Question 160643: A plane flew a distance of 1555 miles in 5 hours. During the first 3 hours of the flight, it flew with a wind a distance of 975 miles. During the remainder of the flight, the plane flew against a wind whose average was 5 mph less than what it had been during the first part of the flight. Find the rate of the plane in still air and the original speed of the wind.
Some helpful equations:
r= the rate in still air
c= the rate of the air current
r+c= rate traveling with the current
r-c= rate traveling against the current
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A plane flew a distance of 1555 miles in 5 hours. During the first 3 hours of
the flight, it flew with a wind a distance of 975 miles. During the remainder
of the flight, the plane flew against a wind whose average was 5 mph less than
what it had been during the first part of the flight. Find the rate of the
plane in still air and the original speed of the wind.
:
Some helpful equations:
r= the rate in still air
c= the rate of the air current
r+c= rate traveling with the current
r-c= rate traveling against the current
:
Summarizing what we know:
:
1st part of the trip 975 mi in 3 hrs, at a speed (r+c)
2nd part of the trip: 1555-975 = 580 mi in 5-3 = 2 hrs at a speed (r - (c-5))
:
Write a distance equation for each part of the trip: (Dist = time * speed)
:
3(r+c) = 975
2(r- (c-5)) = 580; wind given as 5 mph less
:
Simplify: divide the 1st equation by 3, and the 2nd equation by 2 and you have;
r + c = 325
and
r - c + 5 = 290
r - c = 290 - 5
r - c = 285
:
Use these two equation for elimination
r + c = 325
r - c = 285
----------------addition eliminate c, find r
2r = 610
r = 305 mph in still air
:
Find the original speed of the wind using r + c = 325
305 + c = 325
c = 20 mph is the wind on the 1st part of the trip
:
:
Check solution by finding the distances
3(305+20) = 975
2(305-15) = 580 (wind is 5 mph less)
----------------
total dist 1555mi
|
|
|
