document.write( "Question 370418: Two cars each travel 75 miles at a constant rate. one car travels 6 mph faster than the other and arrives 5 minutes before the other arrives. Find the rates of speed of the two cars. \n" ); document.write( "
Algebra.Com's Answer #264144 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! Two cars each travel 75 miles at a constant rate. \n" ); document.write( " one car travels 6 mph faster than the other and arrives 5 minutes before the other arrives. \n" ); document.write( " Find the rates of speed of the two cars. \n" ); document.write( ": \n" ); document.write( "Let s = speed of the slower car \n" ); document.write( "then \n" ); document.write( "(s+6) = speed of the faster car \n" ); document.write( ": \n" ); document.write( "Convert 5 min to hrs 5/60 = 1/12 hr \n" ); document.write( ": \n" ); document.write( "Write a time equation: Time = dist/speed \n" ); document.write( ": \n" ); document.write( "Slow car time - faster car time = 5 min (1/12 hr) \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "Multiply by 12s(s+6) to get rid of the denominators, results: \n" ); document.write( "12(s+6)*75 - 12s(75) = s(s+6) \n" ); document.write( ": \n" ); document.write( "900(s+6) - 900s = s^2 + 6s \n" ); document.write( ": \n" ); document.write( "900s + 5400 - 900s = s^2 + 6s \n" ); document.write( ": \n" ); document.write( "5400 = s^2 + 6s \n" ); document.write( ": \n" ); document.write( "A quadratic equation \n" ); document.write( "0 = s^2 + 6s - 5400 \n" ); document.write( ": \n" ); document.write( "Use the quadratic formula to find s \n" ); document.write( " \n" ); document.write( "do the math, you should get a positive solution s ~ 70.546 mph\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |