Q:

Matt and Anna Killian are frequent fliers on​ Fast-n-Go Airlines. They often fly between two cities that are a distance of 1980 miles apart. On one particular​ trip, they flew into the wind and the trip took 5.5 hours. The return trip with the wind behind​ them, only took about 4.5 hours. Find the speed of the wind and the speed of the plane in still air.

Accepted Solution

A:
Answer:y (plane speed) 400 miles/hrx (wind speed)  40 miles/hrStep-by-step explanation:going into the wind make the plane flew 5,5 hours to get 1980 miles, implies the real travel speed was 1980 / 5,5 = 360 miles/hours, and at the same time is the result of:   speed plane (y)  in direction A to B  plus wind speed (x) in direction B to A. Notice opposites direction speed, so we have: y + (-x) = 360          (1) The return fly  took 4,5 hours so 1980/4,5 = 440 miles/ hours as the result Y + x = 440             (2) We have a two equation with two variables system, It could be solved for any of the procedures. We will use the substitution method. From equation      (1)        y + (-x) = 360         y  - x = 360      y = 360 + x    (1) In the second equation     y + x = 440         then we replace y for its value as function of x (360 + x ) + x = 440 Solving for x        360 + 2x = 440      or    2x = 440 – 360          x   (440 - 360)/2  x = 40 miles/hr  X (wind speed ) = 40 miles/hr And replacing the value of x in eq. (1)        y = 360 + 40 = 400 Y (plane speed) = 400 miles /hr