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
click here to see the step by step solution of the system of equations:
the speed of the wind is 45 mph.