Problema Solution
A twin-engined aircraft can fly 1360 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
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v = the speed of the aircraft in calm air
w = the speed of the win
d = 1360 mi the distance traveled when flying with the wind
t1 = 5 hr the time of the flight with the wind
t2 = 8 hr 0 the time of the flight against the wind
the speed of the plane flying with the wind is: v + w
the speed of the plane flying against the wind is: v - w
since speed = distance/time
when flying with the wind
v + w = d/t1
v + w = 1360/5
v + w = 272
the flying against the wind
v - w = d/t2
v - w = 1360/8
v - w = 170
by solving the system of equations
v + w = 272
v - w = 170
we find
w = 51 mph
click here to see the step by step solution of the system of equations
the speed of the wind is 51 miles per hour.