Two ships P and Q are travelling at night with constant velocities - Edexcel - A-Level Maths Mechanics - Question 7 - 2005 - Paper 1

We start by writing the position vector of P at any time t:

$p = (20i + 10j) + t(3i + 8j) = (20 + 3t)i + (10 + 8t)j.$

For ship Q, since it only moves downwards,

$q = (14i - 6j) + t(3i + 12j) = (14 + 3t)i + (-6 + 12t)j.$

The distance between P and Q can be found using:

$PQ = q - p = [(14 + 3t) - (20 + 3t)]i + [(-6 + 12t) - (10 + 8t)]j$

This simplifies to:

$PQ = (-6i - 16j) + (3ti + 4tj) = (-6 - 16t) + (3t)i + (-16 + 4t)j.$

Calculating the square of the distance: $d^2 = (-6 - 16t)^2 + (3t)^2 + (-16 + 4t)^2$ simplifies to:

$d^2 = 36 + 192t + 256t^2 + 9t^2 + 256 - 128t + 16t^2$

Combining like terms gives: $d^2 = 25t^2 - 92t + 292.$

To determine when P can no longer see Q, we set up the inequality:

$d^2 ext{ must be } \leq 15^2.$ Substituting gives: $25t^2 - 92t + 292 \leq 225.$ Rearranging this, we find: $25t^2 - 92t + 67 \leq 0.$ Finding the roots using the quadratic formula: $t = \frac{-(-92) \pm \sqrt{(-92)^2 - 4(25)(67)}}{2(25)}.$ Calculating yields: $t \approx 2.68,$ which is approximately 2 hours and 41 minutes. Therefore, the observer can see the lights until 2 hours and 41 minutes after midnight.