Vector application

Question

An airplane’s velocity with respect to the air is 580 mph and it is headed N 58° W. The wind, at the altitude of the plane, is from the southwest and has a velocity of 60 mph. What are the resultant speed and direction of the airplane?

Answer 1

Well, the first thing to do when solving vector problems is draw a picture. Unfortunately, I can't upload a picture, so I will do my best to describe it.

We have a plane traveling 580mph N 58° W (means you start facing North, then turn towards the West by an angle of 58 degrees). This tells us we will be in the second quadrant and 58+90=148° off the origin on an xy-axis.

Lets let V be the velocity of the aircraft. Thus

|V| = 580 mph

If we break it down into components we get.

Vx = 580cos(148) = -491.86 mph

Vy = 580sin(148) = 307.35 mph

This checks out as it is in the 2nd quadrant

The wind is originating from the SW, thus it is traveling NE which is 45° off the origin.

Let W be the velocity of the wind. Thus

|W| = 60mph

If we break it down into components we get.

Wx = 60cos(45) = 42.42mph

Wy = 60sin(45) = 42.42mph

Now we can sum the velocities to find the resulting velocity R.

R = V + W = <Vx, Vy> + <Wx, Wy> = <Vx + Wx, Vy + Wy> = < -491.86 + 42.42, 307.35 + 42.42>

Thus

R = <-449.44, 349.77>

|R| = √((-449.44)²+(349.77)²) = 569.5 mph is the resultant speed.

The angle is tanθ = 349.77/-449.44 -> θ = arctan(349.77/-449.44) = -37.89 with respect to the east direction

Thus are resultant velocity is

569.5 mph at N 52° W