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

At the meeting point P, the position vectors of ships S and T must be equal:

For ship T, the position vector at time t is: $ext{r}_T = (6i + j) + (-2i + nj)t$

The position vector of T simplifies to: $ext{r}_T = (6 - 2t)i + (1 + nt)j \text{ km}.$

Setting the position vectors of S and T equal:

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

From the i-components: $-4 + 3t = 6 - 2t \implies 5t = 10 \implies t = 2.$

From the j-components: $2 + 3(2) = 1 + 2n \implies 2 + 6 = 1 + 2n \implies 8 = 1 + 2n \implies 2n = 7 \implies n = 3.5.$

The position vector of ship P at the meeting point is: $ext{r}_P = (6i + j) + (-2i + 3.5j)(2)$

Calculating gives: $ext{r}_P = (6 - 4)i + (1 + 7)j = 2i + 8j.$

The distance OP can be found using the distance formula: