document.write( "Question 22105: River flows at 2 mph. Phil can paddle in still water at 8 mph. If he is 4 miles downstream from a log floating toward him,
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Algebra.Com's Answer #854322 by MathTherapy(10839) $\"\"$ $\"About$

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document.write( "River flows at 2 mph.  Phil can paddle in still water at 8 mph.  If he is 4 miles downstream from a log floating toward him, \r\n" );
document.write( "how long will it take him to reach the log?\r\n" );
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document.write( "Two of the persons who responded don't have a clue, and that's why their answers don't make sense.\r\n" );
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document.write( "Let the time he takes to get to the log, be T\r\n" );
document.write( "Going downstream, his total average speed is his speed in still water, plus the speed of the current, or 8 + 2 = 10 mph\r\n" );
document.write( "Of the 4 miles between he and the log, Phil will cover 10T miles when he gets to the log\r\n" );
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document.write( "Going upstream towards Phil, the log's speed is its speed in still water (0 mph), less the speed of the current, or 0 - 2 = - 2 mph\r\n" );
document.write( "Of the 4 miles between the log and Phil, the log will cover - 2T miles when Phil catches up to it.\r\n" );
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document.write( "A speed of - 2 mph, and a distance of - 2T miles might seem strange and non-sensical, but this actually means that the log is\r\n" );
document.write( "actually travelling/drifting BACKWARDS, or AWAY from Phil.\r\n" );
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document.write( "We now get the following DISTANCE equation: 10T + - 2T = 4\r\n" );
document.write( "                                                                                               8T = 4\r\n" );
document.write( "                               Time it takes Phil to get to the log, or \r\n" );
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document.write( "ANECDOTE**\r\n" );
document.write( "Phil actually traveled 10T, or 10() = 5 miles to get to the log, although he started out just 4 miles from it. With the log travelling \r\n" );
document.write( "backwards, or drifting away from him, he had to travel an extra mile to get to it. Incidentally, the log travelled/drifted - 2() = - 1\r\n" );
document.write( "mile, or 1 mile, BACKWARDS, which is the extra distance Phil ended up making up/travelling to get to it.\r\n" );
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document.write( "Quite INTERESTING, isn't it?

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