A ship S is moving with constant velocity (3i + 3j) km h⁻¹ - Edexcel - A-Level Maths Mechanics - Question 6 - 2013 - Paper 2

Ship T's position at time t is given by the same formula:

$\text{Position vector of T} = (6i + j) + (-2i + rj) \times t$

Which simplifies to:

$(6 - 2t)i + (1 + rt)j$

Setting the position vectors of ships S and T equal to each other since they meet at point P:

$(-4 + 3t)i + (2 + 3t)j = (6 - 2t)i + (1 + rt)j$

Equating the components:

1. For the i-component:
   $-4 + 3t = 6 - 2t$
   Solving gives:
   $5t = 10 \Rightarrow t = 2$

2. For the j-component:
   $2 + 3(2) = 1 + r(2)$
   Substituting t = 2: $2 + 6 = 1 + 2r$
   Thus, $8 = 1 + 2r \Rightarrow 2r = 7 \Rightarrow r = 3.5$

The value of n is therefore:

$n = 3.5$.

To find the distance OP, we first need the position vector of P. Using t = 2 from part (b):

$\text{Position vector of P} = (-4 + 3 \times 2)i + (2 + 3 \times 2)j$
This simplifies to:
$(-4 + 6)i + (2 + 6)j = 2i + 8j$

Now, we calculate the distance OP using the formula for the distance between two points:

$OP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Given the position vector of O is (0i + 0j) and for P, (2i + 8j):

$OP = \sqrt{(2 - 0)^2 + (8 - 0)^2} = \sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} \approx 8.246 \text{ km}$

Thus, the distance OP is approximately:

$OP \approx 8.25 \text{ km}$.