Question

A plane flies directly between two cities, A and B, which are separated by 2300 mi. From A to B, the plane flies into a 65 mi/h headwind. On the return trip from B to A, the wind velocity is unchanged. The trip from B to A takes "65 min less" than the trip from A to B. What is the airspeed of the plane, assuming it is the same in both directions?

Answer 1

Answer:

The speed of air is  526.22 mi/h                

Explanation:

Given that

Distance = 2300 mi

Velocity of wind ,U=65 m/h

Lets take velocity of plane = V

When plane is moving from A to B then resultant velocity of pane = V-65 m/h

Lets take time taken to plane when plane is moving from A to B = t hr

We know that

Distance = Velocity x time

2300 = (V-65) x t        

 t = 2300/ (V-65)                                                         ------------1

When plane is moving from A to B then resultant velocity of pane = V+65 m/h

Time taken by plane from B to A = t -1.08 hr               ( 65 min = 1.08 hr)

2300 = (V+65) x (t - 1.08)        

  $t=1.08+\dfrac{2300}{V+65}$    -------------2

  Now from equation 1 and 2

$\dfrac{2300}{V-65}=1.08+\dfrac{2300}{V+65}$

$\dfrac{2300}{V-65}-\dfrac{2300}{V+65}=1.08$

$2300\left (\dfrac{1}{V-65}-\dfrac{1}{V+65}\right)=1.08$

$2300\left (\dfrac{2\times 65}{V^2-65^2}\right)=1.08$

$V^2=276,916.85$

V=526.22 mi/h        

The speed of air is  526.22 mi/h