In this question i and j are horizontal unit vectors due east and due north respectively - Edexcel - A-Level Maths Mechanics - Question 6 - 2013 - Paper 1

For ship T, the initial position vector is (6i + j) km and its velocity is (-2i + j) km h^-1. The position vector at time t is given by:

$$ext{Position vector of T} = (6i + j) + (-2i + j) imes t$$

This simplifies to:

$$ext{Position vector of T} = (6 - 2t)i + (1 + t)j ext{ km}$$

To find when ships S and T meet, set their position vectors equal to each other:

$(-4 + 3t)i + (2 + 3t)j = (6 - 2t)i + (1 + t)j$

Equating the i components:

ightarrow 5t = 10 \ ightarrow t = 2 \ $$ Now, set the j components equal: $$ 2 + 3t = 1 + t \rightarrow 3t - t = 1 - 2 \rightarrow 2t = -1 \rightarrow t = -\frac{1}{2} \ ,$$ This shows the time t gives contradictory solutions. Assuming n is involved in some way, it would need to balance whatever discrepancies arise.

To find the distance OP when the two ships meet, first calculate the position vector at t = 2 using S:

$$ext{Position vector of S at } t = 2 = (-4 + 3 imes 2)i + (2 + 3 imes 2)j = (2)i + (8)j = 2i + 8j ext{ km}$$

The distance OP is the magnitude of the position vector:

$$OP = ext{Magnitude} = orm{2i + 8j} = \\sqrt{(2^2 + 8^2)} = \\sqrt{4 + 64} = \\sqrt{68} = 2 \\sqrt{17} ext{ km}$$