document.write( "Question 250844: A car leaves the Kansas City at noon traveling west at 45 miles per hour. A second car leaves from the same location 3 hours later traveling west at 60 miles per hour. At what time will the second car catch up with the first? \n" ); document.write( "
Algebra.Com's Answer #184432 by drk(1908)\"\" \"About 
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We need to set up an RTD chart to solve this:\r
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\n" ); document.write( "\n" ); document.write( "west . . . . . RATE . . . . . time . . . . distance
\n" ); document.write( "CAR 1 . . . . 45 mph . . . . . t . . . . . 45t
\n" ); document.write( "CAR 2 . . . . 60 mph . . . . . t-3 . . . . 60t - 180\r
\n" ); document.write( "\n" ); document.write( "we are given car 1 and car 2 rates as 45 and 60 respectively. Car 1 leaves first and car 2 3 hours later. This means that car 2 is on the road for 3 fewer hours, t-3. rate x time = distance. This is where I got 45t and 60t - 180. For the second car to catch up to the first means the distances must be equal.
\n" ); document.write( "So,
\n" ); document.write( "45t = 60t - 180.
\n" ); document.write( "Solving for t, we get
\n" ); document.write( "T = 12 hours.
\n" ); document.write( "Since car 1 left at noon (12 o'clock), the second car would catch up at midnight.\r
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