The nose of an ultralight plane is pointed south, and its airspeed indicator shows 39 m/s m/s . The plane is in a 17 m/s m/s wind blowing toward the southwest relative to the earth. For help with math skills, you may want to review: Vector Addition Resolving Vector Components For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Flying in a crosswind. Part A Letting x be east and y be north, find the components of v ⃗ P/E v→P/E (the velocity of the plane relative to the earth). Express your answers in meters per second separated by a comma. View Available Hint(s) v x vx , v y vy = nothing m/s m/s Submit Part B Find the magnitude of v ⃗ P/E v→P/E . Express your answer in meters per second. View Available Hint(s) v P/E vP/E = nothing m/s m/s Submit Part C Find the direction of v ⃗ P/E v→P/E . Express your answer in degrees. View Available Hint(s) ϕ ϕ = nothing ∘ ∘ south of west Submit Provide Feedback Next

The velocity vector has two components. Depending on the reference system they could be magnitude and direction in the polar coordinates or x-component and y-component in the rectangular coordinates system.

We are given two velocities in the form of magnitude-direction. The plane's velocity goes south at 39 m/s. The zero reference for angles is pointed East, so the south direction has a 270° angle respect to the reference. If the polar coordinates are known, the rectangular coordinates are computed as

[tex]v_{xp}=|v_p|cos\alpha_p[/tex]

[tex]v_{yp}=|v_p|sin\alpha_p[/tex]

[tex]Since\ |v_p|=39 m/s,\ \alpha_p=270^o,[/tex]

[tex]v_{xp}=39cos\ 270^o=0[/tex]

[tex]v_{yp}=39sin\ 270^o=-39[/tex]

Thus, the velocity of the plane is

[tex]\vec v_p=0\hat i-39\hat j[/tex]

The wind is blowing toward the southwest. It means its angle is 225° (3rd quadrant):

[tex]v_{xw}=|v_w|cos\alpha_w[/tex]

[tex]v_{yw}=|v_w|sin\alpha_w[/tex]

[tex]v_{xw}=17cos\ 215^o=-12.02[/tex]

[tex]v_{yw}=17sin\ 215^o=-12.02[/tex]

Thus, the velocity of the wind is

[tex]\vec v_w=-12.02\hat i-12.02\hat j[/tex]

Now we perform the vector addition to compute the plane's final speed

[tex]\vec v_t=\vec v_p+\vec v_w[/tex]

[tex]\vec v_t=0\hat i-39\hat j-12.02\hat i-12.02\hat j[/tex]

[tex]\vec v_t=-12.02\hat i-51.02\hat j[/tex]

A) The magnitude of the total velocity is

[tex]|\vec v_t|=\sqrt{(-12.02)^2+(-51.02)^2}[/tex]

[tex]\boxed{|\vec v_t|=52.42\ m/s}[/tex]

B) The direction angle is given by

[tex]\displaystyle \tan\theta=\frac{-51.02}{-12.02}=4.24[/tex]

[tex]\theta=arctan\ 4.24[/tex]

[tex]\theta=76.74^o[/tex]

This angle is west of south, we must add 180° to express it in due reference, thus

[tex]\boxed{\theta=256.74^o}[/tex]