AQA A-Level Mathematics: Q: Kinematics (7357) – Revision Notes

For full explanations and longer worked examples, use the Kinematics study guide. These notes are for quick recall in the final weeks.

They cover Section Q: Kinematics (Q1 to Q5) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. The specification assesses Section Q on Paper 2. A calculator is required in every 7357 paper.

Practise with the Kinematics practice questions. Course hub: /boards/aqa/a-level/mathematics/. Printable checklist: /checklists/aqa/a-level/mathematics/. Find gaps fast with the free 10-minute diagnostics.

Q1: Key terms

Term	Meaning	Scalar or vector
Position	Where the particle is, measured from a fixed origin O	Vector
Displacement	Change in position	Vector
Distance travelled	Total length of path covered	Scalar
Velocity	Rate of change of displacement	Vector
Speed	Magnitude of velocity	Scalar
Acceleration	Rate of change of velocity	Vector

• Average velocity = total displacement ÷ time.
• Average speed = total distance ÷ time.
• In 1D a vector is shown by its sign. Choose the positive direction and write it down.

Notation from Appendix A: s, u, v, a, t, g; bold r, s, v, a for vectors. Units: m, s, m s⁻¹, m s⁻² (Section P).

Q2: Graph rules

Graph	Gradient	Area under graph
Displacement–time	Velocity	(not used)
Velocity–time	Acceleration	Displacement (below axis = negative)

• Horizontal s–t line: at rest. Horizontal v–t line: constant velocity.
• Distance travelled from a v–t graph: add the sizes of all areas, above and below the axis.
• A v–t graph crossing the t-axis means the particle changes direction there.

Reminder. A v–t graph is a straight line from (0, 0) to (4, 10), then horizontal to (9, 10). Displacement = ½(4)(10) + 5(10) = 20 + 50 = 70 m. Acceleration in the first 4 s = 10/4 = 2.5 m s⁻².

Q3: Constant acceleration

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

Method in steps

1. Draw a diagram and mark the positive direction.
2. List s, u, v, a, t with signs.
3. Pick the formula that leaves out the quantity you neither know nor want.
4. Solve. If a quadratic in t gives two roots, decide which one fits the context.

Deriving them (Q3 says derive). Constant a gives a straight v–t line: gradient a = (v − u)/t gives v = u + at. Area of the trapezium gives s = ½(u + v)t. Substitute for v to get s = ut + ½at², for u to get s = vt − ½at², and for t to get v² = u² + 2as.

Vertical motion under gravity. Take upwards as positive, so a = −g. A ball thrown upwards at 9.8 m s⁻¹ (g = 9.8 m s⁻²) reaches the top when v = 0: t = 9.8/9.8 = 1 s, and height = 9.8²/(2 × 9.8) = 4.9 m. It is back at the start after 2 s by symmetry.

Two stages or two particles

1. Use a separate suvat list for each stage or each particle.
2. Link them: the final v of stage 1 is the u of stage 2; two particles that meet have the same position at the same time.
3. If one particle starts later, write its time as (t − delay) and check the answer is after it starts.

Two dimensions. v = u + at and r = r₀ + ut + ½at² with vectors. Treat i and j separately.

Reminder. u = (2i + 5j) m s⁻¹, a = (i − 3j) m s⁻², starting at O. After 2 s: v = (2 + 2)i + (5 − 6)j = (4i − j) m s⁻¹, and r = (4 + 2)i + (10 − 6)j = (6i + 4j) m.

Q4: Calculus

differentiate:   r  →  v  →  a
integrate:       a  →  v  →  r   (+ constant each time)

Relationship	Use it to
v = dr/dt	find velocity from position
a = dv/dt = d²r/dt²	find acceleration
v = ∫ a dt	find velocity from acceleration
r = ∫ v dt	find position from velocity

Appendix B lists these as formulae you must use without them being provided.

Method for distance travelled with variable velocity

1. Solve v = 0 to find when the particle changes direction.
2. Find the position at the start, at each time of rest inside the interval, and at the end.
3. Add the sizes of the changes.

Reminder. v = 4t − t² from O. At rest at t = 0 and t = 4. x = 2t² − t³/3, so x(4) = 32 − 64/3 = 32/3 m.

Vectors. Differentiate or integrate each component. Use the initial vector to find the constant vector.

Q5: Projectiles

Model: particle, no air resistance, g constant, acceleration −gj (j upwards).

horizontal:  constant velocity          x = (U cos α)t
vertical:    suvat with a = −g          y = (U sin α)t − ½gt²
vector form: r = r₀ + ut − ½gt² j      v = u − gt j

You want	Condition
Greatest height	vertical velocity = 0
Time to land at launch level	vertical displacement = 0, t ≠ 0
Time to land below launch	vertical displacement = −(height)
Range	horizontal velocity × time of flight
Velocity at an instant	both components, then speed and angle

Reminder. A particle is projected horizontally at 12 m s⁻¹ from a point 4.9 m above level ground (g = 9.8 m s⁻²). Vertically: 4.9 = ½(9.8)t², so t = 1 s. It lands 12 × 1 = 12 m from the point directly below its start.

Modelling assumptions

Assumption	What it lets you do	What it ignores
Object is a particle	Treat it as a point	Size, shape and spin
No air resistance	Keep horizontal velocity constant	Drag, which grows with speed
g is constant	Use a = −g j throughout	Small changes of g with location and height
Acceleration constant	Use suvat	Real changes in driving or braking force

If asked to improve a model, name one assumption and say how relaxing it changes the answer (for example, air resistance would reduce the range).

Kinematics feeds straight into the Forces and Newton’s laws revision notes (Section R) and the Moments revision notes (Section S).

Must-know distinctions

This	is not	this
Displacement (signed)		Distance (total path)
Velocity (direction)		Speed (size only)
suvat (constant a only)		calculus (any a)
∫v dt over an interval (displacement)		distance travelled (split at v = 0)
“Parallel to i” (j-component 0)		“at rest” (both components 0)
Deceleration of 3 m s⁻²		a = +3 m s⁻² (it is a = −3 in the direction of motion)

Quick self-test

1. A particle moves 5 m forward then 8 m back. State its displacement and distance travelled.
2. A v–t graph rises in a straight line from (0, 0) to (3, 12). Find the distance travelled.
3. u = 3 m s⁻¹, a = 2 m s⁻², t = 5 s. Find v and s.
4. A car at 20 m s⁻¹ brakes with constant deceleration 2.5 m s⁻². Find its stopping distance.
5. A stone is dropped from rest. Take g = 9.8 m s⁻². Find its speed and the distance fallen after 2 s.
6. x = t³ − 6t² + 9t. When is the particle at rest?
7. v = 3t² + 2 and x = 0 when t = 0. Find x when t = 2.
8. r = t²i + (4t − t²)j. Find the speed at t = 1.
9. u = (3i + 4j) m s⁻¹ and a = (i − 2j) m s⁻². Find v and the speed after 2 s.
10. A particle is projected from level ground with velocity (10i + 19.6j) m s⁻¹. Take g = 9.8 m s⁻². Find the time of flight, the range and the greatest height.
11. r = (2t³ − 9t²)i + (6t − t²)j. Show that the particle is at rest when t = 3.

Answers

1. Displacement −3 m (3 m behind the start); distance 13 m.
2. ½ × 3 × 12 = 18 m.
3. v = 3 + 10 = 13 m s⁻¹; s = 15 + 25 = 40 m.
4. 0 = 400 − 5s, so s = 80 m.
5. v = 19.6 m s⁻¹; s = ½(9.8)(4) = 19.6 m.
6. v = 3t² − 12t + 9 = 3(t − 1)(t − 3) = 0, so t = 1 s and t = 3 s.
7. x = t³ + 2t, so x(2) = 12 m.
8. v = 2t i + (4 − 2t)j; at t = 1, v = 2i + 2j, speed = √8 = 2.83 m s⁻¹.
9. v = 5i + 0j = 5i m s⁻¹; speed 5 m s⁻¹.
10. 19.6t − 4.9t² = 0 gives t = 4 s; range = 10 × 4 = 40 m; greatest height = 19.6²/(2 × 9.8) = 19.6 m.
11. v = (6t² − 18t)i + (6 − 2t)j; at t = 3, v = (54 − 54)i + (6 − 6)j = 0, so it is at rest.

Where marks are usually lost

• Applying suvat to an acceleration that depends on t.
• Integrating v across a change of direction and calling the result the distance.
• Losing the sign of g: with upwards positive, a = −9.8 m s⁻².
• Dropping the constant of integration, or using the wrong initial condition to find it.
• Saying “parallel to j” from a zero i-component without checking the j-component is non-zero.
• Giving a speed when the question asks for a velocity, or giving a direction with no reference line.
• Rounding an intermediate time before using it for a range or a height.
• Keeping the negative root of a quadratic in t, or a root before the motion starts.
• Writing only the final number in a “show that” derivation, with no substitution shown.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018, for A-level exams June 2018 onwards: Section 3.18 Q: Kinematics (Q1 to Q5), with Appendix A: mathematical notation (mechanics) and Appendix B: mathematical formulae and identities (kinematics).