document.write( "Question 262305: A man jogs at 6 mph while a cyclist moves at 18 mph. After how long will the jogger be overtaken if the cyclist leaves one hour after the jogger begins the course? \r
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Algebra.Com's Answer #193251 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
A man jogs at 6 mph while a cyclist moves at 18 mph. After how long will the jogger be overtaken if the cyclist leaves one hour after the jogger begins the course?
\n" ); document.write( "This is the table that I've set up so far, I cannot seem to get beyond this point because I do not know a distance for the two travelers.
\n" ); document.write( "Speed: 6 18
\n" ); document.write( "Distance:
\n" ); document.write( "Time: x x+60
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document.write( "You don't need to change 1 hour to minutes. You wouldn't\r\n" );
document.write( "want to do that, because if you did that you'd have to change'\r\n" );
document.write( "the speeds to miles per minute instead of miles per hour and you\r\n" );
document.write( "wouldn't want to do that!\r\n" );
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document.write( "First, let's do it in our head, then we'll do it by algebra.\r\n" );
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document.write( "Here's how to do it in your head: \r\n" );
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document.write( "When the cyclist starts, the jogger is already 6 miles down the road.\r\n" );
document.write( "So the jogger has a 6-mile head start on the cyclist. The cyclist goes\r\n" );
document.write( "18mph and the jogger goes 6mph, so the cyclist approaches the jogger\r\n" );
document.write( "at a rate of 18mph-6mph or 12mph.  So, approaching the jogger at 12mph,\r\n" );
document.write( "he can make up the jogger's 6 mile head-start in half an hour. \r\n" );
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document.write( "Now here's the algebra way:\r\n" );
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document.write( "Make this chart:\r\n" );
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document.write( "             DISTANCE      RATE      TIME\r\n" );
document.write( "jogger        \r\n" );
document.write( "cyclist        \r\n" );
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document.write( "You want to know the cyclist's time, so put t for that\r\n" );
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document.write( "             DISTANCE      RATE      TIME\r\n" );
document.write( "jogger        \r\n" );
document.write( "cyclist                                t\r\n" );
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document.write( "The jogger's time was 1 hour more so add 1 to t and put\r\n" );
document.write( "t+1 for the jogger's time:\r\n" );
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document.write( "             DISTANCE      RATE      TIME\r\n" );
document.write( "jogger                                t+1\r\n" );
document.write( "cyclist                                t\r\n" );
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document.write( "Now fill in their rates which are given:\r\n" );
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document.write( "             DISTANCE      RATE      TIME\r\n" );
document.write( "jogger                       6        t+1\r\n" );
document.write( "cyclist                     18         t\r\n" );
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document.write( "Now use DISTANCE = RATE x TIME to fill in the\r\n" );
document.write( "distances.\r\n" );
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document.write( "             DISTANCE      RATE      TIME\r\n" );
document.write( "jogger         6(t+1)        6        t+1\r\n" );
document.write( "cyclist         18t         18         t\r\n" );
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document.write( "Now when the cyclist catches up to the jogger,\r\n" );
document.write( "they will have both traveled the same distance,\r\n" );
document.write( "so we set the expressions for their distances equal\r\n" );
document.write( "to each other:\r\n" );
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document.write( "      6(t+1) = 18t\r\n" );
document.write( "      6t + 6 = 18t\r\n" );
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document.write( "Subtract 6t from both sides\r\n" );
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document.write( "           6 = 12t\r\n" );
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document.write( "Divide both sides by 12\r\n" );
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document.write( "          \"6%2F12=12t%2F12\"\r\n" );
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document.write( "Cancel the 12's on the right, and\r\n" );
document.write( "reduce the fraction on the left:\r\n" );
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document.write( "          \"1%2F2=t\"\r\n" );
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document.write( "So the answer is half an hour, just what we got\r\n" );
document.write( "when we did it in our heads.\r\n" );
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document.write( "Edwin
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