document.write( "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.\r
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document.write( "Some helpful equations:
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document.write( "r= the rate in still air
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document.write( "c= the rate of the air current
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document.write( "r+c= rate traveling with the current
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document.write( "r-c= rate traveling against the current \n" );
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Algebra.Com's Answer #118495 by ankor@dixie-net.com(22740)![]() ![]() 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 \n" ); document.write( " the flight, it flew with a wind a distance of 975 miles. During the remainder \n" ); document.write( " of the flight, the plane flew against a wind whose average was 5 mph less than \n" ); document.write( " what it had been during the first part of the flight. Find the rate of the \n" ); document.write( "plane in still air and the original speed of the wind. \n" ); document.write( ": \n" ); document.write( "Some helpful equations: \n" ); document.write( "r= the rate in still air \n" ); document.write( "c= the rate of the air current \n" ); document.write( "r+c= rate traveling with the current \n" ); document.write( "r-c= rate traveling against the current \n" ); document.write( ": \n" ); document.write( "Summarizing what we know: \n" ); document.write( ": \n" ); document.write( "1st part of the trip 975 mi in 3 hrs, at a speed (r+c) \n" ); document.write( "2nd part of the trip: 1555-975 = 580 mi in 5-3 = 2 hrs at a speed (r - (c-5)) \n" ); document.write( ": \n" ); document.write( "Write a distance equation for each part of the trip: (Dist = time * speed) \n" ); document.write( ": \n" ); document.write( "3(r+c) = 975 \n" ); document.write( "2(r- (c-5)) = 580; wind given as 5 mph less \n" ); document.write( ": \n" ); document.write( "Simplify: divide the 1st equation by 3, and the 2nd equation by 2 and you have; \n" ); document.write( "r + c = 325 \n" ); document.write( "and \n" ); document.write( "r - c + 5 = 290 \n" ); document.write( "r - c = 290 - 5 \n" ); document.write( "r - c = 285 \n" ); document.write( ": \n" ); document.write( "Use these two equation for elimination \n" ); document.write( "r + c = 325 \n" ); document.write( "r - c = 285 \n" ); document.write( "----------------addition eliminate c, find r \n" ); document.write( "2r = 610 \n" ); document.write( "r = 305 mph in still air \n" ); document.write( ": \n" ); document.write( "Find the original speed of the wind using r + c = 325 \n" ); document.write( "305 + c = 325 \n" ); document.write( "c = 20 mph is the wind on the 1st part of the trip \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution by finding the distances \n" ); document.write( "3(305+20) = 975 \n" ); document.write( "2(305-15) = 580 (wind is 5 mph less) \n" ); document.write( "---------------- \n" ); document.write( "total dist 1555mi \n" ); document.write( " |