0606/22 Oct/Nov 2025, Question 6: Velocity Vectors and Moving Objects

Part (a) · 2 marks

Find the velocity vector of P given its speed 26 and direction vector (5, −12).

1. 1|(5, −12)| = √(5² + (−12)²) = 13
2. 2Velocity = (26/13)(5, −12) = (10, −24)

Answercolumn vector (10, −24)

Part (b) · 2 marks

Given P passes A with position vector (3, 6) at t = 0, write its position vector at time t.

1. 1Position = initial position + t × velocity
2. 2r = (3, 6) + t(10, −24) = (3 + 10t, 6 − 24t)

Answercolumn vector (3, 6) + t × column vector (10, −24), i.e. (3 + 10t, 6 − 24t)

Part (c) · 3 marks

Show d² = mt² + nt + r for the distance d between P and Q, finding integers m, n, r.

1. 1PQ = (8t − 5, 2 − 25t) − (3 + 10t, 6 − 24t) = (−8 − 2t, −4 − t)
2. 2d² = (8 + 2t)² + (4 + t)²
3. 3= 64 + 32t + 4t² + 16 + 8t + t²
4. 4d² = 5t² + 40t + 80

Answerd² = 5t² + 40t + 80 (m = 5, n = 40, r = 80)

Part (d) · 1 mark

Using the expression for d², show that particles P and Q do not collide.

1. 1d² = 5t² + 40t + 80 = 5(t + 4)²
2. 2d = 0 only when t = −4, which is not allowed since t ≥ 0; so d > 0 for t ≥ 0 and they do not collide

Answerd² = 5(t + 4)² = 0 only when t = −4, but t ≥ 0, so d > 0 and P and Q do not collide

More past-paper questions on velocity vectors and moving objects

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