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December 18, 2014

December 18, 2014

Posted by **jim** on Thursday, December 8, 2011 at 9:45pm.

speed and bearing must the destroyer maintain if it is to reach the aircraft carrier in exactly 2 hours ?

- pre calc -
**Damon**, Thursday, December 8, 2011 at 10:03pm$%^& 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

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