Additional Mathematics – Vectors in two dimensions | e-Consult

Answer 1

1. Represent the boat’s velocity in still water: V_b = 20 km/h due east → components (20, 0).
2. The current is 12 km/h from the north‑east, which means it flows toward the south‑west at 45° to the south of west. Its components:

   • West component = 12 cos 45° = 12 × 0.7071 ≈ 8.49 km/h (negative x‑direction).
   • South component = 12 sin 45° = 8.49 km/h (negative y‑direction).

   So V_c = (‑8.49, ‑8.49).

3. Resultant ground velocity V_g = V_b + V_c = (20‑8.49, 0‑8.49) = (11.51, ‑8.49).
4. Magnitude:

   
   |V_g| = √(11.51² + (‑8.49)²) ≈ √(132.5 + 72.1) ≈ √204.6 ≈ 14.3 km/h.

5. Direction: angle θ south of east = tan⁻¹(|‑8.49| / 11.51) ≈ tan⁻¹(0.738) ≈ 36.5°.

Therefore the boat’s actual speed is approximately 14.3 km/h at a bearing of 36.5° south of east**.