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

These revision notes condense quantities, units and kinematics with constant acceleration for OCR A Level Mathematics A (H240), content references 3.01a to 3.01c and 3.02a to 3.02e, from the OCR specification, Version 3.1 (August 2026). Everything is Stage 1 (common to AS and A Level) except the unit for moment (3.01c) and vector motion in two dimensions (3.02e), which are A Level only. For full explanations and derivations, read the study guide. When you are ready to test yourself, use the practice questions, the course hub, the checklist and the free diagnostics.

Units (3.01a to 3.01c)

Quantity	Unit	Note
Length	metre, m	base quantity
Time	second, s	base quantity
Mass	kilogram, kg	base quantity
Velocity	m s⁻¹	length ÷ time
Acceleration	m s⁻²	velocity ÷ time
Force	newton, N	kg m s⁻²
Weight	N	a force
Moment (A Level only)	N m	force × perpendicular distance

The three base quantities are mutually independent. Add the correct unit to any quantity you are given.

Conversions to SI

From	To	Do this
km h⁻¹	m s⁻¹	divide by 3.6
m s⁻¹	km h⁻¹	multiply by 3.6
g	kg	divide by 1000
cm	m	divide by 100
minutes	s	multiply by 60

Convert first, then substitute.

Checking units. A formula is wrong if the units on its two sides differ. In s = ut + ½at², the terms are m s⁻¹ × s = m and m s⁻² × s² = m, so every term is a length. Writing s = u + at² fails, because u is a velocity but at² is a length. Adding terms with different units is the usual sign of a slip.

Small worked reminder. A runner covers 450 m in 3 minutes. Convert 3 minutes to 180 s, then average speed = 450 ÷ 180 = 2.5 m s⁻¹.

The language of kinematics (3.02a)

Vector	Scalar
position	distance travelled
displacement	speed
velocity
acceleration

• Displacement is change in position. Distance travelled is the length of the path.
• Speed is the magnitude of velocity. It cannot be negative.
• Average speed = distance travelled ÷ time. Average velocity = displacement ÷ time.
• In one dimension, choose a positive direction. Signs carry the direction.

Must-know distinctions

Pair	Difference
Displacement and distance	displacement can be zero or negative; distance cannot
Velocity and speed	velocity has a direction; speed does not
Deceleration and negative acceleration	slowing down gives a negative a only if motion is in the positive direction

Graphs (3.02b and 3.02c)

Graph	Gradient	Area
displacement-time	velocity	none needed
velocity-time	acceleration	displacement

• Horizontal on displacement-time: at rest.
• Horizontal on velocity-time: constant velocity.
• Straight slope on velocity-time: constant acceleration.
• Area below the axis is negative displacement. Add absolute values for distance.

Method in steps for a velocity-time graph

1. Mark every point where the line changes slope.
2. Find any times where v = 0, which are changes of direction.
3. Split the area into rectangles, triangles and trapezia.
4. Attach signs for displacement. Ignore signs for distance.

Small worked reminder. A car’s speed rises uniformly from 0 to 15 m s⁻¹ in 10 s, stays at 15 m s⁻¹ for 20 s, then falls uniformly to rest in 5 s.

accelerations: 15/10 = 1.5, then 0, then -15/5 = -3 (m/s²)
areas: 0.5 × 10 × 15 = 75
       20 × 15 = 300
       0.5 × 5 × 15 = 37.5
total distance = 412.5 m

The accelerations are 1.5, 0 and −3 m s⁻² and the distance is 412.5 m. The car never reverses, so distance and displacement agree.

Constant acceleration (3.02d)

Use only when acceleration is constant and motion is in a straight line.

Formula	No
v = u + at	s
s = ut + ½at²	v
s = ½(u + v)t	a
v² = u² + 2as	t
s = vt − ½at²	u

Method in steps

1. Draw a sketch and choose the positive direction.
2. List s, u, v, a and t. Mark what is given and what is wanted. Use “rest” to fix u = 0 or v = 0.
3. Pick the formula that leaves out the variable you neither know nor want.
4. Substitute with signs, solve, and give a unit.

Choosing a formula from the wording

Wording	What it tells you
starts from rest, or is released	u = 0
comes to rest, or stops	v = 0
constant speed	a = 0
slows down while moving in the positive direction	a < 0
“when it is back at the start”	s = 0
“after 5 s”	t = 5

Once you have three known values and one wanted value, the formula table above picks itself. If two unknowns appear, find one with a formula that leaves out the other, then use a second formula.

Derivations the specification allows

Route	Outline
Graph	gradient of the line gives a = (v − u)/t; trapezium area gives s = ½(u + v)t
Substitution	put v = u + at into s = ½(u + v)t to get s = ut + ½at²
Integration	v = ∫a dt = at + c with c = u, then s = ∫v dt = ut + ½at²

For v² = u² + 2as, eliminate t using t = (v − u)/a in s = ½(u + v)t.

Small worked reminder. A train moving at 12 m s⁻¹ accelerates at 0.5 m s⁻² for 16 s.

v = 12 + 0.5 × 16 = 20
s = 12 × 16 + 0.5 × 0.5 × 256 = 192 + 64 = 256
check: s = 0.5 × (12 + 20) × 16 = 256

It reaches 20 m s⁻¹ and covers 256 m. A second formula gives a quick check.

Quadratic in t. When s = ut + ½at² is solved for t, two positive roots mean the object passes the point twice. The mean of the two roots is the time at which v = 0, where the motion reverses.

Vectors in two dimensions (3.02e, A Level only)

The same four formulae without v² = u² + 2as:

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

• Questions may use column vectors or i, j notation.
• The dot-product form v · v = u · u + 2a · s is excluded from the specification.
• Work with i and j components separately, or with whole vectors.
• Speed = magnitude of v. Position = initial position + displacement.
• Moving parallel to the x-axis means the j-component of v is zero.
• Moving parallel to the y-axis means the i-component of v is zero.

Small worked reminder. u = (1, 4) and a = (−2, 0.5), t = 6.

v = (1, 4) + 6(-2, 0.5) = (-11, 7)
s = 6(1, 4) + 18(-2, 0.5) = (6, 24) + (-36, 9) = (-30, 33)

The velocity is (−11, 7) and the displacement is (−30, 33).

Small worked reminder: when does it move parallel to an axis? u = (3, −2) and a = (−1, 1), starting at position (0, 4).

v = (3 - t, -2 + t)
parallel to x-axis: -2 + t = 0, so t = 2, where v = (1, 0)
position at t = 2: (0, 4) + 2(3, -2) + 2(-1, 1) = (4, 2)
speed at t = 2: 1 m/s

At t = 2 the particle is at (4, 2) with speed 1 m s⁻¹.

Quick self-test

1. Write the newton in terms of kg, m and s.
2. Convert 90 km h⁻¹ to m s⁻¹.
3. Express a mass of 250 g in kilograms.
4. A particle moves 8 m in the positive direction, then 11 m in the negative direction. Find the displacement and the distance travelled.
5. A velocity-time graph is a straight line from (0, 3) to (4, 15). Find the acceleration.
6. A car starts from rest and accelerates at 2.5 m s⁻² for 6 s. Find its final speed and distance travelled.
7. A cyclist moving at 12 m s⁻¹ brakes uniformly to rest over 36 m. Find the acceleration and the braking time.
8. A particle’s velocity increases uniformly from 4 to 10 m s⁻¹ in 5 s. Find the distance covered.
9. A particle has u = (1, 2) m s⁻¹ and a = (2, −1) m s⁻². Find its velocity and displacement after 3 s.
10. Find the speed of a particle with velocity (−6, 8) m s⁻¹.
11. Which constant acceleration formula does not contain u?
12. A force of 20 N acts at right angles to a rod, 0.15 m from the pivot. Give the moment with its unit (A Level only).

Answers

1. kg m s⁻².
2. 25 m s⁻¹.
3. 0.25 kg.
4. Displacement −3 m, distance 19 m.
5. (15 − 3)/4 = 3 m s⁻².
6. v = 2.5 × 6 = 15 m s⁻¹; s = ½ × 2.5 × 36 = 45 m.
7. 0 = 144 + 72a gives a = −2 m s⁻²; t = 12/2 = 6 s.
8. s = ½(4 + 10) × 5 = 35 m.
9. v = (7, −1) m s⁻¹; s = (3 + 9, 6 − 4.5) = (12, 1.5) m.
10. √(36 + 64) = 10 m s⁻¹.
11. s = vt − ½at².
12. 20 × 0.15 = 3 N m.

Where marks are usually lost

• Using a constant acceleration formula when acceleration varies.
• Not converting km h⁻¹ or minutes to SI before substituting.
• Giving a weight in kg or a mass in N.
• Treating total area under a velocity-time graph as distance when part of it is below the axis.
• Losing the sign of a negative acceleration or velocity.
• Giving one root when a quadratic in t gives two valid times.
• Quoting a speed as a vector, or a velocity without its direction.
• Leaving out units on the final answer.
• Not showing the formula and substitution when a question says “show that”.

Official syllabus

These notes follow the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR.