Rowing With and Against the Tide

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.