A harbour lies in a direction 60° South of West from a fort and at a distance 30 km from it, a ship sets out from the harbour at noon and sails due East at 10 km an hour. The time at which the ship will be 70 km from the fort is
Put the fort at the origin, with x pointing east and y pointing north.
The harbour is 30 km away at 60° south of west: \(H = (-30\cos 60°, -30\sin 60°) = (-15, -15\sqrt{3})\).
After \(t\) hours sailing east at 10 km/h, the ship is at \((-15 + 10t, -15\sqrt{3})\).
Distance from the fort = 70: \((10t - 15)^2 + 675 = 4900 \Rightarrow (10t - 15)^2 = 4225 \Rightarrow 10t - 15 = 65\)
So \(t = 8\) hours after noon, that is, 8 PM.
Answer: (B) 8 PM
Reinforce your conceptual grasp with related questions from MAH MCA CET 2015:
Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.
Start Free Mock Test Series →