$\dfrac{t_A}{t_B}=\dfrac{\eta}{\sqrt{\eta^2-1}}=\dfrac{1.2}{\sqrt{1.44-1}}\approx 1.8$
(ii) Problem Statement
"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 $t_A/t_B$ if the velocity of each boat with respect to water is $\eta=1.2$ times greater than the stream velocity."
Reference Answer Verification (from the supplied solution image):
The reference finds $t_A=\dfrac{l}{v'+v_0}+\dfrac{l}{v'-v_0}$ for boat A and $t_B=\dfrac{l}{\sqrt{v'^2-v_0^2}}+\dfrac{l}{\sqrt{v'^2-v_0^2}}=\dfrac{2l}{\sqrt{v'^2-v_0^2}}$ for boat B. Dividing gives $\dfrac{t_A}{t_B}=\dfrac{v'}{\sqrt{v'^2-v_0^2}}=\dfrac{\eta}{\sqrt{\eta^2-1}}=1.8$. Our work matches.
(iii) The Big Idea in Plain Words
Think of a moving walkway at an airport. Walking with it is quick. Walking against it is slow. The time you lose going against it is bigger than the time you gain going with it.
Symbols: $l$ is the distance each boat goes from the buoy. $v_0$ is the speed of the stream. $v'$ is a boat's speed relative to the water. $\eta=v'/v_0$ is the ratio of the two. $t_A$ and $t_B$ are the total times of boats A and B. $s$ is boat B's speed across the river, as seen from the bank.
Boat A goes along the river. It goes downstream, turns, and comes back upstream. Boat B must go straight across and straight back, as seen from the bank. So it aims a little upstream to cancel the push of the stream. Both boats cover the same distance $l$ each way. So only the speeds matter, and the answer does not depend on $l$ or on $v_0$ alone. It depends only on $\eta$.
(iv) The Solution Viewed in Each Reference Frame
Example numbers for a quick check: $v_0=5$ km/h and $v'=6$ km/h, so $\eta=1.2$. Any other $v_0$ gives the same ratio.
1. Ground or Lab Frame (Observer on the Bank)
Origin fixed at the buoyWhy: going with the stream, the stream adds to the boat's speed. Going against it, the stream takes away from it.
Boat A, out: speed $v'+v_0=11$ km/h, so $t_{A1}=\dfrac{l}{v'+v_0}$. Boat A, back: speed $v'-v_0=1$ km/h, so $t_{A2}=\dfrac{l}{v'-v_0}$.
Why: the total time is the sum of the two legs.
$t_A=\dfrac{l}{v'+v_0}+\dfrac{l}{v'-v_0}=\dfrac{2lv'}{v'^2-v_0^2}$.
Why: boat B must stay on a line straight across, so it can have no sideways motion along the bank. It aims upstream so that the sideways part of $v'$ equals $v_0$.
Its speed across is what is left: $s=\sqrt{v'^2-v_0^2}=\sqrt{36-25}\approx 3.32$ km/h. This is the same going out and coming back, so $t_B=\dfrac{2l}{s}=\dfrac{2l}{\sqrt{v'^2-v_0^2}}$.
Why: the ratio removes $l$ and leaves only speeds.
$$\frac{t_A}{t_B}=\frac{2lv'/(v'^2-v_0^2)}{2l/\sqrt{v'^2-v_0^2}}=\frac{v'}{\sqrt{v'^2-v_0^2}}=\frac{\eta}{\sqrt{\eta^2-1}}=\frac{1.2}{\sqrt{0.44}}\approx 1.8$$
Check with numbers: $t_A=l\left(\tfrac{1}{11}+\tfrac{1}{1}\right)=1.091\,l$ and $t_B=\tfrac{2l}{3.317}=0.603\,l$, so $t_A/t_B=1.809$.
2. Water Frame (Medium Frame, Observer Drifting With the Stream)
Origin moves at $v_0$Why: in this frame the water is at rest, so each boat moves at exactly $v'$ relative to the water. The buoy and the banks drift upstream (west) at $v_0$.
Only the speed $v'$ and the drifting buoy matter now.
Why: the boat has to get a distance $l$ from the buoy. The gap grows at the sum of the speeds when they move apart, and shrinks at the difference when one chases the other.
Boat A, out: the boat goes east at $v'$ and the buoy goes west at $v_0$. The gap grows at $v'+v_0$. Boat A, back: both go west. The boat chases the buoy, so the gap shrinks at $v'-v_0$. This gives the same $t_A$ as before.
Why: boat B must stay on the line straight across from the buoy. So it must drift west with the buoy at exactly $v_0$.
Its speed $v'$ splits into $v_0$ (west) and $s$ (across). So $s=\sqrt{v'^2-v_0^2}$ and $t_B=\dfrac{2l}{s}$. The ratio is again $\dfrac{\eta}{\sqrt{\eta^2-1}}\approx 1.8$.
3. Centre-of-Mass (CM) Frame
This frame does not apply. No masses are given, the buoy is anchored to the ground, and a centre of mass of the two boats would add nothing to a question about times.
What Changes and What Stays the Same Across Frames
- What changes: the paths and the velocities. In the water frame the buoy drifts west, and boat A's speeds look like $v'$ while the gap changes at $v'\pm v_0$.
- What stays the same: the distance $l$, the times $t_A$ and $t_B$, and the ratio $t_A/t_B=\eta/\sqrt{\eta^2-1}\approx 1.8$.
(v) The Solution by Each Method
Method 1: Relative-Motion Method (Water Frame)
Fastest- Why: with $v'=1.2v_0$, the gap speeds for boat A are simple multiples of $v_0$.
Out: $v'+v_0=2.2v_0$. Back: $v'-v_0=0.2v_0$. So $t_A=\dfrac{l}{2.2v_0}+\dfrac{l}{0.2v_0}=5.455\,\dfrac{l}{v_0}$. - Why: boat B's speed across is what is left of $v'$ after it cancels the stream.
$s=v_0\sqrt{1.2^2-1}=0.663v_0$. So $t_B=\dfrac{2l}{0.663v_0}=3.015\,\dfrac{l}{v_0}$. - Why: divide to remove $l$ and $v_0$.
$\dfrac{t_A}{t_B}=\dfrac{5.455}{3.015}\approx 1.8$.
Method 2: Graph Method (Velocity Triangle and Timeline)
- Why: for boat B, $\vec v'$, $\vec v_0$ and $\vec v_B$ form a right triangle with $v'$ as the longest side. You can see it in the inset of the simulation.
In units of $v_0$ the sides are $1.2$, $1$ and $\sqrt{1.44-1}=0.663$. So $s=0.663v_0$. - Why: for boat A the speeds simply add or subtract, so the "triangle" is flat. A time is distance divided by speed, so the slow leg takes much longer.
The back leg speed is only $0.2v_0$, so it takes $5\,l/v_0$. The out leg takes only $0.455\,l/v_0$. - Why: on the timeline at the bottom of the simulation, you can read the total lengths of the two bars.
The bar of boat A is $5.455$ units long and the bar of boat B is $3.015$ units long. Their ratio is $\approx 1.8$.
Method 3: Calculus Method
- Why: the distance covered is the integral of the speed over time, and each leg must reach $l$.
Boat A, out: $\displaystyle\int_0^{t_{A1}}(v'+v_0)\,dt=l\Rightarrow t_{A1}=\dfrac{l}{v'+v_0}$. Back: $\displaystyle\int_0^{t_{A2}}(v'-v_0)\,dt=l\Rightarrow t_{A2}=\dfrac{l}{v'-v_0}$. - Why: boat B has the same speed $s$ on both legs.
$\displaystyle\int_0^{t_B/2}s\,dt=l\Rightarrow t_B=\dfrac{2l}{\sqrt{v'^2-v_0^2}}$. - Why: add the legs of A, then divide.
$t_A=\dfrac{2lv'}{v'^2-v_0^2}$, so $\dfrac{t_A}{t_B}=\dfrac{v'}{\sqrt{v'^2-v_0^2}}=\dfrac{\eta}{\sqrt{\eta^2-1}}\approx 1.8$.
Methods 4 and 5: Force and Energy Methods
Force Method: not suited. All speeds are constant, so no net force is involved.
Energy Method: not suited. No masses or work are given.
Fastest: Method 1 (water frame). Most reliable in an exam: Method 3, since writing each time as distance over speed avoids slips. All give $t_A/t_B\approx 1.8$.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|---|---|---|
| $l$ | distance each boat goes from the buoy (cancels out) | km | |
| $v_0$ | stream speed | km/h | |
| $\eta$ | ratio $v'/v_0$ (must be greater than 1) | none | |
| $v'$ | boat speed relative to the water, $\eta v_0$ | km/h | |
| $s$ | boat B's speed across, $\sqrt{v'^2-v_0^2}$ | km/h | |
| $t_{A1},\ t_{A2}$ | boat A's time going with and against the stream | min | |
| $t_A,\ t_B$ | total times of boat A and boat B | min | |
| $t_A/t_B$ | ratio of times (answer) | none |
(vii) Common Exam Traps
- Using $v'$ as boat B's speed across. It is $\sqrt{v'^2-v_0^2}$, because boat B must aim upstream.
- Pointing boat B straight across the water. It would drift downstream and would not move along a line across the river.
- Writing $t_A=\dfrac{2l}{v'}$. The stream does not cancel for boat A. The slow upstream leg costs more time than the fast downstream leg saves.
- Averaging the two speeds of boat A as $\tfrac{(v'+v_0)+(v'-v_0)}{2}=v'$. Average speed over equal distances is not the plain average of speeds.
- Turning $\eta$ upside down. $\eta=v'/v_0$ is bigger than 1, so $\sqrt{\eta^2-1}$ is real.
- Needing $\eta>1$. If the boat is not faster than the stream, boat A cannot come back and boat B cannot cross. The ratio then blows up as $\eta\to 1$.
- Looking for $l$ or $v_0$ on their own. They cancel from the ratio.
Verified: the live numbers, both frames and all three methods give $t_A/t_B\approx 1.81$ for $\eta=1.2$, with all speeds in km/h and times in minutes.