document.write( "Question 516665: A train leaves Lexington for Indianapolis, 200 miles away, at 2:00 PM and averages 60 miles per hour. A second train traveling on an adjacent track leaves Indianapolis for Lexington at 4:30 PM and averages 40 miles per hour. At what time will the trains meet? (Round to the nearest minute.) \n" ); document.write( "
Algebra.Com's Answer #344494 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A train leaves Lexington for Indianapolis, 200 miles away, at 2:00 PM and averages 60 miles per hour. \n" ); document.write( " A second train traveling on an adjacent track leaves Indianapolis for Lexington at 4:30 PM and averages 40 miles per hour. \n" ); document.write( " At what time will the trains meet? (Round to the nearest minute.) \n" ); document.write( ": \n" ); document.write( "Let t = travel time of the Ind/Lex train (2nd train) \n" ); document.write( "then \n" ); document.write( "(t+2.5) = travel time of the Lex/Ind train (1st train) \n" ); document.write( ": \n" ); document.write( "When they meet their travel distances will total 200 mi \n" ); document.write( "Write a distance equation: dist = speed * time \n" ); document.write( ": \n" ); document.write( "60(t+2.5) + 40t = 200 \n" ); document.write( "60t + 150 + 40t = 200 \n" ); document.write( "60t + 40t = 200 - 150 \n" ); document.write( "100t = 50 \n" ); document.write( "t = \n" ); document.write( "t = .5 hrs which is 30 minutes. They meet at 5 PM \n" ); document.write( "; \n" ); document.write( ": \n" ); document.write( "Check this, find the dist each traveled. (1st train travel time: 2.5+.5 = 3 hrs) \n" ); document.write( "60(3) = 180 \n" ); document.write( "40(.5)= 20 \n" ); document.write( "----------- \n" ); document.write( "total: 200 mi \n" ); document.write( " |