Question

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 river, 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.

Answer 1

1.8

Answer 2

Let $v_b$ be the velocity of the boat with respect to water and $v_r$ be the velocity of the river stream. We are given that $v_b = \eta v_r$, where $\eta = 1.2$. Let the equal distance moved by each boat from the buoy be $d$.

1. Motion of Boat A (along the river):

When boat A moves downstream (away from the buoy), its velocity with respect to the ground is $v_{A,down} = v_b + v_r$. Time taken to travel distance $d$ downstream: $t_{A,down} = \frac{d}{v_b + v_r}$.

When boat A moves upstream (back to the buoy), its velocity with respect to the ground is $v_{A,up} = v_b - v_r$. Time taken to travel distance $d$ upstream: $t_{A,up} = \frac{d}{v_b - v_r}$.

The total time for boat A is: $\tau_A = t_{A,down} + t_{A,up} = \frac{d}{v_b + v_r} + \frac{d}{v_b - v_r}$ $\tau_A = d \left( \frac{1}{v_b + v_r} + \frac{1}{v_b - v_r} \right) = d \left( \frac{(v_b - v_r) + (v_b + v_r)}{(v_b + v_r)(v_b - v_r)} \right)$ $\tau_A = d \left( \frac{2v_b}{v_b^2 - v_r^2} \right)$

Substitute $v_b = \eta v_r$: $\tau_A = d \left( \frac{2\eta v_r}{(\eta v_r)^2 - v_r^2} \right) = d \left( \frac{2\eta v_r}{\eta^2 v_r^2 - v_r^2} \right) = d \left( \frac{2\eta v_r}{v_r^2(\eta^2 - 1)} \right)$ $\tau_A = \frac{2\eta d}{v_r(\eta^2 - 1)}$

2. Motion of Boat B (across the river):

To move straight across the river, boat B must head at an angle upstream such that the river's velocity cancels out the upstream component of the boat's velocity relative to water. Let $v_{B,ground}$ be the velocity of boat B with respect to the ground, perpendicular to the river flow. Using the Pythagorean theorem for velocities: $v_b^2 = v_r^2 + v_{B,ground}^2$. So, $v_{B,ground} = \sqrt{v_b^2 - v_r^2}$. This is the effective speed of the boat for moving across the river. Since the boat travels an equal distance $d$ away from the buoy and then returns, the effective speed for both trips is the same. The total time for boat B is: $\tau_B = \frac{2d}{v_{B,ground}} = \frac{2d}{\sqrt{v_b^2 - v_r^2}}$

Substitute $v_b = \eta v_r$: $\tau_B = \frac{2d}{\sqrt{(\eta v_r)^2 - v_r^2}} = \frac{2d}{\sqrt{\eta^2 v_r^2 - v_r^2}} = \frac{2d}{\sqrt{v_r^2(\eta^2 - 1)}}$ $\tau_B = \frac{2d}{v_r\sqrt{\eta^2 - 1}}$

3. Ratio of times $\tau_A / \tau_B$:

Now, we find the ratio of the total times: $\frac{\tau_A}{\tau_B} = \frac{\frac{2\eta d}{v_r(\eta^2 - 1)}}{\frac{2d}{v_r\sqrt{\eta^2 - 1}}}$ $\frac{\tau_A}{\tau_B} = \frac{2\eta d}{v_r(\eta^2 - 1)} \times \frac{v_r\sqrt{\eta^2 - 1}}{2d}$ $\frac{\tau_A}{\tau_B} = \frac{\eta}{\sqrt{\eta^2 - 1}}$

4. Calculation with $\eta = 1.2$:

Substitute the given value $\eta = 1.2$: $\frac{\tau_A}{\tau_B} = \frac{1.2}{\sqrt{(1.2)^2 - 1}}$ $\frac{\tau_A}{\tau_B} = \frac{1.2}{\sqrt{1.44 - 1}}$ $\frac{\tau_A}{\tau_B} = \frac{1.2}{\sqrt{0.44}}$ $\frac{\tau_A}{\tau_B} = \frac{1.2}{\sqrt{\frac{44}{100}}} = \frac{1.2}{\frac{\sqrt{44}}{10}} = \frac{1.2 \times 10}{\sqrt{4 \times 11}} = \frac{12}{2\sqrt{11}}$ $\frac{\tau_A}{\tau_B} = \frac{6}{\sqrt{11}}$

To get a numerical value, $\sqrt{11} \approx 3.3166$. $\frac{\tau_A}{\tau_B} \approx \frac{6}{3.3166} \approx 1.8091$ Rounding off to one decimal place, the ratio is approximately 1.8.

Therefore, the ratio of times of motion of boats $\tau_A/\tau_B$ is approximately 1.8.