A plane traveled 2135 miles with the wind in 3.5 hours and 1855 miles against the wind in the same amount of time. Find the speed of the plane in still air and the speed of wind.

The speed of air in still air = 570 mph and

speed of wind = 40 mph

Step-by-step explanation:

Given that plane travelled 2135 miles in 3.5 hours with the wind.

Let the velocity of plane be x and that of wind be y.

Then with the wind speed = x+y and

against the wind the speed = x-y

Use distance = time x speed

Since travelled 2135 miles in 3.5 hours we have

2135 = 3.5 (x+y) ... i

Similarly with x-y speed it travelled 1855 miles in 3.5 hours

i. e. 1855 = 3.5 (x-y) ... ii

Solving these two we can get x and y

i-ii gives

3.5y+3.5y = 2135-1855 = 280

i. e. 7y = 280

Or y = 40 mph

Substitute in i

3.5 (x+40) = 2135

x+40 = 610

x = 570 mph

Hence the speed of air in still air = x = 570 mph and

speed of wind = 40 mph