Question

An airplane pilot wishes to fly to his destination 500 mi due west, but the wind is blowing at 23.1 mi/hr toward 14.1° north of west. The speed of the plane relative to the air is 166.5 mi/hr. At what angle (south of west) must the pilot orient the plane in order to fly to his destination directly?

If the pilot orients the plane at that angle, what will be the speed of the plane relative to the ground?

Answer 1

Destination is due west. So the resultant velocity should be in west direction.

Velocity of wind with respect to ground, v_wg = 23.1 mph 14.1 degree north of west = 23.1 cos 14.1 west + 23.1 sin 14.1 north = (22.404 west + 5.63 north) mph

Velocity of plane relative to air, v_pa = 166.5 mph at angle theta south of west = (166.5 cos theta west + 166.5 sin theta south) mph

Velocity of plane with respect to ground = velocity of plane with respect to air + velocity of air with respect to ground = v_pa + v_ag = (166.5 cos theta west + 166.5 sin theta south) +  (22.404 west + 5.63 north)

= (22.404 + 166.5 cos theta) west + (5.63 - 166.5 sin theta) north

Since the resultant velocity should be west (plane has to travel west),

so (5.63 - 166.5 sin theta) = 0

=> theta = 1.937 degree

angle (south of west) must the pilot orient the plane in order to fly to his destination directly = 1.937 degree south of west

speed of plane relative to ground =  (22.404 + 166.5 cos theta) west mph =  (22.404 + 166.5 cos 1.937) west = 188.81 mile/hr due west