document.write( "Question 120519: 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 #88339 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A plane flies 720 miles against a steady 30 mph 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 the plane's speed in still air. \n" ); document.write( "-------------------- \n" ); document.write( "Upwind DATA; \n" ); document.write( "Distance = 720 miles ; Rate = p-30; time = d/r = 720/(p-30) hrs. \n" ); document.write( "------------- \n" ); document.write( "Downwind DATa: \n" ); document.write( "Distance = 720 miles ; Rate = p+30: time = d/r = 720/(p+30) hrs \n" ); document.write( "-------------- \n" ); document.write( "EQUATION: \n" ); document.write( "time up + time down = 10 hrs \n" ); document.write( "720/(p-30) + 720/(p+30) = 10 \n" ); document.write( "------------------------ \n" ); document.write( "Divide thru by 10 to get: \n" ); document.write( "72/(p-30) + 72/(p+30) = 1 \n" ); document.write( "-------------- \n" ); document.write( "Multiply thru by (p-30)(p+30)=p^2-900 \n" ); document.write( "--------------- \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-150)(p+6) = 0 \n" ); document.write( "p = 150 mph (plane speed in still air is 150 mph) \n" ); document.write( "================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |