posted by jim .
A navy destroyer is exactly 62 miles south of an aircraft carrier, which is travelling W18°S at 24 mph. What
speed and bearing must the destroyer maintain if it is to reach the aircraft carrier in exactly 2 hours ?
pre calc -
$%^& math teachers do not know navigation. They say "bearing" when they mean "heading" but anyway:
Where will the carrier be in two hours if it starts at the origin at t = 0?
48 miles from origin at 18 degrees S of W
That puts it at x = -48 cos 18 = -45.65
and y = - 48 sin 18 = -14.83
Now the destroyer was at (0, -62) at t = 0
In two hours it must go North 62-14.83 = 47.17 miles North
In two hours it must go 45.85 miles West
distance = sqrt(45.85^2 + 47.17^2)
= 65.78 miles in two hours
so 65.78 /2 = 32.89 knots assuming those miles are nautical miles
tan angle west of north = 45.85/47.17
so angle west of north = 44.2 west of north
360 - 44.2 = 315.8
so steer a heading of 316 degrees true at 32.9 knots