Question

A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 2.79 m/s. The airspeed of the bee (i.e., its speed relative to the air) is 8.75 m/s. In which direction should the bee head in order to fly directly to the flower, due North relative to the ground? Answer in units of ◦ East of North.

Answer 1

Answer:

The bee should fly 17.7º east of north.

Explanation:

Please, see the figure for a description of the problem.

Assuming the hive as the center of our reference system

The vector velocity of the bee is:

v bee = (0, 8.75)

The vector velocity of the wind is:

v wind = (-2.79, 0)

Without course correction, the resulting vector velocity of the bee would be:

v bee + v wind = (0, 8.75) + (-2.79, 0) = (0-2.79, 8.75 + 0) = (-2.79, 8.75)

to anulate the effect of the wind, the bee should move with a velocty vector: v corrected = (2.79, 8.75)  (see figure).

Now, the sum of the vector velocity of the wind and v correctd will give this resulting vector:

v wind + v corrected = (-2.79, 0) + (2.79, 8.75) = (0, 8.75) which is the original vector of the bee heading north.

Using trigonometry we can obtain the angle θ. (see figure)

sin θ = opposite / hypotenuse

then:

sin θ = magnitude of v wind / magnitude of v corrected

or

cos θ = adjacent / hypotenuse

cos θ = magnitude of v bee / magnitude of v corrected

magnitude of v wind = $|v wind| = \sqrt{(-2.79)^{2}+0^{2}} = 2.79$

magnitude of v corrected = $|v corrected| = \sqrt{(2.79)^{2} + (8.75)^{2}}= 9.18$

Then:

sin θ = 2.79 / 9.18

θ = 17.7º

The bee should fly 17.7º east of north.