## Physics Problem

The beavers noticed that swimming from the dam to the grove along the river takes time t₁, and swimming back along the river from the grove to the dam takes time t₂, and t₁ < t₂.

How long will it take a log to float from the grove to the dam?

## Solution

Let's assume the river has a constant current velocity ${\ufffd}_{\ufffd}$, and the beaver swims with a constant velocity ${\ufffd}_{\ufffd}$ relative to the water. When the beaver swims from the dam to the grove (upstream), its effective velocity against the current is ${\ufffd}_{\ufffd}-{\ufffd}_{\ufffd}$, and the time taken is ${\ufffd}_{1}$. When the beaver swims back from the grove to the dam (downstream), its effective velocity with the current is ${\ufffd}_{\ufffd}+{\ufffd}_{\ufffd}$, and the time taken is ${\ufffd}_{2}$.

Let $\ufffd$ be the distance between the dam and the grove. We can write:

$\ufffd=({\ufffd}_{\ufffd}-{\ufffd}_{\ufffd})\cdot {\ufffd}_{1}$ (1) $\ufffd=({\ufffd}_{\ufffd}+{\ufffd}_{\ufffd})\cdot {\ufffd}_{2}$ (2)

We want to find the time $\ufffd$ it takes for a log to float from the grove to the dam. Since the log is carried only by the current, its velocity is ${\ufffd}_{\ufffd}$, and the time taken is $\ufffd=\frac{\ufffd}{{\ufffd}_{\ufffd}}$.

From equations (1) and (2), we can solve for ${\ufffd}_{\ufffd}$ and $\ufffd$:

${\ufffd}_{\ufffd}=\frac{\ufffd}{{\ufffd}_{2}}-{\ufffd}_{\ufffd}=\frac{\ufffd}{{\ufffd}_{1}}+{\ufffd}_{\ufffd}$ $2{\ufffd}_{\ufffd}=\frac{\ufffd}{{\ufffd}_{2}}-\frac{\ufffd}{{\ufffd}_{1}}$ ${\ufffd}_{\ufffd}=\frac{\ufffd}{2}(\frac{1}{{\ufffd}_{2}}-\frac{1}{{\ufffd}_{1}})$

Substituting ${\ufffd}_{\ufffd}$ back into equation (1):

$\ufffd=(\frac{\ufffd}{2}(\frac{1}{{\ufffd}_{2}}-\frac{1}{{\ufffd}_{1}})-{\ufffd}_{\ufffd})\cdot {\ufffd}_{1}$ $2\ufffd=\ufffd(\frac{{\ufffd}_{1}}{{\ufffd}_{2}}-1-\frac{{\ufffd}_{\ufffd}{\ufffd}_{1}}{\ufffd})$ ${\ufffd}_{\ufffd}=\frac{\ufffd}{{\ufffd}_{1}}-\frac{\ufffd}{2}(\frac{1}{{\ufffd}_{2}}+\frac{1}{{\ufffd}_{1}})$ ${\ufffd}_{\ufffd}=\frac{2\ufffd}{{\ufffd}_{1}{\ufffd}_{2}}\cdot \frac{{\ufffd}_{2}-{\ufffd}_{1}}{{\ufffd}_{2}+{\ufffd}_{1}}$

Finally, the time $\ufffd$ it takes for a log to float from the grove to the dam is:

$\ufffd=\frac{\ufffd}{{\ufffd}_{\ufffd}}=\frac{{\ufffd}_{1}{\ufffd}_{2}}{{\ufffd}_{2}-{\ufffd}_{1}}\cdot \frac{{\ufffd}_{2}+{\ufffd}_{1}}{2}=\frac{{\ufffd}_{1}{\ufffd}_{2}}{2}\left(\frac{1}{{\ufffd}_{2}-{\ufffd}_{1}}\right)\left(\frac{{\ufffd}_{2}+{\ufffd}_{1}}{{\ufffd}_{1}{\ufffd}_{2}}\right)$