In this question i and j are horizontal unit vectors due east and due north respectively - Edexcel - A-Level Maths Mechanics - Question 6 - 2013 - Paper 1

To find the value of n where ships S and T meet at point P, we set their position vectors equal to each other.

Position vector of T at time t:

• Initial Position Vector of T: $(6i + j)$ km
• Velocity of T: $(-2i + j)$ km h^-1

At time t, the position vector of T is:

$$ext{Position Vector of T} = (6 - 2t)i + (1 + t)j \text{ km}$$

Set the position vectors of S and T equal:

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

From the i-components:

$$-4 + 3t = 6 - 2t$$

Solving for t:

$$5t = 10 ightarrow t = 2 ext{ hours}$$

Now, substituting $t = 2$ into the j-components:

$$2 + 3(2) = 1 + 2 \rightarrow 2 + 6 = 1 + 2(n) ightarrow 8 = 1 + 2n ightarrow 2n = 7 ightarrow n = 3.5$$

Thus, the value of n is 3.5.

To find the distance OP, we first need the position vector of ship S at the meeting time:

Position vector of S at t = 2:

$$ext{Position Vector of S} = (-4 + 3 \cdot 2)i + (2 + 3 \cdot 2)j = (2)i + (8)j \text{ km}$$

Now, the position vector of T at t = 2:

$$ext{Position Vector of T} = (6 - 2\cdot2)i + (1 + 2)j = (2)i + (3)j \text{ km}$$

Both ships meet at point P, whose coordinates are $(2, 8)$ and $(2, 3)$ respectively. To find the distance OP (from origin O at (0, 0) to point P), we use the formula:

$$ext{Distance} = \\sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } o ext{where} o (x_1, y_1) = (0, 0) ext{ and } (x_2, y_2) = (2, 8).$$

Thus,

$$ext{Distance} = \\sqrt{(2 - 0)^2 + (8 - 0)^2} = \\sqrt{4 + 64} = \\sqrt{68} = 2\sqrt{17} \text{ km}.$$