A ship S is moving with constant velocity (3i + 3j) km h⁻¹ - Edexcel - A-Level Maths Mechanics - Question 6 - 2013 - Paper 1

To find the value of n, we need to establish the position equations for both ships.

For ship S:

• Position vector after time t: $S(t) = (-4 + 3t)i + (2 + 3t)j$

For ship T:

• Position vector after time t: $T(t) = (6 - 2t)i + (1 + t)j$

Setting the two position vectors equal since the ships meet: $(-4 + 3t)i + (2 + 3t)j = (6 - 2t)i + (1 + t)j$

This leads to:

1. For the i component: $-4 + 3t = 6 - 2t$
2. For the j component: $2 + 3t = 1 + t$

Solving these equations gives us the time t and subsequently the required value for n.

The distance OP can be found using the position vectors of points O and P.

If P is represented by the position vector of either ship at the time they meet (let's use S for convenience):

$P = (-4 + 3t, 2 + 3t)$

The distance OP can be calculated using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Where:

• $(x_1, y_1)$ is the origin O (0, 0)
• $(x_2, y_2)$ is the position vector of P.

Thus, substituting we get: $d = \sqrt{((-4 + 3t) - 0)^2 + ((2 + 3t) - 0)^2}$

This simplifies to the final expression for distance.