Question

An airplane flies with a velocity of 60.0 m/s [40o N of W] with respect to the air (this is known as air speed). If the velocity of the airplane according to an observer on the ground is 50.0 m/s [60o N of W], what was the wind velocity?

Answer 1

✪ᴀɴsᴡᴇʀ✪

• The wind is blowing 21.49 m/s, 12.72º North of East.

★ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ★

Let

• unit vector along East be $\hat{i}$
• unit vector along North be $\hat{j}$
• $\vec{V}_{air}$ be velocity w.r.t air
• $\vec{V}_{gr}$ be velocity w.r.t ground
• $\vec{V}_{wind }$ be wind velocity

Velocity w.r.t air

✈︎$\vec{V}_{air} = 60(-cos(40)\hat{i} + sin(40)\hat{j})$

✈︎$\vec{V}_{air} ≈ -45.96\hat{i} + 38.57\hat{j}$

Velocity w.r.t ground

✈︎$\vec{V}_{gr} = 50(-cos(60)\hat{i} + sin(60)\hat{j})$

✈︎$\vec{V}_{gr} ≈ -25\hat{i} + 43.30\hat{j}$

Wind Velocity

✈︎$\vec{V}_{gr} = \vec{V}_{air} + \vec{V}_{wind}$

✈︎$\vec{V}_{wind} = \vec{V}_{gr} - \vec{V}_{air}$

✈︎$\vec{V}_{wind} ≈ (-25\hat{i} + 43.30\hat{j}) - (-45.96\hat{i} + 38.57\hat{j})$

✈︎$\vec{V}_{wind} ≈ 20.96\hat{i} + 4.73\hat{j}$

Magnitude of Wind Velocity

✈︎$|\vec{V}_{wind}| ≈ |20.96\hat{i} + 4.73\hat{j}|$

✈︎$|\vec{V}_{wind}| ≈ \sqrt{20.96² + 4.73²}$

✈︎$|\vec{V}_{wind}| ≈ 21.49$ m/s

Direction of Wind velocity

✈︎$tan(θ) ≈ \frac{4.73}{20.96}$

✈︎$θ ≈ {tan}^{-1}(\frac{4.73}{20.96})$

✈︎$θ ≈ 12.72º$

The wind is blowing 21.49 m/s, 12.72º North of East.