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?

velocity with respect to air as

Vpa = 1.83 m/s

Explanation:

Given dа ta:

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