*Here we present three solutions to the one problem. The vector solution is probably the slickest. The geometry solution here can be simplified by being less rigorous, and the coordinate geometry solution might be made easier by using the formula.*

### LCHL 2007, Q. 2(a)

Ship B is travelling west at 24 km/h. Ship A is travelling north at 32 km/h.

At a certain instant ship B is 8 km north-east of ship A.

(i) Find the velocity of ship A relative to ship B.

(ii) Calculate the length of time, to the nearest minute, for which the ships

are less than or equal to 8 km apart.

*Solution to (i): *We have that and and so

.

### Vector Approach to (ii)

First of all we draw a picture. As we are talking relative to ship B we will put B at the origin. If ship B is 8 km north-east of ship A then ship A is 8 km south-west of ship B.

Where is the initial displacement of ship A relative to ship B, the displacement of ship A relative to ship B, *as a function of time*, is given by

.

Using some trigonometry — vertical and horizontal components of — we have

,

and so

.

We are interested in finding out for how long this displacement of ship A from ship B is less than eight:

.

This is a -quadratic and so is negative *between the roots*, where

.

### Coordinate Geometry Approach to (ii)

We have:

Relative to B, ship A travels along a line —and the set of points a distance eight from comprise a circle, so we have:

The line through has slope equal to (why) and has as a point the initial position of A relative to B,

and so the equation of the line through is:

.

The equation of the circle is . These curves — the line and the circle — intersect at and — and we already know where is.

To find the intersection, substitute into :

.

The sensible thing here is to use the formula but this actually has nice factors. Multiply . Consider . The sum is 172 — too big — but try : bingo .

So we can rewrite

Setting the first factor equal to zero gives . The other gives

.

We can find the corresponding coordinate by substituting back into :

.

Now the distance between these points is found using the distance formula:

.

The speed of ship A relative to B is

,

and so using .

### Geometric/Trigonometric Approach to (ii)

*There are probably a number of geometric/trigonometric approaches. One approach is to drop a perpendicular from to — probably the easiest way. Here is another approach.*

We have:

We have — and the set of points a distance eight from comprise a circle. Therefore we consider the following — ship A is headed to (relative to ship :

Now produce the line segment and the line segment . Note that and . Define .

*(It is possible to get an approximate value for using — and when I saw that the answer was to be approximate (nearest minute) — I thought perhaps I should have done this instead of what follows.)*

Using

,

.

Using the following model triangle:

we have . Now apply the cosine rule to :

;

we know that is non-zero and so .

The speed of ship A relative to B is

,

and so using .

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