A plane is traveling between City A and City B, which are 2000 miles apart. City A is due north of City B, and there is a strong, constant wind blowing due south. At constant speed, it takes a plane 5 hours to go from A to B with the wind at its tail, but 8 hours to go back when facing this headwind. What is the speed of the wind?

Therefore, the speed of the plane when it goes from A to B with the wind is at its tail is (x + y) mph and the speed of the plane when it goes from B to A facing the headwind is (x - y) mph.

Given that,

[tex]x + y = \frac{2000}{5} = 400[/tex] ............ (1) and

[tex]x - y = \frac{2000}{8} = 250[/tex] ............. (2)

Now, from equations (1) and (2) we get

2y = 400 - 250 = 150

⇒ y = 75 mph.

Therefore, the speed of the wind is 75 mph. (Answer)

PiaDeveau PiaDeveau · Answer 2 · 2021-12-08T08:52:42+01:00

Speed of the wind is 75 miles per hour

Given that;

Distance between two city = 2,000 miles

With wind time taken by plane = 5 hours

Against wind time taken by plane = 8 hours

Assume;

Plane speed = a

Wind speed = b

So,

a + b = 2000 / 5

a + b = 400 ........... eq1

a - b = 2000 / 8

a - b = 250 ......... eq2

From eq 1 - eq2

2b = 150

b = 75 mile per hour

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