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
........
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.