document.write( "Question 800757: a train leaves station M and travels west at 35 km/hr. one hour later a faster train leaves station M and also travels west on a parallel track at 40 km/hr. How far from the station M will the faster train catch the slower one? \n" ); document.write( "
Algebra.Com's Answer #483278 by Edwin McCravy(20056)\"\" \"About 
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a train leaves station M and travels west at 35 km/hr. one hour later a faster train leaves station M and also travels west on a parallel track at 40 km/hr. How far from the station M will the faster train catch the slower one?
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document.write( "There are two ways to do it, without algebra, and with algebrs:\r\n" );
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document.write( "Method 1; Without algebra:\r\n" );
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document.write( "You can do it this way in you head.  When the faster train starts an \r\n" );
document.write( "hour after the slower train left, the slower train has 35 km head start.\r\n" );
document.write( "The faster train's catch-up rate  is the difference in their speeds, or \r\n" );
document.write( "40-35 or 5 km/hr. So to catch up the entire 35 km at 5 km/hr will take \r\n" );
document.write( "7 hrs.   \r\n" );
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document.write( "Method 2: With algebra.\r\n" );
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document.write( "Make this chart:\r\n" );
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document.write( "                  Distance    Rate     Time\r\n" );
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document.write( "Slower train     \r\n" );
document.write( "Faster train       \r\n" );
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document.write( "Put in the speeds:\r\n" );
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document.write( "                  Distance    Rate     Time\r\n" );
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document.write( "Slower train                   35\r\n" );
document.write( "Faster train                   40         \r\n" );
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document.write( "Let t = the time it takes the faster train to catch\r\n" );
document.write( "the slower train:\r\n" );
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document.write( "                  Distance    Rate     Time\r\n" );
document.write( "-------------------------------------------\r\n" );
document.write( "Slower train                   35       \r\n" );
document.write( "Faster train                   40        t\r\n" );
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document.write( "Then since the slower train travels 1 hour more than\r\n" );
document.write( "the faster train, we put t+1 for the time of the slower \r\n" );
document.write( "train:\r\n" );
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document.write( "                  Distance    Rate     Time\r\n" );
document.write( "-------------------------------------------\r\n" );
document.write( "Slower train                   35       t+1\r\n" );
document.write( "Faster train                   40        t\r\n" );
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document.write( "Use Distance = Rate × time to indicate the multiplication\r\n" );
document.write( "of the rate times the time.\r\n" );
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document.write( "                  Distance    Rate     Time\r\n" );
document.write( "-------------------------------------------\r\n" );
document.write( "Slower train      35(t+1)      35       t+1\r\n" );
document.write( "Faster train        40t        40        t\r\n" );
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document.write( "Both trains had traveled the same distances when\r\n" );
document.write( "the faster one catches the slower one.  So we\r\n" );
document.write( "set the two distances equal:\r\n" );
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document.write( " 35(t+1) = 40t\r\n" );
document.write( "35t + 35 = 40t\r\n" );
document.write( "      35 = 5t\r\n" );
document.write( "       7 = t\r\n" );
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document.write( "Answer: 7 hours.\r\n" );
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document.write( "Edwin
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