Edexcel · GCSE Maths · 1MA1 · Foundation and Higher
M14 · Graphs in context
Revision notes, worked examples and methods for graphs in context.
Notes and quizzes ready · 50 questions · Video coming soon.
Foundation shows shared notes and questions. Higher includes the shared content and labelled Higher extensions. Changing tier starts the notes and a new quiz set.
Revise the key ideas
Graphs in real contexts
Always identify each axis, its units and its scale. A point (3, 12) might mean 12 kilometres after 3 hours; it does not automatically describe speed.
On a distance-time graph, gradient is speed. A straight rising line shows constant speed; a horizontal section shows no change in distance. A curve becoming steeper shows increasing speed. A distance-time journey
Worked example: A distance increases from 4 km to 16 km between 1 h and 3 h. Speed = 16 − 43 − 1 = 6 km/h.
On a displacement-time graph, gradient is signed velocity. A decreasing graph can mean movement back towards the starting point; cumulative distance travelled cannot decrease.
On a velocity-time graph, a horizontal line means constant velocity. Negative velocity means travel in the opposite direction, rather than negative speed. Velocity increasing at constant acceleration
Use tables and plotted points for an unfamiliar relationship; interpret intersections, intercepts and trends in the situation. A model can be valid only over a stated range.
Worked example: Two taxi fares C = 4 + 2d and C = 7 + 1.5d are equal at their intersection. 4 + 2d = 7 + 1.5d gives d = 6 km and C = £16.
Reciprocal graphs can model a fixed journey: time = distancespeed. Doubling the speed halves the time only when the distance is unchanged.
Higher — acceleration and area
Gradient of a velocity-time graph is acceleration, with units m/s2 when velocity is in m/s and time in s. A straight sloping line means constant acceleration.
Area under a velocity-time graph gives signed displacement. For distance travelled, add the magnitudes of areas above and below the time axis.
Worked example: Velocity increases from 0 to 8 m/s over 4 s. Acceleration = 8 ÷ 4 = 2 m/s2; displacement is triangle area 12 × 4 × 8 = 16 m.
Estimate area under a curved graph using rectangles or trapezia. Smaller strips usually improve the estimate; state that the result is approximate.
For an instantaneous rate at a point on a curve, draw a tangent there and find its gradient using two points on the tangent, not two arbitrary points on the curve.
An average rate over an interval is the gradient of a chord joining its endpoints. A tangent estimates the rate at one instant. These are graphical methods, not differentiation.
For other graphs, interpret the units of gradient and area rather than assuming they always mean acceleration or displacement. A cost-versus-time gradient might represent pounds per hour.
Test yourself
50 questions · Sets of 10 from the selected tier. For fractions, use / when typing; for powers, use superscripts or ^. Follow each question's answer format. These quick checks support revision; practise full written solutions and proofs too.
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