document.write( "Question 422533: Two trains leave a city on parallel tracks, traveling the same direction. The passenger train is going twice as fast as the freight train. After 45 minutes, the trains are 30 miles apart. Find the speed of each train. \n" ); document.write( "

Algebra.Com's Answer #294925 by Edwin McCravy(20059) $\"\"$ $\"About$

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Two trains leave a city on parallel tracks, traveling the same direction. The passenger train is going twice as fast as the freight train. After 45 minutes, the trains are 30 miles apart. Find the speed of each train.
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document.write( "Two ways.  (1) In your head or (2) by algebra:\r\n" );
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document.write( "In your head:\r\n" );
document.write( "Since the faster train goes twice as fast as the slower \r\n" );
document.write( "train, then after 45 minutes, the faster train must be \r\n" );
document.write( "twice as far from the station as the slower train, so \r\n" );
document.write( "the slower train is 30 miles from the station and 30 \r\n" );
document.write( "miles from the faster train, which is 60 miles from the\r\n" );
document.write( "station. So it took the slower train 45 minutes to go \r\n" );
document.write( "30 miles, so it could go another 1/3 of 30 or 10 more \r\n" );
document.write( "miles in 1/3 of 45 or 15 more minutes.  So it could go \r\n" );
document.write( "40 miles in one hour, and the faster train could go 80\r\n" );
document.write( "miles in one hour. So their speeds are 40 mph and 80 mph.\r\n" );
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document.write( "By algebra:\r\n" );
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document.write( "Make this chart:\r\n" );
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document.write( "              Rate     Time    Distance\r\n" );
document.write( "Fast train   \r\n" );
document.write( "Slow train   \r\n" );
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document.write( "Let x = the slow train's rate and 2x = the fast train's rate.\r\n" );
document.write( "Fill those in\r\n" );
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document.write( "              Rate     Time    Distance\r\n" );
document.write( "Fast train    2x      \r\n" );
document.write( "Slow train     x     \r\n" );
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document.write( "Esch went for 45 minutes after leaving the station.  Since\r\n" );
document.write( "45 minutes is 3/4 hour, we put 3/4 for both times.\r\n" );
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document.write( "              Rate     Time    Distance\r\n" );
document.write( "Fast train    2x       3/4      \r\n" );
document.write( "Slow train     x       3/4      \r\n" );
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document.write( "Then we use Distance = Rate × time to fill in the Distances\r\n" );
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document.write( "              Rate     Time    Distance\r\n" );
document.write( "Fast train    2x       3/4      (6/4)x\r\n" );
document.write( "Slow train     x       3/4      (3/4)x\r\n" );
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document.write( "To make the equation:\r\n" );
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document.write( "Fast train's distance = Slow train's distance + 30\r\n" );
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document.write( "             (6/4)x - (3/4)x = 30\r\n" );
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document.write( "                      (3/4)x = 30\r\n" );
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document.write( "Multiply both sides by 4\r\n" );
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document.write( "                          3x = 120\r\n" );
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document.write( "                           x = 40\r\n" );
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document.write( "So the slower train is going 40 mph and the faster\r\n" );
document.write( "train is going twice that speed or 80 mph.\r\n" );
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document.write( "Edwin

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