Ok, I will answer this question.
(i) If u look closely it says that the ship sails 'north east'. hence its unit vector north east will be (i+j)/ square root of 2. And since velocity has both speed and direction, its velocity = 15* square root of 2 * the unit vector which = 15i + 15j
(ii) Since we have established that the velocity in the stated direction is 15i + 15j, the distance it has traveled at 10:30 can be found by multiplying 1.5 into the velocity. which = 22.5i + 22.5j. But since they have asked for the position vector, it will be equal to 22.5i + 22.5j + 2i + 3j = 24.5i + 25.5j
(iii) Similarly for part three; the distance travelled t hours after leaving = (15t)i + (15t)j. Hence in terms of 't' the position will be = (15t + 2)i + (15t + 3)j
(iv) In this subquestion they have simply asked for the velocity of the ship relative to the submarine (Vssu) = Velocity of ship - Velocity of submarine =
(15i + 15j) - (25j). NOTE: 25j is because it is only travelling north. = 15i -10j
Answer to final part in the next post