Question

without the wind, a plane would fly due east at a rate of 150 mph. the wind is blowing southeast at a rate of 50 mph. the wind is blowing at a 45 angle from due east. what is the actual speed of the plane with the wind?

Answer 1

Split the wind vector into south/east components:

east -> 50cos(45)
south -> 50sin(45)

Plane components
east -> 150+50cos(45)
south -> 50 sin(45)

new speed is magnitude of that vector
\[\rightarrow \sqrt{(150+50\cos 45)^2 + (50 \sin 45)^2}\]

Answer 2

Got it.. thank you!

Answer 3

It also asks how far off of the due east path the wind blows it. How would I figure that out?

Answer 4

direction of a vector is given as:
\[\tan \theta = \frac{y}{x}\]
where y is "south" and x is "east" in this problem

Answer 5

Okay great. Thanks a lot! :)

Answer 6

yw