SOLUTION: An airplane flying into a head wind travels the 2400 miles flying distance between two cities in 6 hours. On the return flight, the same distance is traveled in 5 hours. Find the
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Question 1030465: An airplane flying into a head wind travels the 2400 miles flying distance between two cities in 6 hours. On the return flight, the same distance is traveled in 5 hours. Find the air speed of the plane and the speed of the wind, assuming that both remain constant. (The air speed is the speed of the plane if there were no wind.) Write a system of equations and solve Found 2 solutions by mananth, ikleyn:Answer by mananth(16946) (Show Source):
Distance with wind 2400 miles distance against wind 2400
t=d/r against wind (x-y)
2400.00 / ( x - y )= 6.00
6.00 x - -6.00 y = 2400.00 ....................1
with wind (x+y)
2400.00 / ( x + y )= 5.00
5.00 ( x + y ) = 2400.00
5.00 x + 5.00 y = 2400.00 ...............2
Multiply (1) by 5.00
Multiply (2) by 6.00
we get 2.00
30.00 x + -30.00 y = 12000.00
30.00 x + 30.00 y = 14400.00
60.00 x = 26400.00
/ 60.00
x = 440 mph
plug value of x in (1) y
6 x -6 y = 2400
2640 -6 -2640 = 2400
-6 y = 2400
-6 y = -240 mph
y = 40
Plane 440 mph
wind 40 mph
You can put this solution on YOUR website! .
An airplane flying into a head wind travels the 2400 miles flying distance between two cities in 6 hours.
On the return flight, the same distance is traveled in 5 hours. Find the air speed of the plane and the speed of the wind,
assuming that both remain constant. (The air speed is the speed of the plane if there were no wind.)
Write a system of equations and solve
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The airplane speed relative to the Earth surface is = 400 mph, when flying into a head wind.
This speed is the difference u-v of the airplane speed in still air and the wind speed: u - v = 400 mph.
The airplane speed relative to the Earth surface is = 480 mph, when flying with the wind on the return flight.
This speed is the sum u+v of the airplane speed in still air and the wind speed: u + v = 480 mph.
To determine the airplane speed in still air "u" and the wind speed "v", you need to solve this system of two equations:
u - v = 400, (1)
u + v = 480. (2)
Add the equations (1) and (2). You will get
2u = 400 + 480
2u = 880,
u = = 440 miles per hour. It is the airplane speed in still air.
Then from (1) you find v = u - 400 = 440 - 400 = 40 miles per hour. It is the wind speed.
The problem is solved.
It is a typical Travel and Distance problem for a plane flying with and against the wind.
For many other solved similar problems see the lessons
- Wind and Current problems
- Selected problems from the archive on the boat floating Upstream and Downstream
in this site.
