Additional Mathematics – Vectors in two dimensions | e-Consult

Answer 3

1. Velocity of A: v_A = (5, 0) m/s.
   

   Velocity of B: v_B = (0, 3) m/s.
   

   Relative velocity of A with respect to B: v_AB = v_A – v_B = (5, ‑3) m/s.

2. Initial position vector from B to A: r₀ = (‑40, 0) m (A is west of B). Position of A relative to B at time t:

   
   r(t) = r₀ + v_ABt = (‑40 + 5t, ‑3t).

The squared distance: D²(t) = (‑40 + 5t)² + (‑3t)².


Differentiate: d(D²)/dt = 2(‑40 + 5t)(5) + 2(‑3t)(‑3) = 10(‑40 + 5t) + 18t.


Set to zero: 10(‑40 + 5t) + 18t = 0 → ‑400 + 50t + 18t = 0 → 68t = 400 → t = 400/68 ≈ 5.88 s.

3. Insert t ≈ 5.88 s into r(t):

   
   x = ‑40 + 5(5.88) ≈ ‑40 + 29.4 = ‑10.6 m,

   
   y = ‑3(5.88) ≈ ‑17.6 m.
   

   Minimum distance D_min = √[ (‑10.6)² + (‑17.6)² ] ≈ √(112.4 + 309.8) ≈ √422.2 ≈ 20.55 m.

Thus:

• Relative velocity = (5 m/s east, ‑3 m/s south).
• Minimum separation occurs at t ≈ 5.9 s.
• Minimum distance ≈ 20.6 m.