An airplane takes 1 hour longer to go a distance of 551 miles flying against a headwind than on the return trip with a tailwind. if the speed of the wind is 50 miles per hour, find the speed of the plane in still air.

Respuesta :

Let [tex] v_p [/tex] be the speed of the plane in still air. Let [tex] t [/tex] be the time taken by the plane with the wind. Then, from the given information,

[tex] (v_p+50)t=551\\
(v_p-50)(t+1)=252\\ [/tex]

From the above 2 equations,

[tex] (v_p+50)t=551\\
(v_p-50)t=302-v_p\\ [/tex]

Dividing the above 2 equations,

[tex] \frac{v_p-50}{v_p+50} =\frac{302-v_p}{551} \\
551(v_p-50)=(302-v_p)(v_p+50)\\
v_p^2+299v_p-42650=0\\
v_p=105.45
[/tex]

The speed of the plane in still air is [tex] 105.45 \;mi/hour [/tex].

Answer:

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Explanation:

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