document.write( "Question 1116842: A pilot can fly a plane at 125 mph in calm air. A recent trip of 300 mi flying with the wind and 300 mi returning against the wind took 5 h. Find the rate of the wind. \n" ); document.write( "
Algebra.Com's Answer #731756 by solver91311(24713)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Basic formula is \r
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\n" ); document.write( "\n" ); document.write( "For the outbound trip (with the wind), the speed of the aircraft is where is the rate of speed of the wind.\r
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\n" ); document.write( "\n" ); document.write( "For the return trip (against the wind), the speed of the aircraft is \r
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\n" ); document.write( "\n" ); document.write( "Let the time of the outbound trip be , then the time of the return trip must be whatever remains of the total trip time of 5 hours, or .\r
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\n" ); document.write( "\n" ); document.write( "Since the two legs of the trip are equal in distance, namely 300 miles, we have two equations in and :\r
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\n" ); document.write( "\n" ); document.write( "Although we are asked for the speed of the wind, it will be a much neater calculation to solve for first.\r
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\n" ); document.write( "\n" ); document.write( "Add the two equations:\r
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\n" ); document.write( "\n" ); document.write( "Combine the RHS fractions over the common denominator \r
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\n" ); document.write( "\n" ); document.write( "So is either 2 or 3. However, the way we set up the problem with representing the time of the 'with the wind' leg of the trip, must be the smaller of the two values. Hence:\r
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\n" ); document.write( "\n" ); document.write( "and\r
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\n" ); document.write( "\n" ); document.write( "So, since\r
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\n" ); document.write( "\n" ); document.write( "Checking the answer in the return trip equation:\r
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\n" ); document.write( "\n" ); document.write( "Checks\r
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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