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
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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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