Pearson Edexcel · A-Level · Mathematics

    Course version: 9MA0 · Version from Mar 2026

    Understand, use and derive the formulae for constant acceleration for…

    Full specification wording

    Kinematics: 7.3 Understand, use and derive the formulae for constant acceleration for motion in a straight line. Extend to 2 dimensions using vectors.

    For constant acceleration a along a straight line, velocity changes linearly with time: v = u + at.

    Checked against the Pearson Edexcel 9MA0 specification (opens in a new tab) on .

    • Point 3 of 5
    • 3 key terms

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    For constant acceleration a along a straight line, velocity changes linearly with time: v = u + at. Displacement accumulates as s = ut + ½at², and eliminating t gives v² = u² + 2as; the average-velocity form s = ½(u + v)t follows from the trapezium area under a velocity–time graph. Derivation uses a = dv/dt and v = ds/dt with a constant, integrating from t = 0 with v = u and s = 0. In two dimensions each formula applies independently to the x- and y-components, so r = r₀ + u t + ½a t² and v = u + a t become vector equations; a projectile has a = −g ĵ while the horizontal component of velocity stays constant. Choose the equation omitting the unwanted quantity, fix a positive direction, and keep signs consistent.

    What earns the marks

    • States the five constant-acceleration formulae and identifies which variable each omits, so the correct equation can be selected for a given problem.
    • Derives v = u + at by integrating a = dv/dt with constant a and initial condition v = u at t = 0.
    • Derives s = ut + ½at² by integrating v = u + at with s = 0 at t = 0, and obtains v² = u² + 2as by eliminating t.
    • Applies the formulae componentwise in two dimensions, writing r = r₀ + u t + ½a t² and v = u + a t as vector equations.
    • Interprets gradient and area of velocity–time and displacement–time graphs as acceleration and displacement respectively.
    • Maintains consistent sign conventions and units, for example taking upwards as positive so that a = −9.8 m s⁻² under gravity.

    Common mistakes and how to put them right

    • Treating acceleration as variable and applying constant-acceleration formulae outside their range.

      Put it right: Check that a is constant before using them, otherwise use calculus.

    • Mixing up u and v in v² = u² + 2as.

      Put it right: Label initial and final velocities explicitly before substituting.

    • Applying a single scalar equation to a two-dimensional problem.

      Put it right: Resolve into perpendicular components and apply the formulae to each component separately.

    Exam tips

    • List the known and unknown quantities with signs before choosing an equation, so the omitted variable identifies the formula to use.
    • Show the derivation from a = dv/dt and v = ds/dt when a question asks you to derive rather than quote a formula.
    • For vector motion, write the x- and y-equations on separate lines and keep the unit vectors throughout to avoid losing direction marks.

    Try the lesson

    The start of the Kinematics: 7.3 Understand, use and derive the formulae for constant acceleration for motion in a straight line. Extend to 2 dimensions using vectors. lesson, made from checked teaching notes for this specification point.

    Have a go first

    A cyclist starts from rest and accelerates uniformly at 2 m s⁻² along a straight road. How far does the cyclist travel in the first 5 seconds?

    Listen

    For constant acceleration a along a straight line, velocity changes linearly with time: v = u + at. Displacement accumulates as s = ut + ½at², and eliminating t gives v² = u² + 2as; the average-velocity form s = ½(u + v)t follows from the trapezium area under a velocity–time graph. Derivation uses a = dv/dt and v = ds/dt with a constant, integrating from t = 0 with v = u and s = 0. In two dimensions each formula applies independently to the x- and y-components, so r = r₀ + u t + ½a t² and v = u + a t become vector equations; a projectile has a = −g ĵ while the horizontal component of velocity stays constant. Choose the equation omitting the unwanted quantity, fix a positive direction, and keep signs consistent. A strong answer earns marks for these points. States the five constant-acceleration formulae and identifies which variable each omits, so the correct equation can be selected for a given problem. Derives v = u + at by integrating a = dv/dt with constant a and initial condition v = u at t = 0. Derives s = ut + ½at² by integrating v = u + at with s = 0 at t = 0, and obtains v² = u² + 2as by eliminating t. Applies the formulae componentwise in two dimensions, writing r = r₀ + u t + ½a t² and v = u + a t as vector equations. A common slip: Treating acceleration as variable and applying constant-acceleration formulae outside their range. Instead, check that a is constant before using them, otherwise use calculus. Exam tip: List the known and unknown quantities with signs before choosing an equation, so the omitted variable identifies the formula to use.

    Key words

    Say what each word means before you turn the card.

    Constant acceleration as a linear velocity–time model

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    Graphical interpretation of gradient and area

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    Componentwise extension of scalar formulae to vectors

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    See every definition
    Constant acceleration as a linear velocity–time model
    Along a straight line, velocity changes linearly with time, so v = u + at.
    Graphical interpretation of gradient and area
    The gradient and area of velocity–time and displacement–time graphs are interpreted as acceleration and displacement respectively.
    Componentwise extension of scalar formulae to vectors
    In two dimensions each formula applies independently to the x- and y-components, so the formulae become vector equations.

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