document.write( "Question 179618: The Hudson River flows at a rate of 5 miles per hour. A patrol boat travels 40 miles upriver , and returns in a total time of 6 hours. What is the speed of the boat in still water?
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document.write( " I have figured the distance is 40 miles with a speed of r-5 upstream and r in still water, and the total time is 6 hours but I still can not figure out how to write the equation and solve it. \n" );
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Algebra.Com's Answer #134564 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The Hudson River flows at a rate of 5 miles per hour. A patrol boat travels 40 miles upriver , and returns in a total time of 6 hours. What is the speed of the boat in still water? \n" ); document.write( "--------------------------------- \n" ); document.write( "Upriver DATA: \n" ); document.write( "distance = 40 miles ; rate = b - 5 ; time = d/r = 40/(b-5) \n" ); document.write( "----------------------------------- \n" ); document.write( "Downriver DATA: \n" ); document.write( "distance = 40 miles : rate = b + 5 ; time = 40/(b+5) \n" ); document.write( "---------------------- \n" ); document.write( "Equation: \n" ); document.write( "time + time = 6 hrs \n" ); document.write( "40/(b-5) + 40/(b+5) = 6\r \n" ); document.write( "\n" ); document.write( "40(b+5) + 40(b-5) = 6(b^2-25)\r \n" ); document.write( "\n" ); document.write( "80b = 6b^2 - 150 \n" ); document.write( "Rearrange: \n" ); document.write( "3b^2 - 40b -150 = 0 \n" ); document.write( "------------------------------ \n" ); document.write( "positive solution: \n" ); document.write( "b = [40 + sqrt(40^2 - (4*3*-150)]/6\r \n" ); document.write( "\n" ); document.write( "b = 23.05 mph (boat speed in still water) \n" ); document.write( "============================================ \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |