Airlines schedule about 5.5 hours of flying time for an A320 Airbus to fly from Dulles International Airport near Washington, D.C., to Los Angeles International Airport. Airlines schedule about 4.5 hours of flying time for the reverse direction. The distance between these airports is about 2,300 miles. They allow about 0.4 hour for takeoff and landing. a. From this information, estimate (to the nearest 5 mph) the average wind speed the airlines assume in making their schedule. b. What average airplane speed (to the nearest 5 mph) do the airlines assume in making their schedule?

Question
Algebra foundations
asked 2020-11-26
Airlines schedule about 5.5 hours of flying time for an A320 Airbus to fly from Dulles International Airport near Washington, D.C., to Los Angeles International Airport. Airlines schedule about 4.5 hours of flying time for the reverse direction. The distance between these airports is about 2,300 miles. They allow about 0.4 hour for takeoff and landing.
a. From this information, estimate (to the nearest 5 mph) the average wind speed the airlines assume in making their schedule.
b. What average airplane speed (to the nearest 5 mph) do the airlines assume in making their schedule?

Answers (1)

2020-11-27
a. If they allow 0.4 hour for takeoff and landing, then the approximate flight time from Dulles to D.C. is about 5.5−0.4=5.1 hours and the approximate flight time from D.C. to Dulles is 4.5−0.4=4.1 hours.
Let r be the average airplane speed and ww be the average wind speed.
The flight from Dulles to D.C. is scheduled for a longer time because the wind will slow down the plane. The rate of the plane will then be r−w for this trip. Since distance = (rate)(time), if the distance is 2,300 miles, the rate is r−w, and the time is 5.1 hours, then 2,300=5.1(r−w). Dividing both sides by 5.1 gives r−w=451, rounded to the nearest whole number.
For the flight from D.C. to Dulles, the wind will make the plane travel faster so the plane's rate will be r+wr+w. Since the distance is 2,300 miles, the rate is r+w, and the time is 4.1 hours, then 2,300=4.1(r+w). Dividing both sides by 4.1 gives r+w=561, rounded to the nearest whole number.
Add the two equations to eliminate ww and solve for r:
r+w = 561
r−w = 451
2r = 1012
r = 506
Substituting r=506 into r+w=561 gives 506+w=561. Subtracting 506 on both sides gives w=55. The average wind speed to the nearest 5 mph is then 55 mph.
b. From part (a), r=506. Rounding to the nearest 5 mph then gives an average airplane speed of 505 mph.
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