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
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Rowing with and against the Tide: $145$