Question 22105
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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, 
how long will it take him to reach the log?
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Two of the persons who responded don't have a clue, and that's why their answers don't make sense.

Let the time he takes to get to the log, be T
Going downstream, his total average speed is his speed in still water, plus the speed of the current, or 8 + 2 = 10 mph
Of the 4 miles between he and the log, Phil will cover 10T miles when he gets to the log

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
Of the 4 miles between the log and Phil, the log will cover - 2T miles when Phil catches up to it.

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
actually travelling/drifting BACKWARDS, or AWAY from Phil.

We now get the following DISTANCE equation: 10T + - 2T = 4
                                                                                               8T = 4
                               Time it takes Phil to get to the log, or {{{T = 4/8 = highlight(matrix(1,5, 1/2, "hour,", or, 30, minutes))}}}

<font color = red><u>ANECDOTE**</u></font>
Phil actually traveled 10T, or 10({{{1/2}}}) = 5 miles to get to the log, although he started out just 4 miles from it. With the log travelling 
backwards, or drifting away from him, he had to travel an extra mile to get to it. Incidentally, the log travelled/drifted - 2({{{1/2}}}) = - 1
mile, or 1 mile, BACKWARDS, which is the extra distance Phil ended up making up/travelling to get to it.

Quite INTERESTING, isn't it?</font></font></font></b></pre>