Question

An airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 23° south of east. Its direction of motion relative to the earth is 45.0° south of west, while its direction of travel relative to the air is 7° south of west. What is the airplane's speed relative to the air mass in meters per second?

Answer 1

Answer:

velocity with respect to air as

Vpa = 1.83 m/s

Explanation:

Given data:

Vae = 45.0 m/s

\theta_ae (air with respect to earth)  = 23°

\theta_pe  (plane with respect to Earth)  = 45°

\theta_pa  (plane with respect to air) =7°  

We have ,

Vpe = Vpa + Vae

substituting each values with corresponding angle we get

Vpe (cos 45i + sin45 j) = Vpa (cos 7i + sin7 j) +  45 (cos 23i + sin23 j)

where,

Vpe - plane velocity with repect to earth

Vpa - planevelocity with respect to air

Vae  - air velocity with respect to earth

Solving

Vpe (0.52i + 0.85 j) = Vpa (0.73i + 0.65 j) +  45 (-0.53i - 0.84 j)

considering  i terms and j terms separately

Vpe *0.52 =  Vpa (0.73 +45 *(-0.53))

Vpe = 1.40 Vpa -45.86

Vpe *0.85 =  Vpa (0.65 +45 *(-0.84)

Vpe =0.76Vpa - 44.47

putting one value Vpe in either equation we get :

1.40 Vpa -45.86 = 0.76Vpa - 44.47

velocity with respect to air as

Vpa = 1.83 m/s