Answer:
The speed of the airplane is 364,91 mi divided by hr, the direction is 4º40' relative to the line from east to west.
Explanation:
Step 1:
First, we must place the velocity vector of the plane on the horizontal line that joins east and west. This vector has module 334 mi divided by hr and positive direction, taking as reference the positive direction from east to west.
Then we place the wind speed vector, from module 42 mi divided by hr, at 45º downwards with respect to the previous vector, in the southwest direction.
Step 2: Now we must decompose the forces into others that are eastbound to the west and northbound to the south.
Fy: Force in a North-South direction
Fx: Force from East to West
then:
Fx=334 mi/hr + 42 mi/hr * cos (45º) = 363,7 mi/hr
Fy= 42 * without(45º) = 29,7 mi/ hr
Step 3: Once the forces are decomposed, we must calculate the alpha angle of inclination with respect to the east-west direction.
tg (alpha) = Fy / Fx
(alpha) = arctg (Fy / Fx)
(alpha) = arctg (29,7 / 363,7) = 4º 40'
Step 4: Finally we calculate the module of the final velocity vector of the plane.
Fp: Final module of the plane's speed with respect to the ground.
Fp= [tex]\sqrt{(Fx)^{2}+(Fy)^{2} }[/tex]
Then the final answer indicates that the plane will have a speed equal to 364.91 mi / hr and an address from 4º40' downwards with respect to the horizontal line from east to west.
