document.write( "Question 21644: An elevator went from the bottom to the top at 240m tower, remained there for 12 sec, and returned to the bottom in an elapsed time of 2 minutes. I fthe elevator traveled 1 m/s faster on the way down, find its speed going up. \n" ); document.write( "
Algebra.Com's Answer #11812 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Time up + time down = 120 sec -12 sec = 108 sec \n" ); document.write( "Up data: distance = 240 m; rate= x; time= 240/x \n" ); document.write( "Down data: dist = 240 m; rate = x+1; time = 240/(x+1) \n" ); document.write( "EQUATION: \n" ); document.write( " Time up + Time down = 108 sec \n" ); document.write( " 240/x + 240/(x+1) = 108 \n" ); document.write( " 1/x + 1/(x+1) = .45 \n" ); document.write( " 2x +1= .45(x^2+x) \n" ); document.write( " 9x^2-31x - 20 =0 \n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Looks like rate up is 0.55... meters/sec \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |