OCR A Level Mathematics A: Variable acceleration, motion under gravity and projectiles (H240) – Revision Notes

These notes condense content references 3.02f to 3.02i of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Reference 3.02f is Stage 1, shared with AS. References 3.02g, 3.02h and 3.02i are A Level only. For explanations and fuller worked examples, read the study guide. When you are ready to test yourself, use the practice set, the course hub, the checklist and the free diagnostics.

Key results

Idea	Result	Stage
Velocity and acceleration (3.02f)	v = ds/dt, a = dv/dt = d²s/dt²	Stage 1
Reverse (3.02f)	s = ∫v dt, v = ∫a dt, with a constant of integration	Stage 1
Vector form (3.02g)	v = dr/dt, a = dv/dt; r = ∫v dt, v = ∫a dt	A Level only
Gravity (3.02h)	a = −gj = (0, −g), with g = 9.8 m s⁻²	A Level only
Constant acceleration, vectors	v = u + at, r = r₀ + ut + ½at²	A Level only
Launch components (3.02i)	horizontal u cos α, vertical u sin α	A Level only
Time of flight, same level	T = 2u sin α / g	A Level only
Range, same level	R = u² sin 2α / g	A Level only
Greatest height	H = u² sin²α / (2g)	A Level only
Path	y = x tan α − g x² / (2u² cos²α)	A Level only

Method boxes

Variable acceleration in one dimension

1. Decide which direction to go: differentiate from s towards a, or integrate from a towards s.
2. After every integration, add a constant and find it from a given condition.
3. For “at rest”, solve v = 0. For maximum or minimum velocity, solve a = 0.
4. For distance travelled, split the interval where v changes sign and add magnitudes.

Vectors with calculus

1. Treat the i and j components separately.
2. The constant of integration is a vector, found from the initial velocity or position.
3. Speed = √(vₓ² + v_y²). Direction = tan⁻¹(v_y / vₓ), with the reference direction stated.
4. “Parallel to i”: j-component of v is 0. “Parallel to j”: i-component of v is 0.

Projectile problems

1. Split u into components. Draw a sketch with axes and the launch point.
2. Horizontal: constant velocity, x = (u cos α) t.
3. Vertical: acceleration −g, so y = (u sin α) t − ½gt² and v_y = u sin α − gt.
4. Use t to connect the two directions. Find t from the vertical equation first.
5. Reject roots that make no sense, such as negative times.

Worked reminders

A. Distance travelled. A particle has s = 2t³ − 9t² + 12t for 0 ≤ t ≤ 3.

v = 6t² − 18t + 12 = 6(t − 1)(t − 2)      a = 12t − 18
s(0) = 0,  s(1) = 5,  s(2) = 4,  s(3) = 9
distance = 5 + 1 + 5 = 11

It reverses at t = 1 and t = 2, so the distance (11) differs from the displacement (9).

B. Finding a constant. a = 4t − 6 and v = 2 when t = 0.

v = 2t² − 6t + C, C = 2, so v = 2t² − 6t + 2
s = (2/3)t³ − 3t² + 2t + D

C. A projectile at 60°. Launch speed 15 m s⁻¹ from level ground, α = 60°.

components: 7.5 and 13.0 (15 sin 60° = 12.99)
T = 2(12.99)/9.8 ≈ 2.65 s    R = 7.5 × 2.651 ≈ 19.9 m
H = 12.99²/(19.6) ≈ 8.61 m
y = 1.732x − 0.0871x²

D. Range symmetry. For u = 18 m s⁻¹, R = 324 sin 2α / 9.8. The maximum is 33.1 m at α = 45°. Angles α and 90° − α give the same range: 30° and 60° both give 28.6 m. The paths differ in height and in time of flight.

E. Launch from a height. A stone is thrown horizontally at 10 m s⁻¹ from 35 m above level ground.

35 = ½(9.8)t² → t = 2.67 s
horizontal distance = 10 × 2.673 ≈ 26.7 m

Phrases decoded

The question says	You do
“at rest”	solve v = 0 (every component, in two dimensions)
“greatest height” or “highest point”	set the vertical velocity to 0
“returns to the level of projection”	set y = 0 for the second time
“hits the ground”	set the height of the particle to 0, reject the negative time
“moving at 45° below the horizontal”	set v_y = −vₓ
“speed”	find the magnitude of the velocity vector
“passes through the point (p, q)”	substitute into the path equation, or solve x = p then check y = q
“find the range”	find the horizontal distance when the height returns to the stated level

F. Integrating a vector. a = 2t i − 6j, v = 4i + j at t = 0, r = 0 at t = 0.

v = (t² + 4)i + (1 − 6t)j
r = (t³/3 + 4t)i + (t − 3t²)j

G. Vector gravity. A particle starts at 2j with velocity (6i + 10j) and a = −9.8j.

r = 6t i + (2 + 10t − 4.9t²)j     v = 6i + (10 − 9.8t)j

At the highest point 10 − 9.8t = 0, so t = 1.02 s, and the height is 2 + 100/19.6 ≈ 7.10 m.

Must-know distinctions

• Displacement and distance: ∫v dt gives displacement. Distance needs the sign changes of v.
• Speed and velocity: speed is a magnitude, velocity is a vector.
• At rest and at the top: at rest means both components are zero. At the highest point of a projectile only the vertical component is zero, and the horizontal speed is still u cos α.
• Maximum speed and maximum velocity: a = 0 finds a stationary value of v, but check the ends of the interval for the greatest speed.
• Cartesian path and time equations: the path equation has no t. Use it for “passes through a point” and shape questions, not for timing.
• Model and reality: the model ignores air resistance, treats the object as a particle, keeps g constant and ignores wind.

Quick self-test

1. A particle has s = 2t³ − 9t² + 12t. Find its acceleration when t = 4.
2. A particle has a = 6t − 4, v = 3 when t = 0. Find v in terms of t.
3. A particle has position r = 3t²i + (t³ − 2t)j. Find its velocity and acceleration at t = 2.
4. A particle has v = (2t + 1)i − 4tj and starts at the origin. Find its position at t = 3.
5. A ball is thrown from the origin with velocity (3i + 7j) m s⁻¹ in the model a = −9.8j. Find its position and velocity at t = 1.
6. For the ball in question 5, find the greatest height reached.
7. Find the horizontal and vertical components of a launch at 20 m s⁻¹ and 30° above the horizontal.
8. A projectile is launched from level ground at 15 m s⁻¹ at 60°. Find the speed at the moment it lands.
9. State two limitations of the projectile model.
10. A particle moves with v = 3t² − 12t + 5. At what time is the acceleration zero, and what is the velocity then?
11. A stone is thrown horizontally at 10 m s⁻¹ from 35 m above level ground. Find the speed when it lands, to 3 significant figures.

Answers

1. a = 12t − 18, so a = 30 m s⁻².
2. v = 3t² − 4t + 3.
3. v = 6t i + (3t² − 2)j = 12i + 10j. a = 6i + 6tj = 6i + 12j.
4. r = (t² + t)i − 2t²j, so at t = 3: r = 12i − 18j.
5. r = 3i + 2.1j. v = 3i − 2.8j.
6. v_y = 0 at t = 7/9.8, and H = 49/19.6 = 2.5 m.
7. Horizontal 20 cos 30° ≈ 17.3 m s⁻¹, vertical 20 sin 30° = 10 m s⁻¹.
8. The horizontal component stays 7.5 and the vertical component reverses to −12.99 at the same height, so the speed is √(7.5² + 12.99²) = 15 m s⁻¹.
9. Air resistance ignored; object treated as a particle; g taken as constant; wind ignored.
10. a = 6t − 12 = 0 at t = 2, and v = 12 − 24 + 5 = −7 m s⁻¹.
11. v_y = 9.8 × 2.673 ≈ 26.2, so speed = √(10² + 26.2²) ≈ 28.0 m s⁻¹.

Where marks are usually lost

• No constant of integration, or a constant found from the wrong condition.
• Giving s(end) − s(0) when the question asks for distance travelled and the particle turns round.
• Using constant acceleration formulae when a is a function of t.
• Using a, not v, in the condition for a particle at rest.
• Mixing components: using u in a horizontal equation, or g in a horizontal equation.
• Giving an angle with no reference direction, such as “below the horizontal”.
• Rounding g or intermediate values early, so a final 3 significant figure answer drifts.
• Using the range formula for a launch from a height.
• Giving a limitation of the model that is really a limitation of the question, such as “the ball might hit a bird”.

Official syllabus

OCR, A Level in Mathematics A (H240), Version 3.1 (August 2026), for first assessment in 2018, content references 3.02f to 3.02i.