Thread: problem with vector velocity

i'm not a fan a vector velocities and relative velocity..so this is my predicament

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

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

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

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

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.

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

any help would be appreciated

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.

thanks man