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 found using the arctangent function:

$ext{Bearing} = 360 - \arctan\left(\frac{2.5}{6}\right) \approx 337 \text{ degrees}$

First, we calculate the displacement of the ship from 1200 to 1500, which is 3 hours. The position vector at 1500 can be calculated as:

$\text{s} = \text{s}_0 + \text{velocity} \times \text{time} = (16i + 5j) + 3(-2.5i + 6j)$

Calculating gives us:

$\text{s} = (16 - 7.5)i + (5 + 18)j = 8.5i + 23j$

Thus, the position vector of R is ( 8.5i + 23j ).

After 1400, the ship continues moving at the new speed of 5 km h⁻¹. For time ( t ) hours after 1400, the position vector can be expressed as:

$\text{s} = 8.5i + 23j + t(-2.5i + 5j) \text{, for } t \geq 0$

To find when S is due east of R, we set the j-component of the vectors equal. Using the expression from part d, we need to find ( t ) such that:

$23 + 5t = 0$

Solving for ( t ) gives:

$t = 1512 \text{ hours}$

At time 1600, which is 4 hours after 1400, we calculate the position vector of S:

$\text{s} = 8.5i + 23j + 4(-2.5i + 5j) = 8.5 - 10i + 23 + 20j = -1.5i + 43j$

The position vector of R remains (8.5i + 23j). The distance between S and R is calculated using:

$ext{Distance} = \sqrt{(-1.5 - 8.5)^2 + (43 - 23)^2} \approx 4.72 \text{ km}$