Question

A plane is flying to Minnesota with a velocity of 277.73 km/h, N. The plane

Answer 1

Answer:

466 km/h

255°

measured anticlockwise from the east direction

Explanation:

find the resultant velocity of a plane after a crosswind affects its motion. take the positive

x -direction as east, and the positive y -direction as north. The components of the plane's velocity (without wind) are

• upx = 0
• upy =  450 Km/ h

And for the wind.

• uwx = - 120 Km/h
• uwy = 0

The components for the resultant velocity of the plane are given by

• ux =upx + uwx = 0 - 120 Km/h =  - 120 Km/h

• uy = upy + uwy  = - 450 Km/h +0 = - 450 Km/h

The magnitude of the resultant velocity is

$\sqrt{(vx)^2+(vy)^2} =\sqrt{(-120lkm/h^2+(-450lkm/h)^2}$

= 466 km/h

Hope this helps u : )