## Time and Distance To Interception

A boat starts from A
$5 \mathbf{i} + 12 \mathbf{j}$
with a speed of 12 km/h on a bearing of 70 degrees. Another boat starts from the origin with a speed of 15 km/h to intercept the first ship. How long will it be before interception takes place, and where will interception take place?
The diagram illustrates the problem.

The first ship travels a distance 12T and the second ship travels a distance 15T before interception. The angle at A is
$(180-70)+tan^{-1}(\frac{5}{12})=132.6^o$
to 1 decimal place.
The distance of A from the origin is
$\sqrt{5^2+12^2}=13$
km.
The Cosine Rule gives
$(15T)^2=13^2+(12T)^2-2(13)(12T)cos(132.6)$

$225T^2=169+144T^2+211.2T$

$81T^2-211.2T-169=0$

$T=\frac{211.2 \pm \sqrt{(-211.2)^2-4(81)(-169)}}{2 \times 81} = -0.64, \; 3.25$
hours.
Obviously the solution is 3.25 hours hours, and interception takes place 15(3.25)=48.75 km from the origin.