SOLUTION: PROBLEM: DAVE IS FLYING FROM PHILADELPHIA TO CHICAGO. ON THE FLIGHT, FLYING AGAINST THE WIND, THE 780 MILE TRIP TAKES 2 HOURS. THE RETURN TRIP, WITH THE WIND TAKES 1.5 HOURS. FIND

Algebra ->  Equations -> SOLUTION: PROBLEM: DAVE IS FLYING FROM PHILADELPHIA TO CHICAGO. ON THE FLIGHT, FLYING AGAINST THE WIND, THE 780 MILE TRIP TAKES 2 HOURS. THE RETURN TRIP, WITH THE WIND TAKES 1.5 HOURS. FIND      Log On


   

Question 136864: PROBLEM:
DAVE IS FLYING FROM PHILADELPHIA TO CHICAGO. ON THE FLIGHT, FLYING AGAINST THE WIND, THE 780 MILE TRIP TAKES 2 HOURS. THE RETURN TRIP, WITH THE WIND TAKES 1.5 HOURS. FIND THE SPEED OF THE WIND AND FIND THE SPEED OF THE PLANE IN STILL AIR.
PLEASE HELP I DONT EVEN KNOW WHERE TO BEGIN THANKS!!
Found 2 solutions by checkley77, solver91311:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
780/2=390 MPH AGAINST THE WIND
780/1.5=520 MPH WITH THE WIND
(520-390)/2=130/2=65 MPH FOR THE WIND SPED.
PROOF:
520-65=390+65
455=455 THIS IS THE SPEED OF THE PLANE IN STILL AIR.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Do me a favor please. Stop typing your questions in ALL CAPS. It is rude, annoying, hard to read, and is the electronic equivalent of shouting.

The speed of the plane in still air: r
The speed of the wind r%5Bw%5D
The speed of the plane against the wind: r-r%5Bw%5D
The speed of the plane with the wind: r%2Br%5Bw%5D

Distance equals rate times time: d=rt, but this can be solved for rate: r=d%2Ft

For the trip against the wind: r-r%5Bw%5D=780%2F2=390
For the trip with the wind: r%2Br%5Bw%5D=780%2F1.5=520

Add these two equations:
%28r%2Br%5Bw%5D%29%2B%28r-r%5Bw%5D%29=910
2r=910
r=455

Check:
If r is 455, then r[w] has to be 455 - 390 = 520 - 455 = 65.

So, against the wind, the plane goes 780 miles at 390 mph -- trip takes 2 hours
With the wind, the plane goes 780 miles at 520 mph -- trip takes 1.5 hours.