OxfordAQA International AS and A-level Mathematics: Modelling and kinematics with vectors and calculus (9660)

This guide covers sections M2.1 (mathematical modelling) and M2.2 (kinematics) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International A-level Unit M2 (Mechanics), taken with Unit P2 as the mechanics option (the alternative is Unit S2); it is not examined at International AS. Unit M2 expects familiarity with the International AS units and Unit P2, so straight-line motion from Unit PSM1 is your starting point.

Use it with the revision notes and the practice questions. The course hub is /boards/oxfordaqa/a-level/mathematics/ and the printable checklist is /checklists/oxfordaqa/a-level/mathematics/. To find weak spots first, try the free 10-minute diagnostics. For the straight-line version of this work, see the kinematics in a straight line guide; for vector skills, see the vectors guide.

What this unit covers

Ref	What you must be able to do
M2.1	Use assumptions to simplify a real situation, analyse the model and interpret the result
M2.1	Judge the validity of a model and comment on assumptions
M2.1	Explain terms such as particle, light, inextensible string, smooth surface and motion under gravity
M2.1	Recognise when a large moving body can be treated as a point mass
M2.2	Relate position, velocity and acceleration in one, two or three dimensions when acceleration varies
M2.2	Differentiate or integrate r = f(t)i + g(t)j + h(t)k with respect to t
M2.2	Work with ai + bj + ck form or column vectors (either is allowed)

M2.1 Mathematical modelling

A model replaces a real situation by a simpler one you can solve. You make assumptions, solve the maths, then interpret the answer in context and ask whether it is believable. If the prediction and the evidence disagree, you adjust the assumptions and try again.

You must be able to comment on these assumptions.

Term	What it assumes	Effect on the model
Particle	The object is a point with mass but no size	Rotation and size are ignored; every part moves the same way
Large body as a point mass	Size is small compared with the distances travelled	Acceptable for a lorry on a long journey, not for one reversing into a bay
Light (string or rod)	Mass is negligible	Tension is the same throughout the string
Inextensible string	Length cannot change	Both ends have the same speed and acceleration along the string while it is taut
Smooth surface	No friction	The only contact force is perpendicular to the surface
Motion under gravity	Gravity is the only force, with constant g	Air resistance, wind and the variation of g with height are ignored

Worked example 1. A stone is dropped from rest from a bridge 30 m above a river. Model it as a particle under gravity, with g = 9.8 m s⁻². Find the time to reach the water and the speed on impact. The stone is timed at 2.7 s. Comment.

s = ut + ½at²  →  30 = ½ × 9.8 × t²
t² = 60/9.8, so t = 2.47 s (3 s.f.)
v = u + at = 9.8 × 2.4744… = 24.2 m s⁻¹ (3 s.f.)

The model predicts 2.47 s, but the measured time is longer. That fits an assumption that ignores air resistance, which slows the fall. A refinement is to include a resistive force. Timing error is another source of difference, so you would repeat the timing before blaming the model.

M2.2 Position, velocity and acceleration with calculus

For a particle with position vector r relative to a fixed origin O, at time t:

velocity = rate of change of position:   v = dr/dt
acceleration = rate of change of velocity:   a = dv/dt = d²r/dt²
integrating reverses this:   v = ∫ a dt,   r = ∫ v dt

In component form, if r = f(t)i + g(t)j + h(t)k, then v = f′(t)i + g′(t)j + h′(t)k and a = f″(t)i + g″(t)j + h″(t)k. You differentiate each component separately. The same works for column vectors, written (x; y; z) in this guide for one column.

Facts you will use repeatedly:

• Speed is the magnitude of the velocity: |v| = √(vx² + vy² + vz²). It is a scalar.
• The distance of the particle from O is |r|. It is not the distance travelled.
• Velocity is always along the direction of motion. Acceleration need not be.
• The constant acceleration equations (such as v = u + at) apply only if a is constant. With a depending on t, you integrate.
• Integration needs a constant of integration, which is a vector. Find it from a given velocity or position at a known time.

Worked example 2 (differentiating). A particle has position vector r = (t³ − 12t)i + (4t² − 24t)j metres, for t ≥ 0. Find (a) its velocity and acceleration at t = 5, (b) when it moves parallel to i, (c) when it moves parallel to j.

(a)

v = (3t² − 12)i + (8t − 24)j
a = 6t i + 8j
t = 5:  v = 63i + 16j,  a = 30i + 8j
speed = √(63² + 16²) = √4225 = 65 m s⁻¹

(b) For motion parallel to i, the j component of v must be zero: 8t − 24 = 0, so t = 3. Then v = 15i and r = −9i − 36j.

(c) Parallel to j means the i component of v is zero: 3t² − 12 = 0, so t = 2 (t = −2 is rejected because t ≥ 0).

Notice that v is never zero here, because the two components vanish at different times (t = 2 and t = 3). If a question says “at rest”, set every component to zero and check that the same t works for all of them.

Worked example 3 (integrating in three dimensions). A particle moves with acceleration a = (6t; −6; 12t²) m s⁻². When t = 0 its velocity is (2; 5; −1) and its position is (1; 0; 3). Find its velocity and position when t = 2.

Integrate each component and add the constant from the initial conditions:

v = (3t² + c1; −6t + c2; 4t³ + c3)
t = 0: (c1; c2; c3) = (2; 5; −1)
v = (3t² + 2; 5 − 6t; 4t³ − 1)

Integrate again, using r(0) = (1; 0; 3):

r = (t³ + 2t + 1; 5t − 3t²; t⁴ − t + 3)
t = 2:  v = (14; −7; 31),  r = (13; −2; 17)

Check the position by differentiating: d/dt of t⁴ − t + 3 is 4t³ − 1, matching the k component of v. Checking one component this way takes seconds and catches a lost constant.

Worked example 4 (least speed). A particle has velocity v = (2t − 6)i + (8 − 2t)j m s⁻¹. Find the least speed and the time it occurs.

Speed is a square root, so minimise the speed squared instead:

|v|² = (2t − 6)² + (8 − 2t)² = 8t² − 56t + 100
d(|v|²)/dt = 16t − 56 = 0  →  t = 3.5
|v|² = 8(3.5)² − 56(3.5) + 100 = 2
least speed = √2 m s⁻¹ (about 1.41)

The second derivative of |v|² is 16, which is positive, so this is a minimum. You could also complete the square: 8(t − 3.5)² + 2. Both methods give the same result, and the completed square shows the minimum without any differentiation.

Worked example 5 (two particles). Particle A has position vector (2t + 1)i + t²j and particle B has position vector (5t − 5)i + (3t − 2)j, with distances in metres and t in seconds. Do they collide?

Two particles are at the same place at the same time only if both components match for one value of t.

i components:  2t + 1 = 5t − 5  →  t = 2
j components:  t² = 3t − 2  →  t² − 3t + 2 = 0  →  t = 1 or t = 2

The i components agree only at t = 2, and t = 2 is also one of the j solutions. At t = 2 both are at 5i + 4j, so they collide at t = 2 s. If the j equation had given only t = 1 and t = 3, the paths would cross but the particles would miss. Treating each particle as a point is itself a modelling assumption: real objects with size could touch earlier.

Common errors

• Using v = u + at or s = ut + ½at² when acceleration depends on t.
• Forgetting the constant of integration, or adding it as a number rather than a vector.
• Calling |r| the distance travelled. Distance travelled needs the path length, which for straight-line motion means splitting at points where the velocity changes sign.
• Finding speed by differentiating |r|, or by taking |v| of the wrong time.
• Solving the i and j equations for meeting and accepting two different values of t.
• Declaring a particle at rest after one component is zero.
• Describing a model as “wrong” without naming the assumption and its effect. Write which assumption fails and in which direction the prediction moves.

Where next

Move to the revision notes for a compact summary and self-test, then try the practice questions. Revisit the course hub or checklist to tick off M2.1 and M2.2.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA. Sections M2.1 and M2.2 of Unit M2 (International A-level only).