A pilot wants to fly from Dallas to Oklahoma City, a distance of 330 km at an angle of 10.0° west of north. The pilot heads directly toward Oklahoma City with an air speed of 200 km/h. After flying for 1.0 h, the pilot finds that he is 15 km off course to the west of where he expected to be after one hour assuming there was no wind. (a) What is the velocity and direction of the wind? (b) In what direction should the pilot have headed his plane to fly directly to Oklahoma City without being blown off course?

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tiara143 tiara143 · Answer 1 · 2017-08-15T19:27:26+02:00

Ans. (a) It is clear from the question that, the velocity of the wind = 15 km/h and the direction of the wind is westwards.

(b) now we need to find the direction should the pilot have headed his plane to fly directly to Oklahoma City without being blown off course:

The velocity in westward direction (without the wind) = 200 sin (10)

= 34.729 or 34.73 km/hr

Since the wind is supplying a velocity of 15 km/hr , so along with the wind, his flight angle is required to supply the following velocity:

= 34.73 – 15 = 19.73 km/hr

The direction will be in a westward direction.

Now, 200 sin (∅) = 19.73

or ∅ = [tex] sin^{-1} [/tex] (19.73/200)

∅ = 5.6613 or 5.70

Therefore, the direction in which the pilot should head his plane to fly directly to Oklahoma city with being blown off course is: west of North