document.write( "Question 90884: A plane flies 720 mi against a steady 30-mi/h headwind and then returns to the same point with the wind. If the entire trip takes 10 h, what is the plane’s speed in still air? \n" ); document.write( "
Algebra.Com's Answer #65968 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! plane flies 720 mi against a steady 30-mi/h headwind and then returns to the same point with the wind. If the entire trip takes 10 h, what is the plane’s speed in still air? \n" ); document.write( "-------------- \n" ); document.write( "Let the speed of the plane in calm air be \"p\". \n" ); document.write( "--------------------- \n" ); document.write( "Against the wind DATA: \n" ); document.write( "distance = 720 mi ; rate = p-30 ; time = d/r = 720/(p-30) \n" ); document.write( "--------------- \n" ); document.write( "With the wind DATA; \n" ); document.write( "distance = 720 mi ; rate = p+30 ; time = d/r = 720/(p+30) \n" ); document.write( "------------------ \n" ); document.write( "EQUATION: \n" ); document.write( "time + 1ime = 10 hrs. \n" ); document.write( "720/(p-30) + 720/(p+30) = 10 \n" ); document.write( "72/(P-30) + 72/(p+30) = 1 \n" ); document.write( "Multiply thru by P^2-900 to get: \n" ); document.write( "72(p+30) + 72(p-30) = p^2-900 \n" ); document.write( "144p= p^2-900 \n" ); document.write( "p^2-144p-900 = 0 \n" ); document.write( "(p+6)(p-150)=0 \n" ); document.write( "p = 150 mph (speed of the plane in still air) \n" ); document.write( "=============== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |