document.write( "Question 728865: a train, an hour after starting, meets with an accident which detains it an hour .after which it proceed at 3/5 of its former rate arrives 3hrs after time but had the accident happened 50miles farther on the line, it would have arrived 1.5hr sooner.find the length of the journey \n" ); document.write( "
Algebra.Com's Answer #446156 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A train, an hour after starting, meets with an accident which detains it an hour. \n" ); document.write( "After which it proceed at 3/5 of its former rate, arriving 3 hrs late. \n" ); document.write( "But had the accident happened 50miles farther on the line, it would have arrived 1.5hr sooner. \n" ); document.write( "Find the length of the journey \n" ); document.write( ": \n" ); document.write( "let s = normal speed of the train (s also = the dist traveled the 1st hr) \n" ); document.write( "then \n" ); document.write( ".6s = speed after the accident \n" ); document.write( ": \n" ); document.write( "let d = distance of the journey \n" ); document.write( "and \n" ); document.write( "d/s = normal time of the journey \n" ); document.write( ": \n" ); document.write( "Write a time equation for the 1st scenario \n" ); document.write( ": \n" ); document.write( "normal speed time + stopped time + 3/5 speed time - normal time = 3 hrs \n" ); document.write( "1 + 1 + ( \n" ); document.write( "2 + ( \n" ); document.write( "multiply by .6s, resulting in: \n" ); document.write( "2(.6s) + d - s - .6d = .6s(3) \n" ); document.write( "1.2s + d - s - .6d = 1.8s \n" ); document.write( "combine like terms \n" ); document.write( "d - .6d = 1.8s - 1.2s + s \n" ); document.write( ".4d = 1.6s \n" ); document.write( "divide both sides by .4 \n" ); document.write( "d = \n" ); document.write( "d = 4s \n" ); document.write( ": \n" ); document.write( "An equation for the 2nd scenario \n" ); document.write( "\" but had the accident happened 50miles farther on the line, it would have arrived 1.5hr sooner.\" \n" ); document.write( "2 + \n" ); document.write( "2 + \n" ); document.write( "Multiply by .6s \n" ); document.write( ".6s(2) + .6(50) + d - s - 50 - .6d = .6s(1.5) \n" ); document.write( "1.2s + 30 + d - s - 50 - .6d = .9s \n" ); document.write( "Combine like terms \n" ); document.write( "1.2s - s - .9s + d -.6d -50 + 30 = 0 \n" ); document.write( "-.7s + .4d - 20 = 0 \n" ); document.write( "-.7s + .4d = 20 \n" ); document.write( "Replace d with 4s \n" ); document.write( "-.7s + .4(4s) = 20 \n" ); document.write( "-.7s + 1.6s = 20 \n" ); document.write( ".9s = 20 \n" ); document.write( "s = 20/.9 \n" ); document.write( "s = 22.2 mph \n" ); document.write( "then \n" ); document.write( "4(22.2) = 88.8 mi is the journey \n" ); document.write( " |