A ship S is moving with constant velocity $(-2.5i + 6j)$ km h$^{-1}$ - Edexcel - A-Level Maths Mechanics - Question 7 - 2006 - Paper 1

The bearing can be calculated using the arctangent of the velocity components:
$\text{Bearing} = 360 - \arctan\left(\frac{6}{-2.5}\right)$
Which leads to:
$\text{Bearing} = 360 - 67.79 = 337^\circ$

To find the position vector of submerged rock R, we determine the position of S at time 1500. The time difference from 1200 to 1500 is 3 hours. Thus, we can calculate the position vector of S at this time:
$s = (16i + 5j) + 3(-2.5i + 6j) = (16 - 7.5)i + (5 + 18)j = 8.5i + 23j$
Therefore, the position vector of R is $R = 8.5i + 23j$.

After 1400, the ship continues in the same direction until 1500 (i.e., for 1 hour), and then it starts moving towards the new course.
The position vector at 1400 is:
$s = (16i + 5j) + 1(-2.5i + 6j) = (16 - 2.5)i + (5 + 6)j = 13.5i + 11j$
For t hours after 1400, the position becomes:
$s = (13.5 - 2.5 \times (t-1))i + (11 + 5(t-1))j$

For S to be due east of R, the y-coordinate of S must equal the y-coordinate of R.
We set up the equation:
$11 + 5t = 23$
Solving for t gives:
$5t = 12 \Rightarrow t = \frac{12}{5} = 2.4 \text{ hours after 1400}$
So S will be due east of R at 16:24.

At time 1600, the position of S can be computed as continues the new path. The position vector at 1600 is:
$s = (13.5 - 2.5 \times 2) i + (11 + 5 \times 2) j = (13.5 - 5)i + (11 + 10)j = 8.5i + 21j$
Now calculating the distance from R:
$\text{Distance} = \sqrt{(8.5 - 8.5)^2 + (21 - 23)^2} = \sqrt{0 + (-2)^2} = \sqrt{4} = 2 \text{ km}$