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Constant-acceleration motion equations

FREE PHYSICS REVISION · UNIT 3 · B1.3.8

Pearson BTEC Level 3 AAQ Applied Science · Issue 7 (May 2026)

What you will learn

All four stated kinematic equations; signs, rearrangements and validity.

Specification mapping: B1.3.8

Prerequisites: B1.3.6

Physical principles

For straight-line motion with constant acceleration, v = u + at, s = (u+v)t/2, s = ut + ½at² and v² = u² + 2as. Here s is signed displacement, not automatically total path length. These equations are alternative descriptions of the same constant-acceleration model.

Choose the equation that contains the required unknown and the known quantities. Define direction, list signed values and rearrange before substitution. If taking a square root after using v², choose the velocity sign consistent with the motion. If the object reverses direction, split the journey when calculating total distance.

The equations do not apply unchanged to varying acceleration, such as a falling object approaching terminal velocity. An average acceleration does not justify using every constant-acceleration relationship. Check dimensions: ut and at² both have units of length; 2as has units of speed squared. For vertical motion without significant air resistance, choose the sign of g according to the axis.

Scientific diagram and interpretation

From 0 to 2 s velocity rises uniformly from 0 to 8 m s⁻¹; from 2 to 4 s it remains 8 m s⁻¹. Total area is 8 + 16 = 24 m.
From 0 to 2 s velocity rises uniformly from 0 to 8 m s⁻¹; from 2 to 4 s it remains 8 m s⁻¹. Total area is 8 + 16 = 24 m.

Equations and relationships

s = (u+v)t/2; v = u+at; s = ut+½at²; v² = u²+2as

Worked examples and applications

Final velocity

u = 3.0 m s⁻¹, a = 2.0 m s⁻², t = 4.0 s. v = u + at = 3.0 + 2.0 × 4.0 = 11 m s⁻¹.

Displacement

With those values, s = ut + ½at² = 3.0 × 4.0 + 0.5 × 2.0 × 4.0² = 28 m. Check s = (3.0+11)×4.0/2 = 28 m.

Stopping distance

u = 20 m s⁻¹, v = 0, a = −5.0 m s⁻². Rearrange v² = u² + 2as to s = (v²−u²)/(2a) = (0−400)/(−10) = 40 m.

Time from velocities

For u = 4.0 m s⁻¹, v = 12 m s⁻¹ and s = 32 m, t = 2s/(u+v) = 64/16 = 4.0 s.

Common misconceptions

Constant acceleration is a condition of the model, not a guaranteed property of every moving object.

Examination technique

Select an equation that avoids unneeded unknowns and check the sign of the result.

For an explanation, connect the physical cause to its effect. For a calculation, state the model, rearrange before substitution, retain intermediate digits, and finish with units. For a data question, refer to the actual readings and distinguish a trend from proof of a cause.

Check your understanding

Five questions. Choose an option, then Check Answer. Read the feedback and use Try Again to reset. Use Tab between controls, arrow keys within radio groups, and Enter or Space on buttons. These local practice answers are not saved.

Question 1 of 5 · B1.3.8

Which condition is needed for these equations?

Question 2 of 5 · B1.3.8

u = 2, a = 3, t = 4 in SI units give v =…

Question 3 of 5 · B1.3.8

Starting from rest, a = 2 m s⁻² for 3 s gives s =…

Question 4 of 5 · B1.3.8

Which equation avoids time?

Question 5 of 5 · B1.3.8

For a reversing object, s in these equations is…