A ship needs to evacuate a seriously sick sailor. The ship’s captain will not stop the ship nor change course. The ship is traveling on a bearing of 80 at a speed of 47 km/h . When the decision is made to evacuate the sailor, the ship was located 10 km on a bearing of 200 from the helicopter rescue base.
How do you use optimization with this?
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A seriously sick sailor has to be evacuated from a ship. As the ship's captain will neither change the course of the ship nor stop it, a helicopter has to be used to lift the sailor from the ship. When it is decided that this has to be done, the ship is located 10 km at a bearing of 200 from the helicopter base and is traveling at a speed of 47 km/h at a bearing of 80.
Optimization can be used here to minimize the distance that the helicopter has to fly. The least distance that the helicopter has to fly and the time at which the ship reaches there can be determined.
The helicopter is closest to the base when the length of the perpendicular drawn from the base to the line representing the path followed by the ship is the shortest. Let the time when this happens be T.
The length of the perpendicular is given by D where D = 47*T*tan 60 and D^2 + (47*T)^2 = 10^2
=> `(47*T*sqrt 3)^2 + (47*T)^2 = 100`
=> 6627*T^2 + 2209*T^2 = 100
=> T^2 = 100/8836
=> T = 0.106 h
D = 8.66 km
The ship would be closest to the base after 0.106 hours and at a distance of 8.66 km.
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