OxfordAQA International AS and A-level Mathematics: Mathematical modelling and kinematics with vectors and calculus (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover 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); it is not examined at International AS. Distances are in metres and times in seconds unless stated, and where needed take g = 9.8 m s⁻².

Learn the method first in the study guide or the revision notes. Course links: hub, printable checklist and the free 10-minute diagnostics.

Questions

1. Two crates are towed along a level runway by a rope. The crates are modelled as particles, the rope as light and inextensible, and the runway as smooth.

(a) State what “light” tells you about the tension in the rope. [1] (b) State what “inextensible” tells you about the motion of the crates. [1] (c) The runway is in fact rough. State the effect on the model’s predicted acceleration. [1] (d) Give one reason why treating a crate as a particle may be reasonable here. [1]

2. A stone is projected straight up at 21 m s⁻¹ from ground level. Model it as a particle moving under gravity.

(a) Find the greatest height reached and the time taken to return to ground level, to 3 significant figures where needed. [3] (b) The stone is measured reaching 20.3 m. Comment on the model and suggest one improvement. [2]

3. A particle has velocity v = (6t² − 42t + 72; 4t − 12; 2t − 6) for t ≥ 0, and is at the origin when t = 0.

(a) Show that the particle is at rest only when t = 3. [3] (b) Find the position vector of the particle when t = 3. [2] (c) Find the acceleration at t = 3. [1]

4. A particle moves on a smooth horizontal plane. It is at O when t = 0, with velocity −4i + 7j. Its acceleration is 6t i + 4j.

(a) Obtain v as a function of t. [3] (b) Show that its speed when t = 2 is 17 m s⁻¹. [2] (c) Show that it is on the y-axis when t = 2 and find its distance from O. [2]

5. A particle has position vector r = 3cos(2t)i + 3sin(2t)j relative to O.

(a) Find its velocity and acceleration. [2] (b) Show that its speed is constant and find it. [2] (c) Show that a = −4r and state the direction of the acceleration. [2] (d) Describe the path and find the magnitude of the acceleration. [2]

6. A particle moves on a straight line through O. Its displacement from O is x and its velocity is v = 6t − 3t² when t ≥ 0. When t = 0, x = 1.

(a) Find the positive time at which it is instantaneously at rest. [2] (b) Find x in terms of t. [2] (c) Calculate how far it travels altogether during the first 4 seconds. [3]

7. A particle has velocity v = (3t − 10)i + (4t − 5)j and is at 2i + j when t = 0.

(a) Show that the acceleration is constant and find its magnitude. [2] (b) Show that |v|² = 25(t − 2)² + 25. [2] (c) Find the least speed and the time at which it occurs. [2] (d) Find the position vector at that time. [2]

8. Particles A and B have position vectors r_A = (t² + t)i + (5t − 4)j and r_B = (4t − 2)i + t²j.

(a) Find the values of t for which the i components agree, and for which the j components agree. [4] (b) Show that A and B collide and find the position. [2] (c) Find the speed of each particle at the collision. [2] (d) State a modelling assumption behind “collide” meaning “same position”. [1]

9. A buggy P moves on smooth ground with acceleration (6t − 6)i + 4j. At t = 0 it is at O with velocity 3i − 8j.

(a) Find the velocity of P at time t. [3] (b) Find the position vector of P at time t. [2] (c) Show that P moves parallel to i when t = 2, and find its position and speed then. [3] (d) Show that P is next on the x-axis when t = 4 and find its distance from O. [2]

10. A drone has position vector r = (t² − 2t)i + 3tj + (6 − 3t²/2)k, where k is vertically upwards and the ground is the plane k = 0.

(a) Find its velocity and acceleration. [2] (b) Find when it reaches the ground and its position then. [2] (c) Find its speed on reaching the ground. [2] (d) The drone is modelled as a particle. State what this ignores. [1]

Answers

1. (a) The tension is the same throughout the rope. [1] (b) The crates have the same speed and acceleration along the rope while it is taut. [1] (c) Friction opposes motion, so the model overestimates the acceleration. [1] (d) The crate is small compared with the runway length, so its size and rotation can be ignored. [1] Examiner insight: A modelling answer is stronger when it names the effect on the result, as in part (c), rather than only repeating the word.

2. (a) 0 = 21² − 2(9.8)h [1], so h = 22.5 m [1]. Time to return = 2 × 21/9.8 = 4.29 s [1]. (b) The model predicts 22.5 m but 20.3 m was measured, so it overestimates [1]. Air resistance is ignored; include a resistive force [1]. Examiner insight: A comment needs a comparison with the data, a named assumption and a direction, so give all three.

3. (a) i: 6(t − 3)(t − 4) = 0 gives t = 3 or 4 [1]. j: t = 3. k: t = 3 [1]. Only t = 3 is common (at t = 4 the j component is 4), so rest occurs only at t = 3 [1]. (b) r = (2t³ − 21t² + 72t; 2t² − 12t; t² − 6t) [1], so at t = 3, r = (81; −18; −9) [1]. (c) a = (12t − 42; 4; 2) = (−6; 4; 2) at t = 3 [1]. Examiner insight: “Show that” needs every component checked; testing only the i component leaves the argument incomplete.

4. (a) v = (3t² + c)i + (4t + d)j [1]; v(0) gives c = −4, d = 7 [1]; v = (3t² − 4)i + (4t + 7)j [1]. (b) v(2) = 8i + 15j [1], speed = √(64 + 225) = 17 [1]. (c) r = (t³ − 4t)i + (2t² + 7t)j [1]; at t = 2, r = 0i + 22j, so it is on the y-axis, 22 m from O [1]. Examiner insight: When asked to show a value, display the step that produces it (here √289) and then state the value.

5. (a) v = −6sin(2t)i + 6cos(2t)j [1], a = −12cos(2t)i − 12sin(2t)j [1]. (b) |v|² = 36(sin²2t + cos²2t) [1] = 36, so the speed is 6 m s⁻¹ for all t [1]. (c) a = −4(3cos2t i + 3sin2t j) = −4r [1]; the acceleration is directed towards O [1]. (d) |r|² = 9, so a circle of centre O and radius 3 [1]; |a| = 4 × 3 = 12 m s⁻² [1]. Examiner insight: Use the identity sin² + cos² = 1 explicitly; the constant speed is only demonstrated once it appears in your working.

6. (a) 3t(2 − t) = 0 [1], so t = 2 [1]. (b) x = 3t² − t³ + c [1]; x(0) = 1 gives x = 3t² − t³ + 1 [1]. (c) x(2) = 5 and x(4) = −15 [1]. The particle moves forward 5 − 1 = 4 m, then back 5 − (−15) = 20 m [1]; total = 24 m [1]. Examiner insight: Distance travelled differs from x(4) − x(0); a change of direction at t = 2 means the path must be split there.

7. (a) a = 3i + 4j, which has no t in it, so it is constant [1]; |a| = 5 m s⁻² [1]. (b) |v|² = (3t − 10)² + (4t − 5)² = 25t² − 100t + 125 [1] = 25(t − 2)² + 25 [1]. (c) The least value of |v|² is 25 at t = 2 [1], so the least speed is 5 m s⁻¹ at t = 2 [1]. (d) r = (3t²/2 − 10t + 2)i + (2t² − 5t + 1)j [1]; at t = 2, r = −12i − j [1]. Examiner insight: Take the square root at the end of part (c); quoting 25 as the speed confuses |v|² with |v|.

8. (a) i: t² + t = 4t − 2, so t² − 3t + 2 = 0 [1], giving t = 1 or 2 [1]. j: 5t − 4 = t², so t² − 5t + 4 = 0 [1], giving t = 1 or 4 [1]. (b) Only t = 1 is common to both lists, so they collide at t = 1 [1], at 2i + j [1]. (c) v_A = (2t + 1)i + 5j, so at t = 1 it is 3i + 5j and the speed is √34 [1]. v_B = 4i + 2tj, so it is 4i + 2j and the speed is 2√5 [1]. (d) The particles are treated as points with no size [1]. Examiner insight: A collision needs one time common to both component equations; accepting t = 2 or t = 4 from one equation alone would be a false conclusion.

9. (a) v = (3t² − 6t + c)i + (4t + d)j [1]; c = 3, d = −8 from v(0) [1]; v = (3t² − 6t + 3)i + (4t − 8)j [1]. (b) r = (t³ − 3t² + 3t)i + (2t² − 8t)j with zero constants as r(0) = 0 [1]; this is the position vector [1]. (c) The j component of v is 4(t − 2), which is zero at t = 2 [1]; r(2) = 2i − 8j [1]; v(2) = 3i, so the speed is 3 m s⁻¹ [1]. (d) y = 2t(t − 4) = 0 gives t = 4 (t = 0 is the start) [1]; x = 64 − 48 + 12 = 28, so 28 m from O [1]. Examiner insight: State why t = 0 is excluded in part (d), since the phrase “next on the x-axis” refers to a later time.

10. (a) v = (2t − 2)i + 3j − 3tk [1], a = 2i − 3k [1]. (b) k component zero: 6 = 3t²/2, so t² = 4 and t = 2 [1] (t = −2 rejected); r(2) = 6j [1]. (c) v(2) = 2i + 3j − 6k [1]; speed = √(4 + 9 + 36) = 7 m s⁻¹ [1]. (d) It ignores the drone’s size, shape and rotation [1]. Examiner insight: Reject the negative root explicitly, because the model is only defined for t ≥ 0.

Where marks are usually lost

• Using constant acceleration equations when a depends on t.
• Leaving out the vector constant of integration, or finding only one of its components.
• Declaring “at rest” after checking a single component of the velocity.
• Accepting a collision from one component equation, or from two different values of t.
• Confusing |r|, the distance from O, with the distance travelled.
• Giving |v|² as the speed, or leaving a speed as a vector.
• Ignoring a change in the direction of motion when finding total distance.
• Calling a model unrealistic while naming neither the assumption nor its effect on the answer.

Next steps

• Review the revision notes and study guide.
• Tick off M2.1 and M2.2 on the checklist and visit the course hub.
• Take the free 10-minute diagnostics to find gaps.
• Book a free trial class for help with any question you could not finish.

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).