Question

A bird flies north with a velocity of 10.00 meters/second relative to the air. It encounters a crosswind blowing at a velocity of 3.00 meters/second in the easterly direction. What is the magnitude and direction of the resultant velocity of the bird relative to the ground? Hint: A crosswind blows perpendicular to the direction in which the bird is flying.

Answer 1

Answer:

Magnitude = 10.44

Direction = 73.3°

Explanation:

Since the directions are perpendicular you could think of the birth flying north as moving through the "y" and the crosswind moving through the "x" in a coordinate system.

The birth is flying north so "y" is positive and the crosswind is blowing easterly so its moving to the right, then "x" is also positive.

Then you could assume that the vector you need to calculate starts in the origin (0,0) and the second point is (3,10)

Magnitude = $\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}} \\= \sqrt{10^{2}+3^{2}} \\= 10.44$

Direction :

$tan(\theta) = \frac{(y_{2}-y_{1})}{(x_{2}-x_{1})} = \frac{10}{3} \\\theta = tan^{-1}(\frac{10}{3}) \\\theta = 73.3 \°$