$\cos\theta=-\dfrac{1}{\eta}=-\dfrac{1}{2}\;\Rightarrow\;\theta=120^\circ$ from the stream direction, giving $x_{min}=d\sqrt{\eta^2-1}=\sqrt3\,d\approx1.73\,d$
(ii) Problem Statement
"A boat moves relative to water with a velocity which is $n=2.0$ times less than the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting?"
Reference Answer Verification (from the supplied solution image):
The reference writes $\eta=\dfrac{v_0}{v'}=2$ (its $n$) and finds the crossing time $t=\dfrac{d}{v'\sin\theta}$. The drift is $x=(v'\cos\theta+v_0)t=(\cot\theta+\eta\csc\theta)\,d$. Setting $\dfrac{dx}{d\theta}=0$ gives $\cos\theta=-\dfrac{1}{\eta}=-\dfrac{1}{2}$, so $\theta=120^\circ$. Our work matches.
(iii) The Big Idea in Plain Words
A boat crosses a river. The stream pushes it downstream the whole time. This push is called drift. Here the boat is slower than the stream, so some drift cannot be avoided. We want the smallest drift.
Symbols: $d$ is the width of the river. $v_0$ is the speed of the stream. $v'$ is the boat's speed relative to the water. $\eta=v_0/v'$ is how many times slower the boat is. $\theta$ is the angle between the boat's heading and the stream direction. $x$ is the drift.
If you aim straight across, the drift is $\eta d=2d$. If you aim a little upstream, you cancel part of the push. But you also cross more slowly, so the stream works for longer. The best choice balances the two. It is $\theta=120^\circ$, which is $30^\circ$ upstream of straight across.
(iv) The Solution Viewed in Each Reference Frame
Example numbers for a quick check: $d=1$ km, $v_0=4$ km/h and $v'=2$ km/h, so $\eta=2$. Any other values give the same angle.
1. Ground or Lab Frame (Observer on the Bank)
Origin fixed at O on the start bankWhy: the boat's speed over the ground is the stream's velocity plus the boat's velocity in the water. Use $x$ along the stream and $y$ across the river.
$v_x=v'\cos\theta+v_0$ and $v_y=v'\sin\theta$.
Why: the boat reaches the far bank when it has gone the width $d$ across. Only the across part matters for the time.
$t=\dfrac{d}{v'\sin\theta}$.
Why: the drift is how far the boat moves along the stream during that time.
$x=v_xt=(v'\cos\theta+v_0)\dfrac{d}{v'\sin\theta}=(\cot\theta+\eta\csc\theta)\,d$.
Why: the smallest value of $x$ is where its slope with respect to $\theta$ is zero.
$$\frac{dx}{d\theta}=-\,d\,\frac{1+\eta\cos\theta}{\sin^2\theta}=0\;\Rightarrow\;\cos\theta=-\frac{1}{\eta}=-\frac12\;\Rightarrow\;\theta=120^\circ$$
Why: put the angle back to find the least drift and to compare with aiming straight across.
$x_{min}=d\left(\cot120^\circ+2\csc120^\circ\right)=d\left(-\tfrac{1}{\sqrt3}+\tfrac{4}{\sqrt3}\right)=\sqrt3\,d\approx1.73$ km. Straight across ($90^\circ$) gives $2$ km. Crossing time at $120^\circ$ is $\dfrac{1}{2\times0.866}=0.577$ h $\approx34.6$ min.
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 the boat simply moves in a straight line at $v'$ along its heading. The banks, the point O and the point opposite it all move west at $v_0$.
The across speed is still $v'\sin\theta$, so the crossing time is the same: $t=\dfrac{d}{v'\sin\theta}$.
Why: the drift is the gap between the landing spot and the point opposite O. Both have moved in this frame.
The boat moves $v'\cos\theta\,t$ along $x$. The opposite point moves $-v_0t$. The gap is $x=v'\cos\theta\,t+v_0t=(\cot\theta+\eta\csc\theta)\,d$.
Why: it is the same function of $\theta$ as before, so the minimum is in the same place.
$\cos\theta=-\dfrac{1}{\eta}\Rightarrow\theta=120^\circ$ and $x_{min}=\sqrt3\,d$.
3. Centre-of-Mass (CM) Frame
This frame does not apply. There is only one body and no masses are given. In the boat's own frame the boat never moves, so its drift could not even be seen.
What Changes and What Stays the Same Across Frames
- What changes: the path. On the ground the path is a slanted line over the bank. In the water frame it is a straight line along the heading, while the banks slide west.
- What stays the same: the heading angle $\theta$, the crossing time, the drift $x=(\cot\theta+\eta\csc\theta)d$, and the best angle $\theta=120^\circ$.
(v) The Solution by Each Method
Method 1: Relative-Motion Method (Velocity Circle)
Fastest- Why: the speed over ground is $\vec v=\vec v_0+\vec v'$. The path makes an angle $\alpha$ with the bank, and the drift is $x=d\cot\alpha$. A smaller drift means a steeper path, so we want the biggest $\alpha$.
- Why: whatever the heading, the tip of $\vec v$ lies on a circle of radius $v'$ around the tip of $\vec v_0$. The steepest line from O that still touches this circle is the tangent.
A tangent is at a right angle to the radius. So at the best heading, $\vec v\perp\vec v'$. - Why: the right triangle gives $\alpha$ at once.
$\sin\alpha=\dfrac{v'}{v_0}=\dfrac{1}{\eta}=\dfrac12\Rightarrow\alpha=30^\circ$. - Why: $\vec v'$ is turned $90^\circ$ from $\vec v$, measured from the stream direction.
$\theta=90^\circ+\alpha=120^\circ$ and $x_{min}=d\cot30^\circ=\sqrt3\,d$.
Method 2: Graph Method (Drift Against Angle)
- Why: plot $x/d=\cot\theta+2\csc\theta$ against $\theta$. The graph at the bottom of the simulation does this for you.
- Why: a table of values shows where the curve is lowest.
$\theta=90^\circ$: $2.000$. $\ 100^\circ$: $1.855$. $\ 110^\circ$: $1.764$. $\ 120^\circ$: $1.732$. $\ 130^\circ$: $1.772$. $\ 140^\circ$: $1.920$. - Why: the lowest point of the curve is the least drift.
The curve dips at $\theta=120^\circ$ with $x/d\approx1.73$. Slide $\theta$ in the simulation and watch the dot sit at the bottom of the curve there.
Method 3: Calculus Method
- Why: write the drift as one function of the angle.
$f(\theta)=\dfrac{x}{d}=\dfrac{\cos\theta+\eta}{\sin\theta}$. - Why: the minimum is where the slope is zero.
$f'(\theta)=-\dfrac{1+\eta\cos\theta}{\sin^2\theta}=0\Rightarrow\cos\theta=-\dfrac{1}{\eta}=-\dfrac12\Rightarrow\theta=120^\circ$. - Why: check that it is a minimum and not a maximum.
For $\theta<120^\circ$, $1+\eta\cos\theta>0$, so $f'<0$ and the drift falls. For $\theta>120^\circ$, $f'>0$ and it rises. So $120^\circ$ is a minimum, and $f_{min}=\sqrt{\eta^2-1}=\sqrt3$.
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 (velocity circle). Most reliable in an exam: Method 3, because the sign check shows it is a minimum. All give $\theta=120^\circ$.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|---|---|---|
| $d$ | width of the river | km | |
| $v_0$ | stream speed | km/h | |
| $\eta$ | ratio $v_0/v'$ (must be greater than 1) | none | |
| $v'$ | boat speed relative to the water, $v_0/\eta$ | km/h | |
| $\theta$ | heading angle from the stream direction (your slider) | deg | |
| $t$ | crossing time at your $\theta$ | min | |
| $x$ | drift at your $\theta$ | km | |
| $\theta^*$ | best heading angle (answer) | deg | |
| $x_{min}$ | least drift, $d\sqrt{\eta^2-1}$ (answer) | km |
(vii) Common Exam Traps
- Measuring $\theta$ from the wrong line. Here $\theta$ is from the stream direction, not from straight across. $120^\circ$ means $30^\circ$ upstream of straight across.
- Using $\theta=60^\circ$ because $\cos\theta=\tfrac12$. The sign matters: $\cos\theta=-\tfrac1\eta$. At $60^\circ$ the boat leans downstream and the drift grows.
- Thinking that aiming straight across ($90^\circ$) is best. It gives $\eta d=2d$, which is more than $1.73d$.
- Mixing up the fastest crossing with the least drift. The quickest crossing is at $90^\circ$. The least drift is at $120^\circ$.
- Reading "$n$ times less" the wrong way. It means $v'=v_0/n$, so $\eta=v_0/v'=n=2$.
- Trying to cancel the stream fully. That needs $v'\cos\theta=-v_0$, which is impossible when $v'<v_0$.
- Using $\cos\theta=-\tfrac1\eta$ when the boat is faster than the stream ($\eta<1$). Then zero drift is possible at $\cos\theta=-\eta$.
- Looking for $d$ in the angle. It cancels, so $\theta$ does not depend on the width.
Verified: the live numbers, both frames and all three methods give $\theta=120^\circ$ and $x_{min}=\sqrt3\,d\approx1.73\,d$ for $\eta=2$, with speeds in km/h, lengths in km and times in minutes.