document.write( "Question 939601: A train, an hour after starting, meets with an accident which detains it an hour, after which it proceeds at 3/5 of its former rate and arrives three hour after the time; but had the accident happened 50 miles farther on yhe line, it would have arrived one and one-half hour sooner. Find the length of the journey. \n" ); document.write( "
Algebra.Com's Answer #572702 by ankor@dixie-net.com(22740)\"\" \"About 
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A train, an hour after starting, meets with an accident which detains it an hour, after which it proceeds at 3/5 of its former rate and arrives three hour after the time;
\n" ); document.write( "but had the accident happened 50 miles farther on the line, it would have arrived one and one-half hour sooner.
\n" ); document.write( " Find the length of the journey.
\n" ); document.write( ":
\n" ); document.write( "Let s = normal speed of the train
\n" ); document.write( "then
\n" ); document.write( ".6s = speed after accident
\n" ); document.write( ":
\n" ); document.write( "d = length of journey
\n" ); document.write( "then
\n" ); document.write( "\"d%2Fs\" = normal travel time for journey
\n" ); document.write( ":
\n" ); document.write( "\"d%2Fs\" + 2 = travel time with accident (don't count the 1 hr delay)
\n" ); document.write( "1s = distance traveled before the accident
\n" ); document.write( "normal time + 2 hrs = 1 hr + time at .6s
\n" ); document.write( "d/s + 2 = 1 + (d-1s)/.6s
\n" ); document.write( "multiply equation by .6s, cancel the denominators
\n" ); document.write( ".6d + 1.2s = .6s + d - s
\n" ); document.write( ".6d + 1.2s = -.4s + d
\n" ); document.write( "1.2s + .4s = d - .6d
\n" ); document.write( "1.6s = .4d
\n" ); document.write( "divide both sides by .4
\n" ); document.write( "d = 4s
\n" ); document.write( ":
\n" ); document.write( "d/s + 1.5 hr = travel time if accident happened 50 mi further
\n" ); document.write( "d/s + 1.5 = 1 + 50/s + (d - 1s - 50)/.6s
\n" ); document.write( "subtract 1 from both sides
\n" ); document.write( "d/s + .5 = 50/s + (d - 1s - 50)/.6s
\n" ); document.write( "multiply equation by .6s
\n" ); document.write( ".6d + .6s(.5) = .6(50) + d - 1s - 50
\n" ); document.write( ".6d + .3s = 30 + d - 1s - 50
\n" ); document.write( ".3s + 1s = d -.6d - 20
\n" ); document.write( "1.3s = .4d - 20
\n" ); document.write( "replace d with 4s
\n" ); document.write( "1.3s = .4(4s) - 20
\n" ); document.write( "1.3s = 1.6s - 20
\n" ); document.write( "20 = 1.6s - 1.3s
\n" ); document.write( "20 = .3s
\n" ); document.write( "s = 20/.3
\n" ); document.write( "s = 66.67 is the normal speed
\n" ); document.write( "find d
\n" ); document.write( "d = 4(66.67)
\n" ); document.write( "d = 266.667 mi is the distance of the journey
\n" ); document.write( ":
\n" ); document.write( "Check this using the 1st scenario\
\n" ); document.write( "time at normal speed, 1 hr
\n" ); document.write( "time at slower speed: (266.667-66.667)/.6(66.667) = 200/4 = 5 hrs
\n" ); document.write( "------------------------
\n" ); document.write( "total time 6 hrs
\n" ); document.write( "Find the time with no accident, normal arrival
\n" ); document.write( "266.667/66.667 = 4 hrs
\n" ); document.write( "Accident time was 2 hrs longer like it said\r
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