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 
:
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)
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total dist 1555mi