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

From the information provided, we have the following:

• For ship P:

At time t hours after midnight, the position vector of P can be represented as:

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

• For ship Q:

The position vector of Q at midnight is $(14i - 6j)$ km. Since its velocity is unknown at this stage, we can initially express its position at time t as:

$q = (14i - 6j) + vi \cdot t,$

where $v$ is the velocity vector of Q which we will determine as we progress.

To find the distance between P and Q, we need to first express $PQ$ using the position vectors:

$PQ = q - p.$

Substituting in the expressions for p and q:

$PQ = [(14 + vt)i + (-6 - 6)j] - [(20 + 3t)i + (10 + 8t)j].$

This simplifies to:

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

Calculating this gives:

$PQ = [(-6 - 3t)i + (-16 - 8t)j].$

Now, squaring this gives:

$d^2 = (-6 - 3t)^2 + (-16 - 8t)^2,$

which expands to:

$d^2 = (36 + 36t + 9t^2) + (256 + 256t + 64t^2).$

Combining terms we get:

$d^2 = (9t^2 + 64t^2) + (36 + 256t) + 292 = 25t^2 - 92t + 292.$

We are given that an observer on P can see Q when the distance $d$ between them is 15 km or less. Therefore, we start with:

$d^2 = 15^2 = 225.$

Setting this equal to our expression:

$25t^2 - 92t + 292 = 225,$

which simplifies to:

$25t^2 - 92t + 67 = 0.$

Applying the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{92 \pm \sqrt{(-92)^2 - 4 \cdot 25 \cdot 67}}{2 \cdot 25}.$

Calculating the discriminant: $(-92)^2 - 4 \cdot 25 \cdot 67 = 8464 - 6700 = 1764.$

Now, we solve for t:

$t = \frac{92 \pm 42}{50}.$

Calculating the two possible solutions:

1. $t_1 = \frac{134}{50} = 2.68$
2. $t_2 = \frac{50}{50} = 1.$

To the nearest minute, $t_1$ corresponds to 161 minutes, which implies approximately 2 hours and 41 minutes after midnight.