A plane leaves an aircraft carrier and flies
south at 800km/hr while the carrier proceeds in the direction 60 degrees west of north
at 40km/hr. If the plane has enough fuel to fly 6 hrs, how far south can it fly before
returning to the carrier? Find also the distance travelled by the plane during this
flight.
Let a represent the distance the plane
flew due south, b the distance the carrier travelled, and c the distance from the
furthest point south back to the carrier.
The angle between
sides a and b of the triangle is the supplement of 60 degrees, or 120
degrees.
We use the Law of Cosines:
where angle C has measure 120 degrees.
We assume the plane
uses all of the available time, thus the total flight time is 6 hours. Let t be the time
the plane flies due south. Then:
(1) b=(40
km/hr)(6hr)=320km
(2) a=(800 km/hr)(t hr)=800t
km
(3) c=(800 km/hr)((6-t) hr)=(4800-800t)km *time on
return trip is 6hr minus the time flying due south *
Then
. Thus the plane flew due south for about 2.89 hours, and flew for about 3.11 hours
returning to the carrier.
The distances are
. The total distance flown is 4800km, which agrees with a
6hr flight at 800km/hr.
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