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fredd [130]
3 years ago
6

Matt and Anna Killian are frequent fliers on​ Fast-n-Go Airlines. They often fly between two cities that are a distance of 1575

miles apart. On one particular​ trip, they flew into the wind and the trip took 4.5 hours. The return trip with the wind behind​ them, only took about 3.5 hours. Find the speed of the wind and the speed of the plane in still air.
Physics
1 answer:
marin [14]3 years ago
6 0

Answer:

Speed of wind = 50mi/hr, Speed of plane in still air = 400mi/hr

Explanation:

Let the speed of the wind = Vw,

Speed of the plane in still air = Vsa,

The first trip the average speed of the plane = 1575mi/4.5hours = 350mi/hr

The coming trip the wind behind = 1575mi/3.5hrs = 450

Write the motion in equation form

First trip ( the plane flew into the wind)

Vaverage = Vsa - Vw

350 = Vsa - Vw

Second trip the wind was behind

450 = Vsa +Vw

Adding the two equation

800 = 2Vas

Vas = 800/2 = 400mi/hr

Substitute for Vas into equation 1

350mi/hr = 400mi/hr - Vw

Vw = 400-350 = 50mi/hr

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