A boat B is moving with constant velocity - Edexcel - A-Level Maths Mechanics - Question 7 - 2007 - Paper 1

The position vector of boat B at time $t$ hours after noon is given by the initial position plus the change in position due to velocity.

Thus, $b = (3i - 4j) + t(2i + 6j) = (3 + 2t)i + (-4 + 6t)j.$

Boat C's position vector at time $t$ hours after noon is given by: $c = (9i + 20j) + λ(6i + 6j).$

For C to intercept B, their position vectors must be equal: $(3 + 2t)i + (-4 + 6t)j = (9 + 6λ)i + (20 + 6λ)j.$

Comparing components:

1. For the i-component: $3 + 2t = 9 + 6λ \Rightarrow 6λ = 2t - 6 \Rightarrow λ = \frac{2t - 6}{6}.$
2. For the j-component: $-4 + 6t = 20 + 6λ \Rightarrow 6λ = 6t - 24 \Rightarrow λ = \frac{6t - 24}{6} = t - 4.$

Setting the two expressions for $λ$ equal gives: $\frac{2t - 6}{6} = t - 4 \Rightarrow 2t - 6 = 6t - 24 \Rightarrow 4t = 18 \Rightarrow t = \frac{18}{4} = 4.5.$

Substituting to find $λ$ gives: $λ = \frac{2(4.5) - 6}{6} = 1.$

To find the speed of both boats, we calculate the magnitudes of their velocities:

For boat B: $v_B = \sqrt{(2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} ext{ km/h}.$

For boat C: The velocity of C can be found from its position vector. The components related to $λ$ (with $λ = 1$) gives: $c = (9i + 20j) + (6i + 6j) = (15i + 26j) ext{ km.}$

Thus, the velocity of C is: $v_C = \sqrt{(6)^2 + (6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} ext{ km/h.}$

Both speeds are equal as shown. Therefore, before C intercepts B, both boats moving at the same speed.