Module 3 · Forces and Motion

SUVAT and Linear Motion

Specification: OCR A H556 | Sections: 3.1.1 Kinematics and 3.1.2 Linear motion | Focus: motion quantities, constant acceleration, sign conventions, motion graphs and exam-ready problem solving

By the end of this topic you should be able to…

tell the difference between distance, displacement, speed and velocity
use sign conventions confidently in one-dimensional motion
choose and apply the correct SUVAT equation for constant acceleration
link equations of motion to displacement-time and velocity-time graphs
spot when SUVAT is valid and when it is not
handle OCR-style multi-step linear-motion questions without getting lost

Big idea: SUVAT is not a formula list to memorise blindly. It is a constant-acceleration toolkit — and the real skill is choosing the right equation for the information you have.

Part 1

The language of linear motion

Distance is the total path length travelled. It is a scalar, so it has magnitude only.

Displacement is the change in position from start to finish. It is a vector, so direction matters.

Speed is the rate of change of distance. Velocity is the rate of change of displacement.

Acceleration is the rate of change of velocity.

Distance vs displacement

A runner can complete one lap and have travelled 400 m, but have zero displacement because they finish where they started.

Speed vs velocity

Speed cannot be negative. Velocity can be positive or negative depending on direction.

Acceleration

Acceleration can be negative. That does not automatically mean “slowing down” — it depends on the chosen positive direction.

Common mistake

Students often say “negative acceleration means the object is decelerating”. Not always. If the object is moving in the negative direction and acceleration is also negative, the speed is actually increasing.

Part 2

Sign conventions and choosing a positive direction

Before using SUVAT, choose a positive direction and stick to it.

If upward is positive, then free-fall acceleration is a = −9.81 m s⁻².
If downward is positive, then free-fall acceleration is a = +9.81 m s⁻².
Both are correct if used consistently.

OCR habit worth building

Write your positive direction at the start of a vertical motion question. It cuts down sign errors more than almost anything else.

Part 3

The SUVAT equations

The five quantities are:

s = displacement
u = initial velocity
v = final velocity
a = constant acceleration
t = time

1

v = u + at

2

s = 1/2 (u + v)t

3

s = ut + 1/2 at²

4

s = vt − 1/2 at²

5

v² = u² + 2as

Key assumption

SUVAT only applies when acceleration is constant. If acceleration changes, these equations are not valid without splitting the motion into smaller intervals.

Interactive

Motion and graph explorer

Use the sliders to change the starting velocity, acceleration and duration. The moving marker and the three graphs update together, so you can see how one motion creates three linked graph stories.

Displacement–time

Velocity–time

Acceleration–time

Try a negative u with a positive a. The marker will reverse direction, and the displacement–time graph will flatten before turning back up.

Final velocity

8.0 m s⁻¹

Displacement

30.0 m

Average velocity

5.0 m s⁻¹

Motion

speeding up

Part 4

How to choose the right equation

A smart way to pick an equation is to look for the variable that is missing.

No s needed?

Use v = u + at if you need velocity after a time and displacement is not involved.

No time given?

Use v² = u² + 2as if t is the missing quantity.

Average velocity form?

Use s = 1/2 (u + v)t when you know the initial and final velocity and the time.

Exam trap

Do not force every question into the same favourite formula. The best equation is usually the one that avoids unnecessary rearranging.

Part 5

Motion graphs and why they matter

Displacement–time graph

The gradient gives velocity.

Velocity–time graph

The gradient gives acceleration, and the area under the graph gives displacement.

OCR graph link

Linear motion questions often mix graph interpretation with SUVAT. If acceleration is constant, the velocity–time graph is a straight line.

Graph mistake

The area under a velocity–time graph gives displacement, not distance. If part of the graph is below the time axis, that part contributes negative displacement.

Worked Examples

Worked examples

Worked example 1

A car starts from rest and accelerates uniformly at 3.0 m s⁻² for 5.0 s. Find its final velocity.

u = 0, a = 3.0, t = 5.0

Use v = u + at

v = 0 + 3.0 × 5.0 = 15 m s⁻¹

Worked example 2

A cyclist moving at 8.0 m s⁻¹ brakes uniformly to rest in 4.0 s. Find the acceleration.

u = 8.0, v = 0, t = 4.0

Use v = u + at

0 = 8.0 + 4a, so a = −2.0 m s⁻²

Worked example 3

A trolley has initial velocity 2.0 m s⁻¹ and acceleration 1.5 m s⁻² for 6.0 s. Find the displacement.

Use s = ut + 1/2 at²

s = (2.0 × 6.0) + 1/2 × 1.5 × 6.0²

s = 12 + 27 = 39 m

Worked example 4

A stone is thrown vertically upward at 18 m s⁻¹. Take upward as positive and g = 9.81 m s⁻². Find the maximum height reached.

At maximum height, v = 0 and a = −9.81 m s⁻²

Use v² = u² + 2as

0 = 18² + 2(−9.81)s

s = 16.5 m

Worked example 5

A train travelling at 24 m s⁻¹ slows uniformly at 1.5 m s⁻². How far does it travel before stopping?

u = 24, v = 0, a = −1.5

Use v² = u² + 2as

0 = 24² + 2(−1.5)s

s = 192 m

Worked example 6

A ball is dropped from rest and falls 1.80 m. Ignore air resistance. Find the speed just before impact.

Take downward as positive so u = 0, a = 9.81, s = 1.80

Use v² = u² + 2as

v² = 2 × 9.81 × 1.80 = 35.3

v = 5.94 m s⁻¹

Worked example 7

A particle moves with u = −6.0 m s⁻¹ and a = +2.0 m s⁻² for 5.0 s. Describe the motion and find v and s.

Since u is negative and a is positive, the particle first slows down, stops, then reverses direction.

v = u + at = −6.0 + 2.0 × 5.0 = +4.0 m s⁻¹

s = ut + 1/2 at² = (−6.0 × 5.0) + 1/2 × 2.0 × 25

s = −30 + 25 = −5.0 m

Worked example 8

A velocity–time graph shows an object accelerating from 3.0 m s⁻¹ to 11.0 m s⁻¹ in 4.0 s.

a) Find the acceleration.
b) Find the displacement.

a = gradient = (11.0 − 3.0) / 4.0 = 2.0 m s⁻²

s = area under graph = average velocity × time = 1/2 (3.0 + 11.0) × 4.0 = 28 m

Knowledge

Knowledge Check

What is the difference between distance and displacement?

2 marks

Distance is total path length
Displacement is change in position with direction

When is it valid to use SUVAT equations?

1 mark

When acceleration is constant

What does the area under a velocity–time graph represent?

1 mark

Displacement

If upward is positive, what is the sign of g in free fall?

1 mark

Negative

Exam Practice

Exam-Style Questions

State what is meant by a) velocity and b) acceleration.

2 marks

a) Velocity = rate of change of displacement
b) Acceleration = rate of change of velocity

A car increases its velocity uniformly from 12 m s⁻¹ to 30 m s⁻¹ in 6.0 s.

a) Calculate the acceleration. [2 marks]
b) Calculate the displacement in this time. [2 marks]

4 marks

a) a = (30 − 12) / 6.0 = 3.0 m s⁻²
b) s = 1/2 (u + v)t = 1/2 (12 + 30) × 6.0 = 126 m

A student says, “Negative acceleration always means the object is slowing down.” Explain why this statement is not always correct.

3 marks

Acceleration sign depends on the chosen positive direction
If velocity and acceleration have the same sign, speed increases
So negative acceleration could mean speeding up in the negative direction

A ball is thrown vertically upward with speed 14 m s⁻¹. Ignore air resistance.

a) Calculate the time taken to reach maximum height. [2 marks]
b) Calculate the maximum height reached. [3 marks]
c) State the speed of the ball as it returns to its launch point. [1 mark]

6 marks

a) At top, v = 0 and a = −9.81
0 = 14 − 9.81t, so t = 1.43 s
b) Use v² = u² + 2as
0 = 14² + 2(−9.81)s, so s = 10.0 m
c) Speed = 14 m s⁻¹

A velocity–time graph for a runner is a straight line from 2.0 m s⁻¹ at t = 0 to 8.0 m s⁻¹ at t = 5.0 s.

a) Explain why the acceleration is constant. [1 mark]
b) Calculate the acceleration. [2 marks]
c) Calculate the displacement in 5.0 s using the graph. [2 marks]

5 marks

a) Because the graph is a straight line / has constant gradient
b) a = (8.0 − 2.0) / 5.0 = 1.2 m s⁻²
c) displacement = area under graph = 1/2 (2.0 + 8.0) × 5.0 = 25 m

A particle moves in a straight line with initial velocity −4.0 m s⁻¹ and constant acceleration +3.0 m s⁻².

a) Find the velocity after 3.0 s. [2 marks]
b) Find the displacement after 3.0 s. [2 marks]
c) Describe the motion during this interval. [2 marks]

6 marks

a) v = u + at = −4.0 + 3.0 × 3.0 = +5.0 m s⁻¹
b) s = ut + 1/2 at² = (−4.0 × 3.0) + 1/2 × 3.0 × 9 = +1.5 m
c) It initially moves in the negative direction, slows down, stops, then reverses and speeds up in the positive direction

A lift starts from rest, accelerates uniformly at 1.2 m s⁻² for 4.0 s, then continues at constant velocity for 6.0 s, and finally slows uniformly to rest in 3.0 s.

a) Calculate the maximum velocity reached. [2 marks]
b) Calculate the distance travelled during the first 4.0 s. [2 marks]
c) Calculate the total distance travelled. [4 marks]
d) Sketch the velocity–time graph and label key values. [2 marks]

10 marks

a) v = u + at = 0 + 1.2 × 4.0 = 4.8 m s⁻¹
b) s = ut + 1/2 at² = 0 + 1/2 × 1.2 × 16 = 9.6 m
c) Constant-velocity distance = 4.8 × 6.0 = 28.8 m
Braking distance = 1/2 (4.8 + 0) × 3.0 = 7.2 m
Total distance = 9.6 + 28.8 + 7.2 = 45.6 m
d) Straight-line rise from 0 to 4.8 m s⁻¹ over 4 s, flat section for 6 s, straight-line fall to 0 over final 3 s

A student tries to use SUVAT to analyse the motion of a skydiver during the whole jump, from leaving the plane to reaching terminal velocity. Explain why this is not valid for the entire motion.

3 marks

SUVAT requires constant acceleration
The skydiver’s acceleration changes because air resistance changes with speed
So the motion must be split into smaller intervals or analysed differently

Topic Summary

Motion words

Distance and speed are scalars; displacement and velocity are vectors.

SUVAT

Use only when acceleration is constant.

Signs matter

Choose a positive direction first and stay consistent.

Graphs matter too

Velocity-time gradient gives acceleration; area gives displacement.

v = u + at

s = 1/2 (u + v)t

s = ut + 1/2 at²

v² = u² + 2as