Problema Solution

A twin-engine aircraft can fly 1064 miles from city A to city B in 4 hours with the wind and make

the return trip in 7 hours against the wind. What is the speed of the wind?

Answer provided by our tutors

Let

w = the speed of the wind, w>=0

v = the speed of the aircraft in still air, v>=0

d = 1064 mi the distance from city A to city B

t1 = 4 hr the time of the travel with the wind

t2 = 7 hr the time of the travel against the wind

Flying with the wind the speed of the aircraft is: v + w

Flying against the wind the speed of the aircraft is: v - w

Since speed = distance/time we have:

v + w = d/t1

v - w = d/t2

of if we plug the values we get the following system of equations:

v + w = 1064/4

v - w = 1064/7

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click here to see the system of equations solved for v and w

........

v = 209 mph

w = 57 mph

The speed of the wind is 57 mph.