SUVAT Equations Horizontally — Edexcel A-level Mathematics (Applied)

SPEC 7.3 • KINEMATICS

SUVAT Equations: Horizontal Motion

Deriving and applying the equations of constant acceleration

The Five SUVAT Variables

The SUVAT equations describe motion under constant acceleration. They use five key variables:

Displacement

metres (m)

Initial Velocity

m s⁻¹

Final Velocity

m s⁻¹

Acceleration

m s⁻²

Time

seconds (s)

Critical Requirement

The SUVAT equations ONLY apply when acceleration is constant. If acceleration varies, you must use calculus methods (covered in Lesson 7.06).

Deriving the SUVAT Equations

All four SUVAT equations can be derived from the definition of acceleration:

Definition

a = \frac{v - u}{t}

Acceleration = change in velocity ÷ time

Equation 1: Rearranging the definition:

a t = v - u

v = u + a t

✓

Equation 2: Using average velocity:

Average velocity

= \frac{u + v}{2}

Displacement = average velocity × time

s = \frac{( u + v )}{2} \times t

✓

Deriving SUVAT (Continued)

Equation 3: Substituting

v = u + a t

into Equation 2:

s = \frac{( u + v )}{2} \times t = \frac{( u + u + a t )}{2} \times t

s = \frac{( 2 u + a t )}{2} \times t = u t + \frac{1}{2} a t^{2}

✓

Equation 4: Eliminating

t

from Equations 1 and 2:

From

v = u + a t

:

t = \frac{v - u}{a}

Substitute into

s = \frac{( u + v )}{2} \times t

:

s = \frac{( u + v )}{2} \times \frac{( v - u )}{a} = \frac{v ^{2} - u ^{2}}{2 a}

2 a s = v^{2} - u^{2}

, so

v^{2} = u^{2} + 2 a s

✓

Exam Tip

You may be asked to derive these equations. Learn the derivations, not just the final formulae!

The Four SUVAT Equations

v = u + a t

Missing: s (displacement)

s = \frac{( u + v )}{2} \times t

Missing: a (acceleration)

s = u t + \frac{1}{2} a t^{2}

Missing: v (final velocity)

v^{2} = u^{2} + 2 a s

Missing: t (time)

Key Insight

Each equation connects four of the five SUVAT variables. Choose your equation based on which variable you don't need and don't have.

Problem-Solving Strategy

Follow this systematic approach for every SUVAT problem:

5-Step Method

StepAction1List all known values (s, u, v, a, t)2Identify what you need to find3Choose equation with 3 knowns + 1 unknown4Substitute values and solve5Check units and reasonableness of answer

Tip

Always write out s = ?, u = ?, v = ?, a = ?, t = ? at the start. This helps you identify which equation to use.

Worked Example01

A car accelerates uniformly from 5 m s⁻¹ to 25 m s⁻¹ over a distance of 100 m. Calculate the acceleration.

1

List known values: u = 5 m s⁻¹, v = 25 m s⁻¹, s = 100 m, a = ?, t = not needed

2

Choose equation without t:

v^{2} = u^{2} + 2 a s

3

Rearrange for a:

a = \frac{v ^{2} - u ^{2}}{2 s}

4

Substitute:

a = \frac{2 5 ^{2} - 5 ^{2}}{2 \times 100} = \frac{625 - 25}{200} = \frac{600}{200}

Answer: a = 3 m s⁻²

Worked Example02

A train decelerates uniformly at 0.5 m s⁻² from 30 m s⁻¹ until it stops. Calculate the time taken.

1

List known values: u = 30 m s⁻¹, v = 0 m s⁻¹ (stops), a = −0.5 m s⁻² (deceleration), t = ?

2

Choose equation without s:

v = u + a t

3

Rearrange for t:

t = \frac{v - u}{a}

4

Substitute:

t = \frac{0 - 30}{- 0.5} = \frac{- 30}{- 0.5} = 60

Answer: t = 60 s

Worked Example03

A cyclist starts from rest and accelerates at 2 m s⁻² for 8 seconds. Calculate the distance travelled.

1

List known values: u = 0 m s⁻¹ (from rest), a = 2 m s⁻², t = 8 s, s = ?

2

Choose equation without v:

s = u t + \frac{1}{2} a t^{2}

3

Substitute:

s = (0) (8) + \frac{1}{2} (2) (8)^{2}

4

Calculate:

s = 0 + \frac{1}{2} (2) (64) = 64

Answer: s = 64 m

Sign Conventions

In horizontal motion problems, signs indicate direction:

Sign Rules

QuantityPositive (+)Negative (−)Displacement (s)In chosen positive directionOpposite to positive directionVelocity (u, v)Moving in positive directionMoving in negative directionAcceleration (a)Speeding up (same direction)Slowing down (deceleration)

Important

Deceleration = negative acceleration. If an object is slowing down, enter acceleration as a negative value. Be consistent with your sign convention throughout the problem!

Worked Example04

A car travelling at 20 m s⁻¹ brakes and comes to rest in 40 m. Calculate the deceleration.

1

List known values: u = 20 m s⁻¹, v = 0 m s⁻¹, s = 40 m, a = ?

2

Choose equation without t:

v^{2} = u^{2} + 2 a s

3

Substitute:

0^{2} = 2 0^{2} + 2 a (40)

4

Solve:

0 = 400 + 80 a

, so

80 a = - 400

,

a = - 5

Answer: a = −5 m s⁻² (deceleration of 5 m s⁻²)

Worked Example05

A particle is projected along a rough horizontal surface with initial velocity 12 m s⁻¹. Due to friction, it decelerates uniformly at 3 m s⁻². Find the time taken for the particle to return to its starting point.

1

The particle moves forward, slows down, stops, reverses, and returns. When it returns to start: s = 0

2

List known values: s = 0, u = 12 m s⁻¹, a = −3 m s⁻² (opposes motion throughout), t = ?

3

Use:

s = u t + \frac{1}{2} a t^{2}

4

Substitute:

0 = 12 t + \frac{1}{2} (- 3) t^{2} = 12 t - 1.5 t^{2}

5

Factorise:

0 = t (12 - 1.5 t)

, so t = 0 or t = 8

6

t = 0 is the start time, so t = 8 s is when it returns

Answer: t = 8 seconds

Key Insight

When s = 0, the equation

s = u t + \frac{1}{2} a t^{2}

gives two solutions: t = 0 (start) and a later time when the particle returns. The friction always opposes motion, so a remains negative throughout.

Two-Stage Motion Problems

Some problems involve multiple phases of motion. Apply SUVAT to each phase separately.

Key Principles

Phase 1 final velocity = Phase 2 initial velocity

Apply SUVAT separately to each phase

Total displacement = s₁ + s₂

Total time = t₁ + t₂

Strategy

Draw a diagram showing the phases. Label each phase with its own set of SUVAT values. Work through each phase in order.

Worked Example06

A car accelerates from rest at 3 m s⁻² for 10 s, then travels at constant velocity for 200 m. Find the total distance.

1

Phase 1: u = 0, a = 3 m s⁻², t = 10 s

2

Find s₁:

s_{1} = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} (3) (10)^{2} = 150

m

3

Find final velocity:

v = u + a t = 0 + 3 (10) = 30

m s⁻¹

4

Phase 2: Constant velocity, so s₂ = 200 m (given)

5

Total distance = s₁ + s₂ = 150 + 200 = 350 m

Answer: Total distance = 350 m

Worked Example07

A particle travels 45 m while accelerating uniformly at 2 m s⁻² for 5 seconds. Find its initial velocity.

1

List known values: s = 45 m, a = 2 m s⁻², t = 5 s, u = ?

2

Choose equation without v:

s = u t + \frac{1}{2} a t^{2}

3

Substitute:

45 = u (5) + \frac{1}{2} (2) (5)^{2}

4

Simplify:

45 = 5 u + \frac{1}{2} (2) (25) = 5 u + 25

5

Solve:

5 u = 20

, so

u = 4

Answer: u = 4 m s⁻¹

Worked Example08

A lorry is travelling at 15 m s⁻¹ when the driver sees an obstruction 50 m ahead. The driver's reaction time is 0.8 s. If the lorry decelerates at 4 m s⁻² after braking begins, determine whether the lorry stops in time.

1

Reaction phase: Constant velocity at 15 m s⁻¹ for 0.8 s

2

Distance during reaction:

s_{1} = 15 \times 0.8 = 12

m

3

Braking phase: u = 15 m s⁻¹, v = 0, a = −4 m s⁻²

4

Braking distance:

v^{2} = u^{2} + 2 a s

→

0 = 225 + 2 (- 4) s_{2}

→

s_{2} = 28.125

m

5

Total stopping distance = 12 + 28.125 = 40.125 m

Answer: Total = 40.125 m < 50 m ✓ Yes, the lorry stops in time

Practice

Question 1: A motorcycle accelerates from 8 m s⁻¹ to 20 m s⁻¹ over 6 seconds. Calculate the distance travelled.

Known: u = 8 m s⁻¹, v = 20 m s⁻¹, t = 6 s, s = ?

Use:

s = \frac{( u + v )}{2} \times t = \frac{( 8 + 20 )}{2} \times 6 = \frac{28}{2} \times 6 = 14 \times 6

Answer: s = 84 m

Practice

Question 2: A train travelling at 25 m s⁻¹ decelerates uniformly and stops in 125 m. Calculate the deceleration.

Known: u = 25 m s⁻¹, v = 0 m s⁻¹, s = 125 m, a = ?

Use:

v^{2} = u^{2} + 2 a s

→

0 = 625 + 2 a (125)

→

0 = 625 + 250 a

250 a = - 625

→

a = - 2.5

Answer: a = −2.5 m s⁻² (deceleration of 2.5 m s⁻²)

Practice

Question 3: A car starts from rest and accelerates at 2.5 m s⁻² for 12 s, then decelerates at 1.5 m s⁻² until it stops. Find the total distance travelled.

Phase 1: u = 0, a = 2.5 m s⁻², t = 12 s

s_{1} = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} (2.5) (144) = 180

m

Final velocity:

v = u + a t = 0 + 2.5 (12) = 30

m s⁻¹

Phase 2: u = 30 m s⁻¹, v = 0, a = −1.5 m s⁻²

v^{2} = u^{2} + 2 a s

→

0 = 900 + 2 (- 1.5) s_{2}

→

s_{2} = 300

m

Answer: Total distance = 180 + 300 = 480 m

Summary

Key Concepts

ConceptDetails

SUVAT Variabless, u, v, a, t — only valid for constant acceleration

Four Equationsv = u + at, s = ½(u+v)t, s = ut + ½at², v² = u² + 2as

Equation SelectionChoose based on which variable is missing/not needed

Sign ConventionDeceleration is negative; be consistent throughout

Multi-Stage ProblemsApply SUVAT separately to each phase

Exam Skills

Always list s, u, v, a, t at the start

Show the equation you're using before substituting

Include units in your final answer

Check your answer is physically reasonable