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

The expression for the position vector of B at time t hours after noon can be given as:

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

This simplifies to:

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

To find λ, we equate the positions of boats B and C.

Given the position vector of C:

$c = (9i + 20j) + (6i + 6j) \cdot t$

The position vectors will be equal when:

$b = c$

Substituting the earlier expressions:

$(3 + 2t)i + (-4 + 6t)j = (9 + 6t)i + (20 + 6t)j$

Equating the i components:

$3 + 2t = 9 + 6t \Rightarrow -4t = 6 \Rightarrow t = -\frac{3}{2}$

Next, for the j components:

$-4 + 6t = 20 + 6t$

This implies there is no further information obtained from j-component as it is satisfied for any t.

Thus, with the value found, λ can adjust in relation to these values based on their constant parameters.

The speed of boat B is calculated as:

$|v_B| = \sqrt{(2)^2 + (6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$

The speed of boat C can be given from its components:

$|v_C| = \sqrt{(6)^2 + (6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$

Since the questions suggest their relationship under constant parameters, adjust for comparative speeds under intercept assumptions, verifying constants suitably to demonstrate equal motion effectively.