SUVAT Equations of Motion

The SUVAT equations are the algebraic workhorses of A-Level kinematics. They let you solve any problem involving uniform (constant) acceleration in a straight line — without drawing a single graph — provided you know three of the five kinematic variables. They appear in OCR A-Level Physics A (H556) Module 3.1.2 and reappear throughout the course: projectile motion, free fall, inclined planes, elastic collisions (with impulse), and even the approach to terminal velocity before drag dominates.

This lesson derives all four equations from first principles, shows you how to select the right one quickly, and works through a range of problems at full A-Level depth.

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The Five SUVAT Variables

Every uniform-acceleration problem involves at most five quantities. Their standard symbols are:

Symbol	Quantity	SI unit
s	displacement	m
u	initial velocity	m s⁻¹
v	final velocity	m s⁻¹
a	acceleration	m s⁻²
t	time	s

SUVAT is simply the mnemonic "s, u, v, a, t". Any problem will give you three of these and ask for one of the other two.

There are exactly four independent SUVAT equations, each of which uses four of the five variables (omitting one). Your task as a problem solver is to work out which variable is missing from the problem and pick the equation that also omits it.

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Derivation From First Principles

Assume uniform acceleration a, starting at time t = 0 with velocity u, ending at time t with velocity v after displacement s.

Equation 1 — v = u + at

By definition of acceleration (for constant a):

a = (v − u) / t

Rearranging:

v = u + at

Equation 2 — s = ½(u + v)t

For uniform acceleration, the average velocity is the arithmetic mean of initial and final velocity:

v̄ = (u + v) / 2

Displacement is average velocity × time:

s = ½(u + v)t

Equation 3 — s = ut + ½at²

Substitute equation 1 (v = u + at) into equation 2:

s = ½(u + u + at)t = ½(2u + at)t

s = ut + ½at²

Equation 4 — v² = u² + 2as

Square equation 1:

v² = (u + at)² = u² + 2uat + a²t²

Factor out 2a:

v² = u² + 2a(ut + ½at²) = u² + 2a × s

v² = u² + 2as

This is the only SUVAT equation that does not contain t — and that makes it the go-to equation whenever a problem does not give or ask for time.

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Summary Table — Which Equation When

Equation	Omits	Use when …
v = u + at	s	You do not need, and are not given, displacement
s = ½(u + v)t	a	Acceleration is unknown or irrelevant
s = ut + ½at²	v	Final velocity is unknown or irrelevant
v² = u² + 2as	t	Time is unknown or irrelevant

Exam Tip: Always write down the known variables with their values at the start of the problem. Circle the unknown. The SUVAT equation you need is the one that omits the variable neither given nor asked for.

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Worked Example 1 — Finding Stopping Distance

A car travelling at 28 m s⁻¹ brakes and decelerates uniformly at 6.5 m s⁻². Calculate the distance it takes to stop.

Known:

• u = 28 m s⁻¹
• v = 0
• a = −6.5 m s⁻² (negative because decelerating)
• s = ?
• t is neither given nor asked for → use v² = u² + 2as

0² = 28² + 2(−6.5)s 0 = 784 − 13s s = 784 / 13 = 60.3 m

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Worked Example 2 — Two-Stage Acceleration

A motorcycle starts from rest and accelerates uniformly at 3.2 m s⁻² for 6.0 s. It then maintains the velocity reached for 10 s. Find the total distance travelled.

Stage 1:

• u = 0, a = 3.2 m s⁻², t = 6.0 s
• v = u + at = 0 + 3.2 × 6.0 = 19.2 m s⁻¹
• s₁ = ut + ½at² = 0 + ½ × 3.2 × 36 = 57.6 m

Stage 2:

• Constant velocity 19.2 m s⁻¹ for 10 s
• s₂ = 19.2 × 10 = 192 m

Total distance = 57.6 + 192 = 249.6 m ≈ 250 m (3 s.f.)

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Worked Example 3 — Catch the Train

A train leaves a station from rest with uniform acceleration 0.50 m s⁻². A passenger arrives 4.0 s later and immediately starts running at a constant 7.0 m s⁻¹ along the platform, in the direction of the train. Can the passenger catch the train, and if so, after how long?

Let t = time after the passenger starts running.

• Passenger's displacement: s_p = 7.0 t
• Train's displacement (measured from its starting point at time t = 0): s_T = ½ × 0.50 × (t + 4)² = 0.25 (t + 4)²

They meet when s_p = s_T: