## 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 ${�}_{�}$, and the beaver swims with a constant velocity ${�}_{�}$ relative to the water. When the beaver swims from the dam to the grove (upstream), its effective velocity against the current is ${�}_{�}-{�}_{�}$, and the time taken is ${�}_{1}$. When the beaver swims back from the grove to the dam (downstream), its effective velocity with the current is ${�}_{�}+{�}_{�}$, and the time taken is ${�}_{2}$.

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

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

We want to find the time $�$ 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 ${�}_{�}$, and the time taken is $�=\frac{�}{{�}_{�}}$.

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

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

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

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

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

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