Question 1061742: It took 3 hours for a plane, flying against the wind, to travel 900 miles from Alabama to Minnesota. The "ground speed" of the plane is 300 miles per hour. On the return trip, the flight took only 2 hours with a ground speed of 450 miles per hour. During both flights the speed and the direction of the wind were the same. Find the speed of the plane in still air and the speed of the wind.
Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! It took 3 hours for a plane, flying against the wind, to travel 900 miles from Alabama to Minnesota. The "ground speed" of the plane is 300 miles per hour. On the return trip, the flight took only 2 hours with a ground speed of 450 miles per hour. During both flights the speed and the direction of the wind were the same. Find the speed of the plane in still air and the speed of the wind.
----
With wind:: p + c = 450
Against::: p - c = 300
----------------------------
Add and solve for p::
2p = 750
plane = 375 mph (plane rate in still air)
-----
Solve for "c":
p + c = 450
c = 450-375 = 75 mph (rate of the wind)
-----------------
Cheers,
Stan H.
-----------------
Answer by ankor@dixie-net.com(22740)  (Show Source):
You can put this solution on YOUR website! It took 3 hours for a plane, flying against the wind, to travel 900 miles from Alabama to Minnesota.
The "ground speed" of the plane is 300 miles per hour.
On the return trip, the flight took only 2 hours with a ground speed of 450 miles per hour.
During both flights the speed and the direction of the wind were the same.
Find the speed of the plane in still air and the speed of the wind.
:
let s = the plane speed in still air
let w = the speed of the wind
then
(s+w) = ground speed with the wind
and
(s-w) = ground speed against the wind
:
s - w = 300
s + w = 450
-------------Addition eliminates w, find s
2s + 0 = 750
s = 750/2
s = 375 mph speed in still air
and
375 + w = 450
w = 450 - 375
w = 75 mph wind speed
|
|
|
