Question

An airplane flies due north at 225 km/h relative to the air. There is a wind blowing at 60 km/h to the northeast relative to the ground. What are the plane's speed and direction relative to the ground?

Answer 1

Answer:

270.78 m/s and 9 degrees north of east

Explanation:

Suppose that the wind direction is 45 degree relative to the eastern direction. Let east and north be the positive horizontal ($\hat{i}$) and vertical ($\hat{j}$) directions here. We can calculate the horizontal and vertical components of wind velocity

$v_v = v_h = 60 cos45^0 = 60*0.707 = 42.43 km/h$

The velocity of the wind can be rewritten as the following

$\vec{v_w} = 42.43 \hat{i} + 42.43 \hat{j}$

The velocity of the airplane that is 225 km/h due north is

$\vec{v_a} = 225 \hat{j}$

The plane velocity relative to the ground is its velocity relative to the wind plus the wind velocity relative to the ground

$\vec{v} = \vec{v_w} + \vec{v_a} = 42.43 \hat{i} + 42.43 \hat{j} + 225 \hat{j} = 42.43 \hat{i} + 267.43 \hat{j}$

The magnitude and direction of this velocity vector can be calculated

$v = \sqrt{v_x^2 + v_y^2} = \sqrt{42.43^2 + 267.43^2} = \sqrt{1800.3049 + 71518.8049} = \sqrt{73319.1098} = 270.78$

$tan\alpha = \frac{v_x}{v_y} = \frac{42.43}{267.43} = 0.16$

$\alpha = tan^{-1}0.16 = 0.16 rad \approx 9.02 degrees$  north of east