The nose of an ultralight plane is pointed south, and its airspeed indicator shows 44 m/s. The plane is in a 18 m/s wind blowing toward the southwest relative to earth. a. letting x be east and y be north, find the components of \vec v_{\rm P/E} (the velocity of the plane relative to the earth.
b. Find the magnitude of \vec v_{\rm P/E}.
c. Find the direction of \vec v_{\rm P/E}.

a) vx = -12.7 m/s vy = -56.7m/s

b) v= 58.1 m/s

c) θ = 77.4º S of W

Explanation:

a)

In order to get the components of the velocity of the plane relative to the earth, we need just to get the components of both velocities first:Since the nose of the plane is pointing south, if we take y to be north, and positive, this means that the velocity of the plane can be written as follows:

       

Since the wind is pointing SW, it's pointing exactly 45º regarding both directions, so we can find its components as follows (they are equal each other in magnitude)

       

       

The component of v along the x-axis is simply (2), as the plane has no component of velocity along this axis:

       

The component of v along the -y axis is just the sum of (1) and (3)

b)

We can find the magnitude of the velocity vector, just applying the Pythagorean Theorem to (4) and (5):

       

c)

Taking the triangle defined by vx, vy and v, we can find the angle that v does with the negative x-axis, just using the definition of tangent, as follows:

       

Taking tg⁻¹ from (7), we get:

        tg⁻¹ θ = tg⁻¹ (4.46) = 77.4º S of W. (8)

The answer is -6hope this