What is the speed of the airplane in still air in the following case:An airplane covers a distance of 1,500 miles in 3 hours when it flies with the wind and 3 and 1/3 hours when it flies against...

What is the speed of the airplane in still air in the following case:

An airplane covers a distance of 1,500 miles in 3 hours when it flies with the wind and 3 and 1/3 hours when it flies against the wind.

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given that the speed of the plane with the wind = 1500/3 hours.

==> The speed = 1500/3 = 500 mph.

Let (s) be the apeed of the plane in still air.

Letthe speed of the wind be (sw).

==> 500 = s + sw...............(1)

Also, we are given that the speed against the wind = 1500/(3 1/3).

==> 1500/ (10/3) = 1500*3/10 = 450 mph.

==> 450 = s - sw..................(2)

Now we will add equation (1) and (2);

==> 2s = 950

==> s = 950/2 = 475.

Then, the speed of the plane in still air is 475 mph.

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The airplane covers a distance of 1,500 miles in 3 hours when it flies with the wind and 3 and 1/3 hours when it flies against the wind. We have to find the speed of the plane in still air.

Let the speed of the wind be W and the speed of the airplane be A. Speed is equal to distance/time.

When the airplane travels with the wind, the two speeds are added together.

W + A = 1500/3...(1)

When the airplane travels against the wind, the speed is given by A - W.

A - W = 1500/ (3 + 1/3)...(2)

Now (2) + (1) gives

A - W + W + A = 1500/ (3 + 1/3) + 1500/3

=> 2A = 1500/ (3 + 1/3) + 1500/3

=> 2A = 1500/(10/3) + 1500/3

=> 2A = 150*3 + 500

=> 2A = 450 + 500

=> 2A = 950

=> A = 950/2

=> A = 475

Therefore the speed of the airplane in still air is 475 miles/hr.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Let x and y be the speed of the airplane and wind.

Then while flying with wind, the airplane covers 1500miles in 3 hours. So the speed with wind is x+y = 1500/3 = 500m/h.

=> x+y = 500..............(1).

Therefore x-y = 450....(2).

Eq (1) - eq (2) gives: x+y+x-y = 500+450 = 950.

=> 2x= 950.

=> x = 950/2 = 475m/h.

Eq (1)- eq (2) gives: x+y-(x-y) = 500-450 = 50.

=> 2y = 50.

=> y = 50/2 = 25 m/h.

Therefore the speed of the airplane x = 475 m/h and the speed of the wind y = 25m/h.

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