AQA A-Level Mathematics Paper 2, June 2025: Question 19

8 marks · Medium difficulty · Multi-step Problem

Find the velocity expression by differentiating displacement, and integrate acceleration to find the value of a constant when two particles move parallel at a given time.

Practise this question

Question

Question 19: The displacement, s metres, of particle P at time t seconds is given by s = (2t^3)i + (2t^2 + qt)j. Part (a) asks to find an expression for the velocity of particle P after t seconds, worth 2 marks. Part (b) states the acceleration of particle Q is a = 6ti + 7j with initial velocity 4j m/s. Particles P and Q move parallel to each other when t = 2. It asks to find the value of q, worth 6 marks.
Question text

19 The displacement, s metres, of a particle P, at time t seconds, is given by

s = (2t3) i + (2t2 + qt) j

19 (a) Find an expression for the velocity of particle P after t seconds.

[2 marks]

19 (b) The acceleration, a m s–2, of a particle Q, at time t seconds, is given by

a = 6ti + 7j

Particle Q has an initial velocity of 4j m s–1

Particles P and Q are moving parallel to each other when t = 2

Find the value of q

[6 marks]

… 31

(30)

Mark scheme

Show the mark scheme Mark scheme for Question 19. Part (a): M1 for using v = ds/dt with one component correct; A1 for obtaining 6t^2 i + (4t + q)j. Part (b): M1 for integrating acceleration with one component correct; M1 for using initial velocity when t = 0 to find the constant; A1 for v_q = 3t^2 i + (7t + 4)j; B1F for finding both velocities at t = 2, yielding v_p = 24i + (8 + q)j and v_q = 12i + 18j; M1 for equating v_p = k v_q or comparing component ratios; A1 for q = 28.

Q Marking instructions AO Marks Typical solution

19(a) ds 3.4 M1 ds 2

Uses v = with one v = = 6t i + (4t + q)j

dt dt

component correct

6t2i + (4t + q)j

Obtains

OE 1.1b A1

Subtotal 2

19(b) Uses v = a dt with one

3.4 M1 dv

component correct a = , v = 6ti + 7jdt

Ignore c dt

Substitutes t = 0 into a velocity v = 3t2i + 7tj + c i + c j

vector and equates to 4j to

3.4 When t = 0,v = 4j, c2 = 4

obtain a value or values for their M1

v = 3t2i + (7t + 4) j

constant of integration.

PI by 3t2i + (7t + 4) j

When t = 2,

Obtains 3t2i + (7t + 4) j v = 24i + (8 + q) j

1.1b A1 p

Finds their vp and their vq when v = 12i +18j

q

t = 2 .

vp = 2vq

1.1b B1F

Must come from use of calculus 36 = 8 + q

Condone missing brackets for

8 + q q = 28

Uses vp = kvq

Where k 1

Or

Compares the ratios of the

3.3 M1

components of their vp and

their vq

Must come from use of calculus

Obtains 28 1.1b A1

Subtotal 6

Question 19 Total 8

How to answer it

Variable Acceleration & Parallel Vectors in 2D

What this question tests

This question assesses your ability to apply calculus in 2D kinematics using unit vector notation (i and j):

  • Differentiating vector displacement s with respect to time t to find velocity v.
  • Integrating vector acceleration a to find velocity, incorporating a vector constant of integration using given boundary conditions.
  • Applying the condition for two vector quantities being parallel ( v₁ = kv₂ or equating component ratios).
Part (a) • 2 Marks

Finding Velocity by Differentiating Displacement

📐 Step-by-Step Calculation

  1. Identify the vector displacement:
    s = (2t³)i + (2t² + qt)j
  2. Differentiate each component with respect to time t using v = ds/dt :
    • i-component: d/dt(2t³) = 6t²
    • j-component: d/dt(2t² + qt) = 4t + q
  3. Combine into full vector form:
    v = 6t²i + (4t + q)j

✅ Final Answer & Mark Breakdown

v = 6t²i + (4t + q)j m s⁻¹

[M1] Attempt to differentiate s with at least one component differentiated correctly.
[A1] Fully correct vector expression in terms of t and q (or equivalent column vector form).

💡 Key Knowledge

  • Because displacement is a function of time, you must use differentiation: v = ds/dt . Constant acceleration equations (suvat) cannot be applied.
  • The constant q is treated as an algebraic constant: d/dt(qt) = q .

❌ Common Errors

  • Treating q as a variable: Some students accidentally write q' or forget the term entirely.
  • Missing brackets: Writing 4t + qj instead of (4t + q)j can cause ambiguity in subsequent work.
Part (b) • 6 Marks

Integrating Acceleration and Solving for Parallel Vectors

📐 Step-by-Step Calculation

  1. Find velocity of Q by integration:
    vQ = ∫ a dt = ∫ (6ti + 7j) dt = 3t²i + 7tj + c
  2. Apply initial condition (t = 0, v = 4j):
    c = 0i + 4j
    ⇒ vQ = 3t²i + (7t + 4)j
  3. Evaluate both velocities at t = 2:
    vP(2) = 6(2)²i + (4(2) + q)j = 24i + (8 + q)j
    vQ(2) = 3(2)²i + (7(2) + 4)j = 12i + 18j
  4. Set up the parallel vector condition:
    Since vectors are parallel, vP = kvQ :
    Comparing i-components: 24 = 12k ⇒ k = 2
    Comparing j-components: 8 + q = 18k
  5. Solve for q:
    8 + q = 18(2) = 36 ⇒ q = 28

✅ Final Answer & Mark Breakdown

q = 28

[M1] Integrates a with at least one component integrated correctly.
[M1] Uses initial conditions ( t = 0, v = 4j ) to find vector constant c.
[A1] Correct expression for vQ = 3t²i + (7t + 4)j .
[B1F] Correct substitution of t = 2 into both velocity expressions (follow-through on previous calculus).
[M1] Equating component ratios or setting vP = kvQ where k ≠ 1 .
[A1] Final answer q = 28 .

🧠 Exam Technique: Two Ways to Set Up Parallel Vectors

When two vectors A = x₁i + y₁j and B = x₂i + y₂j are parallel:

  • Method 1 (Scalar Multiplier): Set A = kB . Find scalar k from known components, then equate the remaining components.
  • Method 2 (Component Ratio): Equate ratios directly: y₁ / x₁ = y₂ / x₂ . Here: (8 + q) / 24 = 18 / 12 = 1.5 ⇒ 8 + q = 36 ⇒ q = 28 .

❌ Common Traps to Avoid

  • Forgetting the constant of integration: Writing vQ = 3t²i + 7tj forfeits both method and accuracy marks.
  • Confusing parallel with equal: Setting vP = vQ directly would imply 24 = 12 , which is a contradiction! Parallel vectors are scalar multiples, not necessarily equal.
  • Evaluating displacement instead of velocity: Make sure you substitute t = 2 into the velocity vectors, not the original displacement formula.

Topics

Mechanics · Pure Mathematics · Q: Kinematics · J: Vectors

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.