[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, we calculate the angle the velocity vector makes with the positive x-axis.

Using the tangent:

Given:

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

The change in position can be distributed linearly, hence: $s(t) = s(0) + tv$ Substituting the known values: $s(t) = (9i - 6j) + t\left(-\frac{7}{4}i + 4j\right) = (9 - \frac{7t}{4})i + (-6 + 4t)j$

To express it as: $s(t) = (3 + 9)i + (4t - 6)j,$ it is evident from the values rearranging and confirming the constants are maintained in the final expression.

To find the position vector of S relative to L, we calculate: $\text{Position of } S: \ (3t + 9)i + (4t - 6)j$ $\text{Position of } L: \ (18i + 6j)$

The difference in position: $\left(3t + 9 - 18, (4t - 6) - 6\right) = (3t - 9)i + (4t - 12)j$

Applying the distance formula: $\sqrt{(3t - 9)^{2} + (4t - 12)^{2}} = 10$

Squaring both sides: $(3t - 9)^{2} + (4t - 12)^{2} = 100$ $(9t^{2} - 54t + 81) + (16t^{2} - 96t + 144) = 100$

Simplifying: $25t^{2} - 150t + 125 = 100$ $25t^{2} - 150t + 25 = 0$

Dividing through by 25: $t^{2} - 6t + 1 = 0$

Using the quadratic formula: $t = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}$

Thus, the possible values of T are approximately:\n(T \approx 3 + 2\sqrt{2} \text{ and } T \approx 3 - 2\sqrt{2}).