Rowing With and Against the Tide

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Problem

If, during ebb tide, a wherry should set out from Haverhill, to come down the river,
and, at the same time, another should set out from Newburyport, to go up the river,
allowing the difference to be $18$ miles;
suppose the current forwards one and retards the other $1 \frac 1 2$ miles per hour;
the boats are equally laden, the rowers equally good,
and, in the common way of working in still water, would proceed at the rate of $4$ miles per hour;
when, in the river, will the two boats meet?


Solution

$5 \frac 5 8$ miles upstream from Newburyport.


Proof

The rowers are moving at $8$ miles per hour relative to each other.

Hence they will meet in $\dfrac {18} 8 = \dfrac 9 4$ hours.

During that time, the rower from Newburyport is rowing at a speed of $4 - 1 \frac 1 2 = 2 \frac 1 2 = \dfrac 5 2$ miles per hour relative to land.

Thus he gets as far is it takes to travel for $\dfrac 9 4$ hours, that is: $\dfrac 9 4 \times \dfrac 5 2 = \dfrac {45} 8 = 5 \frac 5 8$ miles from Newburyport.

$\blacksquare$


Historical Note

David Wells reports in his Curious and Interesting Puzzles of $1992$ that this problem first appeared in an American textbook in $1788$.

He omits to inform us which.


Sources