document.write( "Question 65329: A plane flies 720 miles against a steady 30 mi/hr headwind and then returns to the same point with the wind. If the entire trip takes 10 hours, what is the plane’s speed in still air? \n" ); document.write( "
Algebra.Com's Answer #45883 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! --------------- \n" ); document.write( "A plane flies 720 miles against a steady 30 mi/hr headwind and then returns to the same point with the wind. If the entire trip takes 10 hours, what is the plane’s speed in still air? \n" ); document.write( "--------------- \n" ); document.write( "Let p be plane speed \n" ); document.write( "Against the wind DATA: \n" ); document.write( "distance= 720 miles ; Rate= p-30 mph ; time = d/r= 720/(p-30) hrs. \n" ); document.write( "----------------- \n" ); document.write( "With the wind DATA: \n" ); document.write( "distance =720 miles ; rate = p+30 mph ; time = d/r = 720/(p+30) hrs. \n" ); document.write( "---------- \n" ); document.write( "EQUATION: \n" ); document.write( "time with + time against = 10 hrs \n" ); document.write( "720/(p+30) + 720/(p-30)= 10 \n" ); document.write( "Divide thru by 720 to get: \n" ); document.write( "1/(p+30) + 1/(p-30)= 1/72 \n" ); document.write( "Multiply thru by (p+30)(p-30) to get: \n" ); document.write( "p-30+p+30=(1/72)(p^2-900) \n" ); document.write( "Simplify the left side to get: \n" ); document.write( "2p=(1/72)(p^2-900) \n" ); document.write( "Multiply both sides by 72 to get: \n" ); document.write( "144p=p^2-900 \n" ); document.write( "Rearrange: \n" ); document.write( "p^2-144p-900=0 \n" ); document.write( "Use the quadratic formula to get: \n" ); document.write( "p=[144+-sqrt(144^2-4*-900)]/2 \n" ); document.write( "p=[144+-156]/2 \n" ); document.write( "Select the positive solution to get: \n" ); document.write( "p=[144+156]/2=300/2=150 mph \n" ); document.write( "Cheers, \n" ); document.write( "Stan H.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |