In this question, i is a unit vector due east and j is a unit vector due north.

At 0900 hours a ship sails from the point P with position vector (2i + 3j) km relative to an origin 0. The ship sails north-east with a speed of 15 km/h

problem statement is poorly written ... for part (ii) to work out, the statement "The ship sails north-east with a speed of 15 km/h" should say that the ship sails 15 km/hr east and 15 km/hr north. As written, it gives one the impression that the ship's velocity is 15 km/hr in the NE direction.

(i) Find, in terms of i and j, the velocity of the ship.

v = 15i+ 15j

(ii) Show that the ship will be at the point with position vector (24.5i + 25.5j) km at 1030 hours.

r(1.5) = (2i+ 3j) + (15i+ 15j)(1.5)

(iii) Find, in terms of i, j and t, the position of the ship t hours after leaving P.

r(t) = (2i+ 3j) + (15i+ 15j)t

At the same time as the ship leaves P, a submarine leaves the point Q with position vector (47i — 27j) km. The submarine proceeds with a speed of 25 km/h due north to meet the ship.

(iv) Find, in terms of i and j, the velocity of the ship relative to the submarine.

(sub velocity) + (relative velocity) = ship velocity

relative velocity = (ship velocity) - (sub velocity)

relative velocity = (15i+ 15j) - (25j) = (15i- 10j) km/hr

(v) Find the position vector of the point where the submarine meets the ship.

(47i- 27j) + (25j)t = (2i+ 3j) + (15i+ 15j)t

solve for t ... then determine their common position.