document.write( "Question 1018427: A boat travelled 336 miles downstream and back. The trip downstream took 12 hours. The trip back took 14 hours.
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document.write( "What is the speed of the boat in still water? What is the speed of the current? \n" );
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Algebra.Com's Answer #634530 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A boat travelled 336 miles downstream and back. \n" ); document.write( " The trip downstream took 12 hours. The trip back took 14 hours. \n" ); document.write( " What is the speed of the boat in still water? \n" ); document.write( " What is the speed of the current? \n" ); document.write( ": \n" ); document.write( "let s = boat speed in still water \n" ); document.write( "let c = rate of the current \n" ); document.write( "then \n" ); document.write( "(s+c) = effective speed downstream \n" ); document.write( "and \n" ); document.write( "(s-c) = effective speed upstream \n" ); document.write( ": \n" ); document.write( "Write a distance equation for each way. dist = time * speed \n" ); document.write( "12(s+c) = 336 \n" ); document.write( "14(s-c) = 336 \n" ); document.write( "we are in luck; we can greatly simplify both these equations \n" ); document.write( "divide the first by 12, divide the 2nd by 14, and you have \n" ); document.write( "s + c = 28 \n" ); document.write( "s - c = 24 \n" ); document.write( "--------------Addition eliminates c, find s \n" ); document.write( "2s + 0 = 52 \n" ); document.write( "s = 52/2 \n" ); document.write( "s = 26 mph in still water \n" ); document.write( "and \n" ); document.write( "26 + c = 28 \n" ); document.write( "c = 28 - 26 \n" ); document.write( "c = 2 mph is the rate of the current \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this using the 1st original equation \n" ); document.write( "12(26 + 2) = 336 \n" ); document.write( "12(28) = 336 \n" ); document.write( "You can check this for yourself in the 2nd original equation \n" ); document.write( " |