Question

a plane travels at a speed of 190 mph in still air. Flying with a tailwind, the plane is clocked over a distance of 500 miles. Flying against a headwind, it takes 1 hour longer to complete the return trip. What was the wind velocity?

Answer 1

The velocity of wind is 34.88 miles per hour

Explanation:

Given as :

The velocity of plane = 190 miles per hour

Let The velocity of wind = w miles per hour

The distance cover by plane = 500 miles

The time taken to cover 500 miles with the wind = t hours

The time taken to cover 500 miles against the wind = ( 1 + t ) hours

Now , Speed =
`(\terxtrm Distance)/(\textrm Time)`

Now, With the wind

190 + w =
`(\terxtrm 500)/(\textrm t)`

or, t =
`(\terxtrm 500)/(\textrm 190 + w)`

And Against the wind

190 - w =
`(\terxtrm 500)/(\textrm ( t + 1 ))`

Solving the equation

I.e ( 190 - w ) ( t + 1 ) = 500

or, ( 190 - w ) (
`(\terxtrm 500)/(\textrm 190 + w)` + 1 ) = 500

Or, ( 190 - w ) [ 500 + ( 190 + w ) ] = 500 ( 190 + w )

Or, 500 × 190 + 190 × ( 190 + w ) - 500 w - w × ( 190 + w ) = 500 × 190 + 500 w

or, 190 × ( 190 + w ) - 500 w - w × 190 - w² = 500 w

or, 190² + 190 w - 500 w - 190 w - w² = 500 w

or, - w² - 1000 w + 190² = 0

Or , w² + 1000 w - 36100 = 0

Solving this quadratic equation

w =
`\frac{-b\pm \sqrt{b^(2)-4* a* c}}{2* a}`

Or, w =
`\frac{-1000\pm \sqrt{1000^(2)-4* 1* (-36100)}}{2* 1}`

Or, w =
`(-1000\pm √(1144400))/(2)`

∴ w =
`(-1000 + 1069.76)/(2)` ,
`(-1000 - 1069.76)/(2)`

or, w = 34.88 mph , - 1034.88 mph

So, The wind velocity = w = 34.88 mph

Hence The velocity of wind is 34.88 miles per hour Answer