document.write( "Question 136068This question is from textbook algebra structure and method
\n" ); document.write( ": Tracy leaves for work at 8:a.m traveling at 36 mph. Sarah finds that Tracy has forgotten her briefcase, so she leaves 8:05 chasing after her at 50 mph. At what time will Sarah catch tracy? \n" ); document.write( "
Algebra.Com's Answer #99696 by ankor@dixie-net.com(22740)\"\" \"About 
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Tracy leaves for work at 8:a.m traveling at 36 mph. Sarah finds that Tracy has forgotten her briefcase, so she leaves 8:05 chasing after her at 50 mph. At what time will Sarah catch tracy?
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\n" ); document.write( "Change 5 min to hrs: 5/60 = \"1%2F12\" hrs\r
\n" ); document.write( "\n" ); document.write( "Let t = travel time (in hrs), for T to catch up with S
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\n" ); document.write( "When T catches up with S, there will have traveled the same distance
\n" ); document.write( "Write a distance equation from this fact: Dist = speed * time
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\n" ); document.write( "T's dist = S's distance
\n" ); document.write( "50t = 36(t+\"1%2F12\")
\n" ); document.write( "50t = 36t + \"36%2F12\"
\n" ); document.write( "50t - 36t = 3
\n" ); document.write( "14t = 3
\n" ); document.write( "t = \"3%2F14\"
\n" ); document.write( "change to minutes and fractional minutes to seconds
\n" ); document.write( "60 * \"3%2F14\" = 12.857 min = 12 min 51 sec
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\n" ); document.write( "8:05 + 12 min 51 sec = 08:17:51 AM, when T catches up with S
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\n" ); document.write( "Check solution on calc by finding the distances
\n" ); document.write( "36(\"3%2F14+%2B+1%2F12%29\") =
\n" ); document.write( "36 * .297 = 10.7 mi
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\n" ); document.write( "50 * \"3%2F14\" = 10.7 mi
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