Two boats, A and B, move away from a buoy anchored at the middle of a river along the mutually perpendicular straight lines: the boat A along the rive, and the boat B across the river. Having moved off an equal distance from the buoy the boats returned. Find the ratio of times of motion of boats `tau_A//tau_B` if the velocity of each boat with respect to water is `eta=1.2` times greater than the stream velocity.

Explanation

To solve the problem of finding the ratio of times of motion of boats A and B, we will follow these steps:

### Step 1: Define the Variables
Let:
- \( v_r \) = stream velocity of the river
- \( v_a \) = velocity of boat A with respect to the river
- \( v_b \) = velocity of boat B with respect to the river
- \( \eta = 1.2 \) = the factor by which the velocity of the boats is greater than the stream velocity. Therefore, we can express the velocities as:
\[ v_a = \eta v_r = 1.2 v_r \]
\[ v_b = \eta v_r = 1.2 v_r \]

### Step 2: Determine the Effective Velocities
For boat A moving along the river:
- The effective velocity of boat A with respect to the ground is simply \( v_a = 1.2 v_r \).

For boat B moving across the river:
- The effective velocity of boat B can be calculated using the Pythagorean theorem since it moves perpendicular to the river's current:
\[ v_b = \sqrt{(v_r)^2 + (1.2 v_r)^2} = \sqrt{v_r^2 + 1.44 v_r^2} = \sqrt{2.44 v_r^2} = \sqrt{2.44} v_r \approx 1.56 v_r \]

### Step 3: Calculate the Time Taken by Each Boat
Both boats travel the same distance \( D \) away from the buoy and then return.

For boat A:
- The time taken \( \tau_A \) to travel distance \( D \) and return is:
\[ \tau_A = \frac{D}{v_a} + \frac{D}{v_a} = \frac{2D}{1.2 v_r} = \frac{5D}{6v_r} \]

For boat B:
- The time taken \( \tau_B \) to travel distance \( D \) and return is:
\[ \tau_B = \frac{D}{v_b} + \frac{D}{v_b} = \frac{2D}{1.56 v_r} = \frac{D}{0.78 v_r} \]

### Step 4: Find the Ratio of Times
Now we can find the ratio of the times taken by boats A and B:
\[ \frac{\tau_A}{\tau_B} = \frac{\frac{5D}{6v_r}}{\frac{D}{0.78 v_r}} = \frac{5}{6} \cdot \frac{0.78}{1} = \frac{5 \cdot 0.78}{6} = \frac{3.9}{6} = 0.65 \]

### Final Step: Simplify the Ratio
To express the ratio in a more understandable form:
\[ \frac{\tau_A}{\tau_B} = \frac{39}{60} = \frac{13}{20} \]

### Conclusion
Thus, the ratio of the times of motion of boats A and B is:
\[ \frac{\tau_A}{\tau_B} = \frac{13}{20} \]