SOLUTION: A plane left an airport and flew with the wind for 4 hours, covering 2000 miles. It then returned over the same route to the airport against the same wind in 5 hours. Find the rate
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: A plane left an airport and flew with the wind for 4 hours, covering 2000 miles. It then returned over the same route to the airport against the same wind in 5 hours. Find the rate
Log On
Question 553313: A plane left an airport and flew with the wind for 4 hours, covering 2000 miles. It then returned over the same route to the airport against the same wind in 5 hours. Find the rate of the plane in still air and the speed of the wind. Found 2 solutions by mananth, Alan3354:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! Plane speed =x mph
wind speed speed =y mph
against wind x-y 5.00 hours
with wind x+y 4.00 hours
Distance = same= 2000 miles
t=d/r
2000 / ( x - y )= 5.00
5 ( x - y ) = 2,000.00
5 x -5 y = 2000 ....................1
2000 / ( x + y )= 4.00
4.00 ( x + y ) = 2000
4.00 x + 4.00 y = 2000 ...............2
Multiply (1) by 4.00
Multiply (2) by 5.00
we get
20 x + -20 y = 8000
20 x + 20 y = 10000
40 x = 18000
/ 40
x = 450 mph
plug value of x in (1)
5 x -5 y = 2000
2250 -5 y = 2000
-5 y = 2000 -2250
-5 y = -250
y = 50 mph
You can put this solution on YOUR website! A plane left an airport and flew with the wind for 4 hours, covering 2000 miles. It then returned over the same route to the airport against the same wind in 5 hours. Find the rate of the plane in still air and the speed of the wind.
--------------
2000/4 = 500 mi/hr
2000/5 = 400 mi/hr
----
The plane's airspeed is the average, 450 mi/hr
The windspeed is the difference, = 50 mi/hr
