document.write( "Question 729392: If a slow engine leaves the tracks 1 1/2 hours earlier than a faster train that travels 1 1/2 times faster than the slow train, how long will it take for the fast train to catch up to the slow train?? \n" ); document.write( "
Algebra.Com's Answer #445834 by ankor@dixie-net.com(22740)\"\" \"About 
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If a slow engine leaves the tracks 1 1/2 hours earlier than a faster train that
\n" ); document.write( " travels 1 1/2 times faster than the slow train,
\n" ); document.write( " how long will it take for the fast train to catch up to the slow train??
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\n" ); document.write( "let t = travel time of the fast train
\n" ); document.write( "then
\n" ); document.write( "(t+1.5) = the travel time of the slow train
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\n" ); document.write( "let s = speed of the slow train
\n" ); document.write( "then
\n" ); document.write( "1.5s = speed of the fast train
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\n" ); document.write( "Write a distance equation, dist = speed * time,
\n" ); document.write( "When the fast catches the slow, they will have traveled the same distance.
\n" ); document.write( "1.5st = s(t+1.5)
\n" ); document.write( "1.5st = st + 1.5s
\n" ); document.write( "Divide thru by s
\n" ); document.write( "1.5t = t + 1.5
\n" ); document.write( "1.5t - t = 1.5
\n" ); document.write( ".5t = 1.5
\n" ); document.write( "mult by 2
\n" ); document.write( "t = 3 hrs for the fast to catch the slow
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\n" ); document.write( "Check this, let's assume the slow train speed is 40 mph, then the fast = 60
\n" ); document.write( "They should have traveled the same dist
\n" ); document.write( "40 * 4.5 = 180 mi
\n" ); document.write( "60 * 3 = 180 mi
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