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?

Respuesta :

Answer:

The velocity of wind is 34.88 miles per hour

Step-by-step 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 = [tex]\dfrac{\terxtrm Distance}{\textrm Time}[/tex]

Now, With the wind

190 + w =  [tex]\dfrac{\terxtrm 500}{\textrm t}[/tex]

or, t = [tex]\dfrac{\terxtrm 500}{\textrm 190 + w}[/tex]

And Against the wind

190 - w = [tex]\dfrac{\terxtrm 500}{\textrm ( t + 1 )}[/tex]

Solving the equation

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

or, ( 190 - w ) ( [tex]\dfrac{\terxtrm 500}{\textrm 190 + w}[/tex] + 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 = [tex]\frac{-b\pm \sqrt{b^{2}-4\times a\times c}}{2\times a}[/tex]

Or, w = [tex]\frac{-1000\pm \sqrt{1000^{2}-4\times 1\times (-36100)}}{2\times 1}[/tex]

Or, w = [tex]\frac{-1000\pm \sqrt{1144400}}{2}[/tex]

∴ w = [tex]\frac{-1000 + 1069.76}{2}[/tex] ,  [tex]\frac{-1000 - 1069.76}{2}[/tex]

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