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Welcome to GCSE Edexcel Maths revision.

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Topic M 14: Graphs in context.

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This video covers Higher tier.

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It includes the shared content and the labelled Higher extensions.

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Always identify each axis, its units and its scale.

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A point open bracket 3, 12 close bracket might mean 12 kilometres after 3 hours; it does not automatically describe speed.

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On a distance-time graph, gradient is speed.

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A straight rising line shows constant speed; a horizontal section shows no change in distance.

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A curve becoming steeper shows increasing speed.

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A distance-time journey

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Worked example: A distance increases from 4 kilometres to 16 kilometres between 1 hour and 3 hours.

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Speed equals the fraction with numerator open bracket 16 minus 4 close bracket and denominator open bracket 3 minus 1 close bracket , end fraction equals 6 kilometres per hour.

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On a displacement-time graph, gradient is signed velocity.

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A decreasing graph can mean movement back towards the starting point; cumulative distance travelled cannot decrease.

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On a velocity-time graph, a horizontal line means constant velocity.

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Negative velocity means travel in the opposite direction, rather than negative speed.

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Velocity increasing at constant acceleration

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Use tables and plotted points for an unfamiliar relationship; interpret intersections, intercepts and trends in the situation.

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A model can be valid only over a stated range.

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Worked example: Two taxi fares C equals 4 plus 2 d and C equals 7 plus 1.5 d are equal at their intersection. 4 plus 2 d equals 7 plus 1.5 d gives d equals 6 kilometres and C equals 16 pounds.

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Reciprocal graphs can model a fixed journey: time equals the fraction with numerator open bracket distance close bracket and denominator open bracket speed close bracket ,

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end fraction .

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Doubling the speed halves the time only when the distance is unchanged.

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Gradient of a velocity-time graph is acceleration, with units metres per second squared when velocity is in metres per second and time in s.

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A straight sloping line means constant acceleration.

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Area under a velocity-time graph gives signed displacement.

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For distance travelled, add the magnitudes of areas above and below the time axis.

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Worked example: Velocity increases from 0 to 8 metres per second over 4 seconds.

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Acceleration equals 8 divided by 4 equals 2 metres per second squared ; displacement is triangle area 1 over 2 multiplied by 4 multiplied by 8 equals 16 metres.

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Estimate area under a curved graph using rectangles or trapezia.

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Smaller strips usually improve the estimate; state that the result is approximate.

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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.

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An average rate over an interval is the gradient of a chord joining its endpoints.

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A tangent estimates the rate at one instant.

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These are graphical methods, not differentiation.

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For other graphs, interpret the units of gradient and area rather than assuming they always mean acceleration or displacement.

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A cost-versus-time gradient might represent pounds per hour.

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That completes Graphs in context.

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Revisit the notes and test yourself on the revision website.