OCR A Level Mathematics A: Quantities, units, and kinematics with constant acceleration (H240)

This guide covers quantities, units and kinematics with constant acceleration in OCR A Level Mathematics A (H240), content references 3.01a to 3.01c and 3.02a to 3.02e of the OCR specification, Version 3.1 (August 2026). Most of it is Stage 1, common to AS and A Level. Two items are A Level only: the unit for moment (3.01c) and the extension of the constant acceleration formulae to two dimensions with vectors (3.02e). The specification places Mechanics in Paper 3. Take the revision notes for recall and the practice set for questions. The course hub and the checklist show where this sits in the course, and the free diagnostics help you find weak spots.

What this unit covers

Reference	You must be able to	Stage
3.01a	Use the SI base quantities and units: length (m), time (s), mass (kg), and understand that they are mutually independent	Stage 1
3.01b	Use derived quantities and units: velocity (m s⁻¹), acceleration (m s⁻²), force (N), weight (N); add the right unit to a given quantity	Stage 1
3.01c	Use the unit for moment (N m)	A Level only
3.02a	Use the language of kinematics, including the vector and scalar natures of the quantities	Stage 1
3.02b, 3.02c	Use and interpret graphs for straight-line motion, with gradients and areas	Stage 1
3.02d	Use and derive the constant acceleration formulae in a straight line	Stage 1
3.02e	Extend those formulae to two dimensions using vectors	A Level only

Variable acceleration, motion under gravity and projectiles are covered elsewhere.

Quantities and units (3.01a to 3.01c)

The SI system rests on three base quantities that you need: length in metres (m), time in seconds (s) and mass in kilograms (kg). They are mutually independent. You cannot write one as a combination of the other two.

Every other mechanics unit is built from these. The specification lists these derived units.

Quantity	Unit	Built from
Velocity	m s⁻¹ (or m/s)	length ÷ time
Acceleration	m s⁻² (or m/s²)	velocity ÷ time
Force	newton, N	mass × acceleration, so kg m s⁻²
Weight	N	it is a force, not a mass
Moment (A Level only)	N m	force × perpendicular distance

If a question gives a quantity without a unit, you should be able to supply one. If it gives a quantity in other units, convert to SI before you substitute into any formula.

Convert	Method
km h⁻¹ to m s⁻¹	multiply by 1000 ÷ 3600, so divide by 3.6
m s⁻¹ to km h⁻¹	multiply by 3.6
g to kg, cm to m	divide by 1000, divide by 100
minutes to seconds	multiply by 60

Example 1. A cyclist travels at 21.6 km h⁻¹. A spanner is 35 cm long and a force of 48 N acts at right angles to it at its end. Express the speed in m s⁻¹ and give the moment about the other end.

21.6 km/h = 21.6 ÷ 3.6 = 6 m/s
35 cm = 0.35 m
moment = 48 × 0.35 = 16.8 N m

The speed is 6 m s⁻¹ and the moment is 16.8 N m. Notice that the unit of moment is a product, newtons times metres. It is not a newton per metre.

The language of kinematics (3.02a)

Kinematics describes motion without asking what causes it. You need precise terms.

Term	Meaning	Vector or scalar
Position	where the object is, measured from a fixed origin	vector
Displacement	change in position, from start to finish in a straight line	vector
Distance travelled	total length of the path actually covered	scalar
Velocity	rate of change of displacement	vector
Speed	rate of change of distance travelled; the magnitude of velocity	scalar
Acceleration	rate of change of velocity	vector

In one dimension a vector is a signed number. Choose a positive direction and keep to it. A velocity of −3 m s⁻¹ means 3 m s⁻¹ in the negative direction. Speed is never negative.

An equation of motion is an equation that links the quantities describing the motion. The constant acceleration formulae later in this guide are equations of motion.

Average speed is total distance travelled divided by total time. Average velocity is displacement divided by time. They can differ in size and in sign.

Example 2. A particle starts at O on a straight line. It moves 12 m in the positive direction, then 5 m in the negative direction, then 9 m more in the negative direction. The whole journey takes 20 s. Find its final position, the distance travelled, its average speed and its average velocity.

positions: 0 -> 12 -> 7 -> -2
displacement = -2 m
distance = 12 + 5 + 9 = 26 m
average speed = 26 / 20 = 1.3 m/s
average velocity = -2 / 20 = -0.1 m/s

The particle ends 2 m on the negative side of O, having travelled 26 m. The average speed is 1.3 m s⁻¹ and the average velocity is −0.1 m s⁻¹.

Graphs in kinematics (3.02b and 3.02c)

For motion in a straight line, three facts do most of the work.

On a graph of	Gradient gives	Area under it gives
displacement against time	velocity	not used
velocity against time	acceleration	displacement

A horizontal line on a displacement-time graph means the object is at rest. A straight sloping line means constant velocity. A horizontal line on a velocity-time graph means constant velocity, and a straight sloping line means constant acceleration.

Areas below the time axis count as negative displacement. For total distance travelled, add the sizes of the areas and ignore the signs.

Example 3. A particle’s velocity increases uniformly from 0 to 12 m s⁻¹ in the first 5 s, stays at 12 m s⁻¹ until t = 11, then decreases uniformly to −4 m s⁻¹ at t = 15. Find the acceleration in each stage, the time it changes direction, the displacement at t = 15 and the distance travelled.

stage 1: a = 12 / 5 = 2.4 m/s²
stage 2: a = 0
stage 3: a = (-4 - 12) / 4 = -4 m/s²
velocity is zero when 12 - 4(t - 11) = 0, so t = 14
areas: 0.5 × 5 × 12 = 30
       6 × 12 = 72
       0.5 × 3 × 12 = 18
       -0.5 × 1 × 4 = -2
displacement = 30 + 72 + 18 - 2 = 118 m
distance = 30 + 72 + 18 + 2 = 122 m

The accelerations are 2.4, 0 and −4 m s⁻². It changes direction at t = 14. The displacement is 118 m and the distance travelled is 122 m.

Sketch the graph first, then split it into triangles, rectangles and trapezia.

The constant acceleration formulae (3.02d)

When acceleration is constant and motion is in a straight line, five quantities describe it.

Symbol	Meaning
s	displacement from the start
u	initial velocity
v	final velocity
a	acceleration
t	time taken

The specification lists these formulae.

Formula	Missing variable
v = u + at	s
s = ut + ½at²	v
s = ½(u + v)t	a
v² = u² + 2as	t
s = vt − ½at²	u

Each formula leaves out exactly one variable. To choose one, list the three quantities you know and the one you want, then pick the formula missing the fifth.

You may be asked to derive these formulae, and the specification names three routes.

By a velocity-time graph. A straight line from (0, u) to (t, v) has gradient a, so v = u + at. The area under it is a trapezium, which gives s = ½(u + v)t.

By substitution. Put v = u + at into s = ½(u + v)t:

s = ½(u + u + at)t = ½(2u + at)t = ut + ½at²

To get s = vt − ½at², replace u by v − at in s = ut + ½at²:

s = (v - at)t + ½at² = vt - at² + ½at² = vt - ½at²

For v² = u² + 2as, find t = (v − u)/a from the first formula and substitute it into s = ½(u + v)t. This gives 2as = (u + v)(v − u) = v² − u².

By integration. Constant a gives v = ∫a dt = at + c. Since v = u when t = 0, c = u. Integrating again gives s = ut + ½at², taking s = 0 when t = 0.

Example 4. A van moving at 8 m s⁻¹ accelerates uniformly at 1.6 m s⁻² over a distance of 60 m. Find its final speed and the time taken.

u = 8, a = 1.6, s = 60, want v: use v² = u² + 2as
v² = 64 + 2 × 1.6 × 60 = 256, so v = 16
time: use v = u + at, so 16 = 8 + 1.6t, t = 5

The final speed is 16 m s⁻¹ and the time is 5 s.

Example 5. A particle is at O moving along a line with velocity 6 m s⁻¹ and acceleration −2 m s⁻². Find when it is 5 m from O in the positive direction, when it returns to O, and the distance travelled up to the return.

s = 6t - t²
s = 5: t² - 6t + 5 = 0, so t = 1 or t = 5
s = 0: t(6 - t) = 0, so t = 6
v = 0 when 6 - 2t = 0, so t = 3, where s = 18 - 9 = 9
distance up to t = 6 is 9 + 9 = 18 m

It is 5 m from O at t = 1 and t = 5, returns at t = 6, and has travelled 18 m. A quadratic in t often gives two answers. The particle passes the same point going out and coming back.

Two dimensions with vectors (3.02e, A Level only)

For constant acceleration in a plane, apply the formulae to vectors. The specification gives these.

v = u + at
s = ut + ½at²
s = ½(u + v)t
s = vt - ½at²

Questions may use column vectors or i, j notation. The vector version of v² = u² + 2as, with dot products, is excluded. The i and j components are independent, so each component obeys the one-dimensional formulae with its own u, v and a.

Speed is the magnitude of the velocity vector. If the object does not start at the origin, add its initial position vector to s to get its position.

Example 6. A particle has initial velocity (2, −3) m s⁻¹ and constant acceleration (0.5, 1.5) m s⁻². Find its velocity, speed and displacement after 4 s, and the time at which it moves parallel to the x-axis.

v = (2, -3) + 4(0.5, 1.5) = (4, 3)
speed = sqrt(4² + 3²) = 5
s = 4(2, -3) + ½ × 16 × (0.5, 1.5) = (8, -12) + (4, 12) = (12, 0)
check: ½(u + v)t = ½(6, 0) × 4 = (12, 0)
parallel to x-axis: -3 + 1.5t = 0, so t = 2

After 4 s the velocity is (4, 3) m s⁻¹, the speed is 5 m s⁻¹ and the displacement is (12, 0) m. It moves parallel to the x-axis at t = 2, when its velocity is (3, 0) m s⁻¹.

Common errors

• Using the formulae when acceleration is not constant. State the assumption before you start.
• Mixing km h⁻¹ and m s⁻¹ in one equation.
• Treating distance as displacement. They differ whenever the motion reverses.
• Taking the area under a velocity-time graph as distance when part of it lies below the axis.
• Dropping the sign of a deceleration. If motion is in the positive direction and the object slows down, a is negative.
• Missing that “starts from rest” means u = 0, and “comes to rest” means v = 0.
• Quoting a final answer without a unit, or giving a speed as a negative number.

Next steps

Work through the revision notes, then attempt the practice questions without looking at the answers. Use the checklist to record what you have covered and the course hub to move on.

Official syllabus

This guide follows the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR.