velocity, rest and reversal to the height and tangent gradient of a displacement–time graph.
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.2. Motion is restricted to one dimension. Choose a positive direction; distance and speed are scalars, while displacement, velocity and acceleration carry a direction. Deceleration means decreasing speed, not necessarily negative acceleration. Calculus uses Paper 1 techniques only. Constant-acceleration formulae require constant acceleration on the stated interval; use a new interval and its initial velocity when conditions change. Where vertical motion occurs, use the supplied g = 10 m s⁻² and the stated no-air-resistance model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Calculate the gradient of a straight line.
- Differentiate a quadratic using Paper 1 methods.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What velocity does a straight displacement graph from (0 s, 2 m) to (3 s, 11 m) show?
What you will learn
- Interpret a displacement–time graph’s gradients and horizontal sections.
- Sketch a curve with correct endpoints, turning points and motion intervals.
The height gives position and the gradient gives velocity
On a displacement–time graph, the vertical coordinate is directed displacement from the chosen reference, not distance travelled. The tangent gradient at a time gives instantaneous velocity. An upward slope means positive velocity, a downward slope means negative velocity, and a horizontal section means rest throughout that interval. The height of the graph alone does not establish which way the particle is moving.
For s = 12t − t² on 0 ≤ t ≤ 8, the velocity is ds/dt = 12 − 2t. The curve starts at (0, 0), has a horizontal tangent at (6, 36) and ends at (8, 32). It rises before 6 s and falls after 6 s, even though s stays positive. The tangent’s slope becomes progressively smaller because acceleration is −2 m s⁻². Mark these features before joining the points with a smooth concave-down curve.
A chord describes an average, while a tangent describes an instant
Between two points, the chord gradient Δs/Δt is average velocity. For the example’s whole 8 s interval it is 32/8 = 4 m s⁻¹, whereas instantaneous velocity at 8 s is −4 m s⁻¹. A chord need not share the end tangent’s slope. At the maximum, v is momentarily zero, but a is non-zero, so the particle reverses rather than remaining at rest.
For a piecewise straight graph, calculate each segment’s slope separately. A horizontal segment has zero velocity despite a possibly non-zero position. Add absolute changes in s to calculate distance across a reversal; the final graph height gives only net displacement when the initial reference is zero. At a sharp corner the idealised graph does not give a single tangent velocity, so do not assign a unique instantaneous velocity there without a smoother model.
Pause and explain
For s = 12t − t², what happens at t = 6 s?
Worked example
Sketch s = 12t − t² m for 0 ≤ t ≤ 8 s and identify its reversal and final velocity.
Show the worked solution
- Differentiate to obtain v = 12 − 2t and solve v = 0, giving t = 6 s.
- Plot (0, 0), the maximum (6, 36), and endpoint (8, 32), joining with a concave-down curve.
- The slope changes from positive to negative at 6 s; final velocity is 12 − 16 = −4 m s⁻¹.
Answer Maximum s = 36 m at 6 s; reversal at 6 s; final velocity −4 m s⁻¹.
Try a topic test
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Test this topic →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.
Sketch s = 5 + 3t m on 0 ≤ t ≤ 4 s, and find velocity, displacement and distance over that interval.
Give me a hint
Its initial height is 5 m and its gradient is constant.
Compare my reasoning
- Plot (0, 5) and (4, 17) and join with a straight rising line.
- Velocity is the gradient +3 m s⁻¹, with no reversal.
- Displacement is 17 − 5 = 12 m, and distance is also 12 m.
Straight line from (0, 5) to (4, 17); velocity +3 m s⁻¹; displacement and distance 12 m.
Look for these in your work
- Retain the non-zero initial position in the sketch.
- Use the constant gradient and distinguish position from journey length.
For s = 2 + 8t − t² m on 0 ≤ t ≤ 6 s, locate the turning point and calculate net displacement and total distance.
Give me a hint
Solve ds/dt = 0 and split the journey at that time.
Compare my reasoning
- Velocity is 8 − 2t, so the turning time is 4 s and position there is 18 m.
- The initial position is 2 m and final position is 14 m, giving displacement +12 m.
- Distance is (18 − 2) + (18 − 14) = 20 m.
Maximum at (4 s, 18 m); displacement +12 m; distance 20 m.
Look for these in your work
- Check the derivative changes sign at the turning time.
- Add both path lengths rather than using the net graph-height change.
A displacement graph consists of straight segments through (0, 2), (3, 11), (5, 11) and (7, 3), with seconds and metres as axes. Give velocities on each open interval, whole-journey displacement, distance and average velocity.
Give me a hint
Read each segment’s slope, including the horizontal middle segment.
Compare my reasoning
- Slopes are +3 m s⁻¹ for 0 < t < 3, zero for 3 < t < 5, and −4 m s⁻¹ for 5 < t < 7.
- Displacement is 3 − 2 = +1 m and distance is 9 + 0 + 8 = 17 m.
- Average velocity is +1/7 m s⁻¹; corner instants do not have one defined tangent velocity in this idealised graph.
Segment velocities +3, 0 and −4 m s⁻¹; displacement +1 m; distance 17 m; average velocity +1/7 m s⁻¹.
Look for these in your work
- Interpret the horizontal interval as rest at a non-zero position.
- State intervals and avoid assigning one tangent at a sharp corner.
Common mistakes
- A positive graph height does not imply positive velocity.
- Use a tangent for instantaneous velocity and a chord for average velocity.
- A momentary horizontal tangent is not an interval of rest.
What can you explain now?
Can a displacement–time graph be above the time axis while the particle moves in the negative direction?
Compare with the explanation
Yes, if its gradient is negative.
The height describes position, while motion direction is determined by the gradient.
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Make it stick
Read displacement from height and velocity from gradient; locate rest and reversals before calculating total distance.
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