[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 along a straight line with constant velocity - Edexcel - A-Level Maths Mechanics - Question 7 - 2010 - Paper 1

To find the direction in which S is moving, we need to calculate the angle $heta$ using the tangent function:

tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{16}{-7} \implies \theta = \tan^{-1}\left(\frac{16}{-7}\right).$$ This angle gives us a bearing; therefore: The bearing is calculated as: $$\text{bearing} = 180 + \theta \approx 180 + 36.87 \approx 217°.$$

From the positions at t=0 and t=4, we have:

At t=0: $s = 9i - 6j$
At t=4: $s = 2i + 10j.$

We can express s in terms of t:

Taking the average velocities:

$s = \left(\frac{3 + 9}{4}t + (9 - 6)j\right) \\ = (3 + 9)i + (4t - 6)j.$

To find the distances between the ship and the lighthouse, we first calculate the position of S relative to L:

The position vector of S can be expressed as:

$s = (3T + 9)i + (4T - 6)j.$

The position vector of L is given by $(18i + 6j)$. The distance d is:

$d = \sqrt{(3T + 9 - 18)^2 + (4T - 6 - 6)^2} = 10$

Now,

$\sqrt{(-3T - 9)^2 + (4T - 12)^2} = 10,$

Squaring both sides gives:

$(-3T - 9)^2 + (4T - 12)^2 = 100.$

Expanding gives:

$9T^2 + 54T + 81 + 16T^2 - 96T + 144 = 100.$

Combining terms, we have:

$25T^2 - 42T + 125 = 0.$

Using the quadratic formula:

$T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{42 \pm \sqrt{(-42)^2 - 4(25)(125)}}{50} = 1.5.$ Thus, the possible value of T is T = 1.5.