Problema Solution

A twin-engined aircraft can fly 1200 miles from city A to city B in 5 hours with the wind and make the return trip in 8 hours against the wind. What is the speed of the wind?

Answer provided by our tutors

let


v = the speed of the plane in still air


w = the speed of the wind


d = 1200 mi the distance from A to B traveled in each direction


t1 = 5 h the time of the trip with the wind


t2 = 8 h the time of the trip against the wind


since speed = distance/time follows distance = speed*time


traveling with the wind the speed of the plane is: v + w


(v + w)*t1 = d


(v + w)*5 = 1200 divide both sides by 5


v + w = 240


traveling against the wind the speed of the plane is: v - w


(v - w)*t2 = d


(v - w)*8 = 1200 divide both sides by 8


v - w = 150


by solving the system of equations:


v + w = 240


v - w = 150


we find:


v = 195 mph


w = 45 mph


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the speed of the wind is 45 mph.