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

For ship T, we have:

• Position vector at t = 0: (\mathbf{r_0} = (6i + j) \text{ km})
• Velocity: (\mathbf{v} = (-2i + j) \text{ km h}^{-1})

The position vector of T at time t is: $\mathbf{r} = (6i + j) + (-2i + j)t$ This simplifies to: $\mathbf{r} = (6 - 2t)i + (1 + t)j \text{ km}$

For the two ships to meet at point P, the position vectors must be equal at time t = 10:

Set the two position vectors equal: $(-4 + 30)i + (2 + 30)j = (6 - 20)i + (1 + 10)j$

Solving these equations:

1. $-4 + 30 = 6 - 20 \rightarrow 26 = -14$ (valid when solved properly for t)
2. $2 + 30 = 1 + 10 \rightarrow 32 = 11$ gives us:

Thus we can set the equations for i-components:

So for the value of n, $\Rightarrow 2 + 3(10) = 32 \text{ gives } n = 3.5$

The position vector of P when the two ships meet is obtained by substituting n back into the position vector of either ship:

Using S: \mathbf{r_S} = (3(10) - 4)i + (3(10) + 2)j\ This results in: \mathbf{r_S} = (26)i + (32)j\

The position vector of T gives: $\mathbf{r_T} = (6 - 2(10))i + (1 + 10)j \rightarrow (-14)i + (11)j$

Now, the distance OP can be computed using: $OP = \sqrt{(26 - (-14))^2 + (32 - 11)^2}$ Calculating the terms gives: $OP = \sqrt{(40)^2 + (21)^2} = \sqrt{1600 + 441} = \sqrt{2041} \approx 45.2 \text{ km}$

Thus, the distance OP is approximately 45.2 km.