In what direction must the pilot head the plane in order to reach her destination?

Question

A plane has an airspeed of 400 mph. The pilot wishes to reach a destination 900 mi due east, but a wind is blowing at 50.0 mph in the direction 60.0 degrees north of east. In what direction must the pilot head the plane in order to reach her destination? How long will the trip take?

Answer 1

The pilot needs to head in a direction south of east to offset the north east wind that is blowing.

Use vectors.

For the wind the bearing is 60° north of east.
w = 50(icos60° + jsin60°) = 50(i/2 + j√3/2) = 25(i + j√3)

For the plane the bearing is unknown south of east. Use θ.
p = 400(icosθ + jsinθ)

The resultant path of the plane in the wind is p + w and is due east. This means the j component nets out to zero.

p + w = 25(i + j√3) + 400(icosθ + jsinθ)
p + w = (25 + 400cosθ)i + (25√3 + 400sinθ)j
p + w = 25(1 + 16cosθ)i + 25(√3 + 16sinθ)j

√3 + 16sinθ = 0
16sinθ = -√3
sinθ = -√3/16
θ = arcsin(-√3/16) ≈ -6.2146286°

The pilot must head 6.2146286° south of east.
___________________

cosθ = √(1 - sin²θ) = √(1 - 3/256) = √(253/256) = √253/16

Plug in for sinθ and cosθ.

p + w = 25(1 + 16cosθ)i + 25(√3 + 16sinθ)j
p + w = 25[1 + 16√253/16]i + 25[√3 - 16(√3/16)]j
p + w = 25[1 + √253]i + 0j
p + w = 25[1 + √253]i

The length of time to travel 900 miles due east is:

t = 900 / {25[1 + √253]} = 36 / (1 + √253) ≈ 2.1294248 hours
t ≈ 2 hours, 7 minutes, 45.929342 seconds

Answer 2

I will take the plane up higher to flight level 15,000 feet above the cross wind and keep heading East. ETA..2 hours and 10 min.

Answer 3

Head down. Down the runway. He'll never reach his destination sitting at the gate pondering side-slip due to wind.

Answer 4

2hrs 10 min, and she needs to head east