[In this question i and j are horizontal unit vectors due east and due north respectively and position vectors are given with respect to a fixed origin.] A ship S is moving with constant velocity (−12i + 7.5j) km h^{-1} - Edexcel - A-Level Maths Mechanics - Question 6 - 2012 - Paper 1

The position vector of S after t hours can be expressed as:

$s = 40i - 6j + t(-12i + 7.5j)$

Simplifying gives:

$s = (40 - 12t)i + (-6 + 7.5t)j$

Substituting ( t = 3 ) into the equation for s, we have:

$s = (40 - 12(3))i + (-6 + 7.5(3))j$

Calculating this leads to:

$s = (40 - 36)i + (-6 + 22.5)j = 4i + 16.5j$

The position vector of B is ( (7i + 12.5j) ). The distance between S and B is given by:

$SB = \sqrt{(4 - 7)^2 + (16.5 - 12.5)^2} = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ km}$

To determine when S is due north of B, the x-component of S must be equal to the x-component of B. Set:

$40 - 12t = 7$

Solving, we get:

$12t = 33 \implies t = \frac{33}{12} = 2.75 \text{ hours}$

Now substitute ( t = 2.75 ) into the expression for s:

$s = (40 - 12(2.75))i + (-6 + 7.5(2.75))j$

Calculating this gives:

$s = (40 - 33)i + (-6 + 20.625)j = 7i + 14.625j$

Finally, the distance of S from B is:

$SB = \sqrt{(7 - 7)^2 + (14.625 - 12.5)^2} = \sqrt{0 + (2.125)^2} = 2.125 \text{ km}$