How to Choose a Boat Heading to Land Directly Opposite

A boat points straight at a landing marker on the opposite riverbank. The motor runs at a constant speed, but the boat arrives well downstream of the marker. Where should the skipper point the bow on the next attempt?

Aim upstream, not directly at the marker. The boat needs enough upstream velocity through the water to cancel the river’s downstream current. For a steady, uniform current, the required heading is

\[ \theta=\sin^{-1}\left(\frac{u}{v}\right) \]

where \(u\) is the current’s speed relative to the riverbank, \(v\) is the boat’s speed relative to the water, and \(\theta\) is measured upstream from the direction straight across the river.

This works when \(u<v\). The boat’s direction of travel relative to the bank will then point directly towards the opposite marker, even though its bow points upstream.

There’s a trade-off. Predict whether this upstream heading makes the crossing faster or slower than pointing straight across. It makes the crossing slower: some of the boat’s velocity through the water is being used to cancel the current rather than move towards the far bank.

01Which observer decides whether the boat lands directly opposite?

The answer depends on the reference frame.

Imagine two people watching the crossing. One stands on the riverbank. The other floats downstream on a small raft, moving with the water.

The bank observer sees the boat moving across the river and being carried downstream. The drifting observer sees the boat moving through the surrounding water in whichever direction its motor propels it.

Both observations can be correct because they measure velocity relative to different objects.

We’ll use three labels:

• \(B\): the boat.
• \(W\): the water.
• \(G\): the ground, including the stationary riverbanks.

The boat itself is another possible reference frame. A passenger travelling with the boat sees the fixed landing marker move in the opposite direction to the boat’s ground velocity.

For our problem, the bank observer’s frame is the one that matters. The destination is fixed to the ground, so the boat must have zero downstream velocity relative to the ground.

The velocities combine as vectors:

\[ \vec v_{B/G} = \vec v_{B/W}+\vec v_{W/G} \]

This is vector addition, not addition of two scalar speeds.

For example, a boat travelling at 5.0 m/s relative to the water in a 3.0 m/s current doesn’t necessarily have a ground speed of 8.0 m/s or 2.0 m/s. Its ground velocity depends on the direction it points.

An instrument’s speed reading is only useful here if we know its reference frame. A measurement of ground speed is not interchangeable with a measurement of speed through the water.

02How to cancel the downstream motion

Choose two directions before calculating anything:

• \(+x\): downstream, parallel to the riverbanks.
• \(+y\): straight across the river, towards the destination.

Let the boat point at an angle \(\theta\) upstream from \(+y\). We take a positive \(\theta\) to mean the bow points towards the upstream side of the straight-across direction.

Assume the boat travels at constant speed \(v\) through the water, in the direction its bow points. The current has constant speed \(u\) in the \(+x\) direction.

The boat’s velocity through the water has two components:

• \(v\sin\theta\) upstream.
• \(v\cos\theta\) across the river.

The water contributes another velocity, \(u\), downstream.

So the boat’s ground-velocity components are

\[ \begin{aligned} v_{x,B/G}&=u-v\sin\theta\\ v_{y,B/G}&=v\cos\theta \end{aligned} \]

Here, \(v_{x,B/G}\) and \(v_{y,B/G}\) are the boat’s downstream and across-river velocity components, measured by the bank observer, in m/s.

To land directly opposite in a uniform current, the downstream component must vanish:

\[ u-v\sin\theta=0 \]

Therefore,

\[ v\sin\theta=u \]

and

\[ \theta=\sin^{-1}\left(\frac{u}{v}\right) \]

Notice that we’re not trying to cancel the entire current with the boat’s entire velocity.

We’re cancelling one component.

The remaining across-river ground velocity is

\[ v_y=v\cos\theta \]

where \(v_y\) is positive towards the far bank.

If the river is \(d\) metres wide, the crossing time is

\[ t=\frac{d}{v\cos\theta} \]

where \(t\) is measured in seconds.

For a heading that exactly cancels a uniform current, Pythagoras also gives

\[ v_y=\sqrt{v^2-u^2} \]

so

\[ t=\frac{d}{\sqrt{v^2-u^2}} \]

The second expression is useful because it gives the crossing time without first calculating the heading.

03How a missed crossing can reveal the correct heading

What if the question doesn’t tell you the current speed?

You may be given the result of an earlier crossing instead. That failed attempt can be treated as a measurement.

Consider an illustrative river trial, not a report of an actual experiment.

Two fixed marker posts, A and B, are placed directly opposite each other on the near and far banks. The river width is 96 m.

The boat’s speed through the water has been independently established as 4.0 m/s. A camera fixed relative to the bank records where the boat arrives, and a stopwatch measures the crossing time.

On the first trial, the skipper points straight across towards B.

What should this design reveal?

If the river has no current, the boat should reach B. If water flows downstream, the boat should land downstream, and the distance it misses by can be used to estimate the current.

Suppose the recorded results are:

Why the first measurement isn’t enough to prove the model

The initial trial gives an estimate of the current under the stated assumptions. It doesn’t establish that the current is uniform everywhere.

A stronger flow halfway across, changing motor speed, or a steering error could affect the observed landing point.

A revised investigation could add floating markers whose positions are tracked against fixed bank references, together with a record of the boat’s heading and position at several points across the river. The floating markers would provide information about water motion, while the boat tracking would reveal changes in its ground velocity.

The corrected heading could then be tested in a second crossing.

This is the important experimental distinction: a successful prediction supports the constant-current model for the conditions tested. One matching landing point doesn’t prove that every part of the river has the same current.

04When is landing directly opposite impossible?

The equation

\[ \sin\theta=\frac{u}{v} \]

has a physical limit.

The largest possible upstream component the boat can produce is its full through-water speed, \(v\).

Three cases follow.

If \(u=v\), the required angle is \(90^\circ\) upstream. In the ideal model, the boat then has zero ground velocity and never reaches the far bank.

If \(u>v\), the ratio \(u/v\) is greater than one, so there is no real angle satisfying the cancellation equation.

This impossibility applies to a steady, uniform current.

If the current changes during the crossing, the boat might drift downstream in one region and recover upstream in another. Then the correct question isn’t necessarily whether the downstream velocity is zero at every instant. It’s whether the total downstream displacement is zero when the boat reaches the opposite bank.

For a constant heading in a uniform current, that displacement is

\[ x=(u-v\sin\theta)t \]

with

\[ t=\frac{d}{v\cos\theta} \]

These two equations are especially useful when a question includes an observed miss, a crossing-time restriction, or a changing current.

The simple method also assumes straight, parallel banks, current parallel to the banks, constant boat speed relative to the water, and negligible wind and sideways slip. In an actual crossing, changing currents and steering corrections can make the ground track curved.

05Questions and solutions

The questions become progressively less straightforward. Keep \(+x\) downstream and \(+y\) across the river unless a question states otherwise. All numerical scenarios are illustrative.

Question 1

A boat travels at 5.0 m/s relative to the water across a river 96 m wide. The uniform current flows downstream at 1.4 m/s.

Calculate:

a) The upstream heading required to land directly opposite the starting point.

b) The crossing time with this heading.

c) How far downstream the boat would land if it pointed straight across instead.

Solution 1

The boat must aim \(16.3^\circ\) upstream and takes 20.0 s to reach the opposite point. Pointing straight across would instead leave it 26.9 m downstream of that point.

a) Heading

The upstream component must cancel the current:

\[ \begin{aligned} 5.0\sin\theta&=1.4\\ \sin\theta&=0.280\\ \theta&=16.3^\circ \end{aligned} \]

b) Crossing time

The across-river ground velocity is

\[ \begin{aligned} v_y&=\sqrt{5.0^2-1.4^2}\\ &=4.80\,\mathrm{m\,s^{-1}} \end{aligned} \]

Therefore,

\[ t=\frac{96}{4.80}=20.0\,\mathrm{s} \]

c) Downstream displacement

With a straight-across heading, the across-river speed is 5.0 m/s, so

\[ t=\frac{96}{5.0}=19.2\,\mathrm{s} \]

The current carries the boat downstream by

\[ \begin{aligned} x&=ut\\ &=1.4(19.2)\\ &=26.9\,\mathrm{m} \end{aligned} \]

Pointing straight at B produces the shorter crossing time, but not the required landing position. That’s the distinction this question is testing.

Question 2

A boat can travel at 4.0 m/s relative to the water. The river is 90 m wide, with a uniform current of 2.0 m/s downstream.

The skipper must reach the far bank within 25.0 s.

The boat may choose any one constant heading between straight across and directly upstream, but it cannot change heading or motor speed during the crossing.

a) Can the boat land directly opposite within the time limit?

b) If not, determine the heading that gets it as close as possible to the opposite marker while meeting the time limit.

c) Calculate its downstream miss.

Solution 2

The boat cannot land directly opposite within 25.0 s. The closest permitted landing is 6.41 m downstream, achieved with a heading of approximately \(25.8^\circ\) upstream.

a) Test the exact-opposite heading.

The heading required to cancel the current is

\[ \begin{aligned} \sin\theta&=\frac{2.0}{4.0}\\ \theta&=30.0^\circ \end{aligned} \]

That heading would give an across-river velocity of

\[ v_y=4.0\cos30.0^\circ =3.46\,\mathrm{m\,s^{-1}} \]

So the crossing would take

\[ \begin{aligned} t&=\frac{90}{3.46}\\ &=26.0\,\mathrm{s} \end{aligned} \]

More precisely, it takes approximately 25.98 s. This exceeds the 25.0 s deadline.

The tempting answer is to use \(90/4.0=22.5\) s. But 4.0 m/s is the total speed through the water, not the across-river component when the boat points upstream.

b) Find the largest upstream heading allowed by the deadline.

To cross within 25.0 s, the boat needs at least

\[ v_y=\frac{90}{25.0} =3.60\,\mathrm{m\,s^{-1}} \]

across the river.

Thus,

\[ \begin{aligned} 4.0\cos\theta&\ge3.60\\ \cos\theta&\ge0.900\\ \theta&\le25.8^\circ \end{aligned} \]

More precisely, the limit is \(25.84^\circ\).

For headings between \(0^\circ\) and \(30^\circ\), increasing the upstream angle reduces the final downstream displacement. This follows from the displacement expression

\[ x=\frac{90(2.0-4.0\sin\theta)} {4.0\cos\theta} \]

which decreases over that interval.

The best permitted heading is therefore the largest one that still meets the deadline: \(25.84^\circ\) upstream.

c) Find the remaining drift.

At this angle,

\[ \sin\theta=\sqrt{1-0.900^2} =0.4359 \]

The remaining downstream ground velocity is

\[ \begin{aligned} v_x&=2.0-4.0(0.4359)\\ &=0.2564\,\mathrm{m\,s^{-1}} \end{aligned} \]

Over 25.0 s, the displacement is

\[ \begin{aligned} x&=v_xt\\ &=0.2564(25.0)\\ &=6.41\,\mathrm{m} \end{aligned} \]

The fastest heading isn’t necessarily the best heading, and the exact-opposite heading isn’t always permitted by the problem’s other constraints.

Here, the skipper must accept a small downstream miss to satisfy the deadline.

Question 3

A 120 m wide river is modelled as two equal-width bands of moving water.

The bank-side 60 m flows downstream at 0.80 m/s. The far-side 60 m flows in the opposite direction, at 2.0 m/s upstream.

A boat travels at 4.0 m/s relative to the local water and must keep one constant compass heading throughout the crossing.

a) Determine the heading that makes the boat land directly opposite its starting point.

b) Calculate its displacement parallel to the banks when it first reaches the midpoint of the river.

c) Explain why its instantaneous downstream velocity does not need to be zero in either band.

Solution 3

The boat must aim \(8.63^\circ\) towards the positive \(x\)-direction, which is downstream at the starting bank. It will travel approximately 21.2 m downstream during the first half, then recover that displacement during the second half and finish at B.

a) Find the required heading.

Treat downstream as positive. The two water velocities are

\[ u_1=+0.80\,\mathrm{m\,s^{-1}} \]

and

\[ u_2=-2.0\,\mathrm{m\,s^{-1}} \]

Let \(\theta\) be positive upstream from the original straight-across direction. A negative result therefore indicates a heading towards \(+x\).

Because the boat maintains the same speed and compass heading, its across-river velocity \(v\cos\theta\) is unchanged between bands.

It spends equal time in the two equal-width bands.

The total downstream displacement must be zero:

\[ (u_1-v\sin\theta)t_1 + (u_2-v\sin\theta)t_2=0 \]

Since \(t_1=t_2\), this simplifies to

\[ \begin{aligned} u_1+u_2&=2v\sin\theta\\ 0.80-2.0&=8.0\sin\theta\\ \sin\theta&=-0.150\\ \theta&=-8.63^\circ \end{aligned} \]

The negative sign means aiming \(8.63^\circ\) towards the downstream direction defined at launch.

b) Find the midpoint displacement.

The across-river speed is

\[ \begin{aligned} v_y&=4.0\cos8.63^\circ\\ &=3.95\,\mathrm{m\,s^{-1}} \end{aligned} \]

The time spent in the first 60 m band is

\[ \begin{aligned} t_1&=\frac{60}{4.0\cos8.63^\circ}\\ &=15.17\,\mathrm{s} \end{aligned} \]

The boat’s downstream component relative to the water is \(+0.600\) m/s.

In the first band, its downstream ground velocity is therefore

\[ v_{x1}=0.80+0.600 =1.40\,\mathrm{m\,s^{-1}} \]

Hence,

\[ \begin{aligned} x_1&=v_{x1}t_1\\ &=1.40(15.17)\\ &=21.2\,\mathrm{m} \end{aligned} \]

The boat is 21.2 m downstream of the direct crossing line at the midpoint.

c) Explain the changing ground track.

In the second band, the current reverses direction.

The boat’s downstream ground velocity becomes

\[ \begin{aligned} v_{x2}&=-2.0+0.600\\ &=-1.40\,\mathrm{m\,s^{-1}} \end{aligned} \]

Since the second band takes the same 15.17 s to cross,

\[ x_2=-1.40(15.17) =-21.2\,\mathrm{m} \]

The total downstream displacement is

\[ x_{\text{total}}=21.2-21.2=0\,\mathrm{m} \]

This is the trap: for a uniform current, landing opposite means cancelling downstream velocity throughout the journey. Here, the boat can drift downstream first and upstream later.

The landing condition is zero total downstream displacement, not zero downstream velocity at every instant.

Question 4

A river is 120 m wide, and its current flows uniformly downstream at 2.0 m/s.

The boat’s motor changes its speed through the water as it crosses:

• In the first 60 m, it travels at 6.0 m/s relative to the water.
• In the second 60 m, it travels at 3.0 m/s relative to the water.

The skipper must choose one compass heading before launching and cannot change it.

a) Find the heading that lands the boat directly opposite.

b) Calculate its displacement parallel to the banks at the halfway point and its total crossing time.

c) Another student averages the speeds to get 4.5 m/s and uses \(\sin\theta=2.0/4.5\). Calculate the landing error this would produce and explain why averaging the speeds is incorrect.

Solution 4

The boat must maintain a heading of \(30.0^\circ\) upstream. It moves about 11.5 m upstream during the first half and exactly recovers that displacement during the second half. Its total crossing time is 34.6 s. Using the arithmetic mean speed would cause a 7.44 m downstream miss.

a) Determine the fixed heading.

Let the first and second through-water speeds be \(v_1=6.0\) m/s and \(v_2=3.0\) m/s.

The times spent in the two regions are

\[ t_1=\frac{60}{6.0\cos\theta} \]

and

\[ t_2=\frac{60}{3.0\cos\theta} \]

The downstream displacement in each region is the downstream ground velocity multiplied by the time in that region.

To finish directly opposite,

\[ (2.0-6.0\sin\theta)t_1 + (2.0-3.0\sin\theta)t_2=0 \]

Substitute the time expressions:

\[ \frac{10(2-6\sin\theta)}{\cos\theta} + \frac{20(2-3\sin\theta)}{\cos\theta} =0 \]

Multiplying through by \(\cos\theta\),

\[ \begin{aligned} 20-60\sin\theta +40-60\sin\theta&=0\\ 60&=120\sin\theta\\ \sin\theta&=0.500\\ \theta&=30.0^\circ \end{aligned} \]

b) Find the midpoint displacement and crossing time.

During the first half, the upstream boat component is

\[ 6.0\sin30^\circ=3.0\,\mathrm{m\,s^{-1}} \]

The downstream ground velocity is

\[ v_{x1}=2.0-3.0 =-1.0\,\mathrm{m\,s^{-1}} \]

The first crossing time is

\[ \begin{aligned} t_1&=\frac{60}{6.0\cos30^\circ}\\ &=11.55\,\mathrm{s} \end{aligned} \]

Therefore,

\[ x_1=(-1.0)(11.55) =-11.55\,\mathrm{m} \]

The boat reaches the midpoint approximately 11.5 m upstream of the direct crossing line.

In the slower second half,

\[ v_{x2}=2.0-3.0\sin30^\circ =0.500\,\mathrm{m\,s^{-1}} \]

The second crossing time is

\[ \begin{aligned} t_2&=\frac{60}{3.0\cos30^\circ}\\ &=23.09\,\mathrm{s} \end{aligned} \]

So

\[ x_2=0.500(23.09) =+11.55\,\mathrm{m} \]

The displacements cancel.

The total crossing time is

\[ \begin{aligned} t_{\text{total}}&=t_1+t_2\\ &=11.55+23.09\\ &=34.64\,\mathrm{s} \end{aligned} \]

c) Test the arithmetic-mean approach.

The student’s proposed average speed is

\[ v_{\text{mean}}=\frac{6.0+3.0}{2} =4.5\,\mathrm{m\,s^{-1}} \]

This gives

\[ \begin{aligned} \sin\theta&=\frac{2.0}{4.5}\\ \theta&=26.4^\circ \end{aligned} \]

But the actual displacements must still be calculated using the speed in each region.

With \(\sin\theta=2/4.5\), the total displacement becomes

\[ x= \frac{60-120\sin\theta} {\cos\theta} \]

Substituting,

\[ \begin{aligned} x&=\frac{60-120(2/4.5)} {\cos26.4^\circ}\\ &=+7.44\,\mathrm{m} \end{aligned} \]

The boat lands approximately 7.44 m downstream.

The crucial detail is time. The boat spends twice as long crossing the slower half, so the current has twice as long to act there.

Equal crossing distances do not mean equal time intervals. Averaging the two speeds gives the wrong balance between current displacement and upstream compensation.

Question 5

Two idealised rivers are each 100 m wide. Both have straight, parallel banks and two 50 m wide bands across the river.

In each river, one band has still water, and the other has a strong downstream current of 5.0 m/s.

A motorboat travels at a constant 4.0 m/s relative to the local water. The skipper may change heading at any point, but must always make positive progress towards the far bank.

The two arrangements are:

• River A: The strong current occupies the first 50 m; the final 50 m is still water.
• River B: The first 50 m is still water; the final 50 m has the strong current.

A fixed marker B lies directly opposite starting point A.

There is also a navigation restriction. Throughout the crossing, the boat must remain inside a corridor extending from the straight line A-B to a parallel line 38.0 m downstream.

In coordinates, the boat must satisfy

\[ 0\le x\le38.0\,\mathrm{m} \]

at all times.

a) If the corridor restriction is removed, can the skipper land directly opposite in both rivers using a single constant heading? Calculate that heading and the maximum distance the boat travels away from the A-B line.

b) With the 38.0 m corridor restriction, determine which river, if either, allows a successful crossing. Provide a feasible steering strategy and calculate the important displacements.

c) An engineer proposes narrowing the permitted corridor to 37.0 m. Would either river remain crossable under the same rules? Justify your conclusion even if the skipper can vary the heading continuously.

Solution 5

Without the corridor, a single heading of \(38.7^\circ\) upstream works in both rivers. The boat reaches approximately 40.0 m from the A-B line during the crossing. With the 38.0 m corridor, only River A is possible. If the corridor is narrowed to 37.0 m, neither river is possible.

The key is that the same final displacement doesn’t guarantee the same path, and a current stronger than the boat creates an unavoidable downstream movement while the boat crosses that region.

a) Ignore the corridor and find a constant heading.

The two bands have equal widths.

Because the skipper maintains one compass heading and the through-water speed is constant, the boat has the same across-river velocity in both bands. It therefore spends equal time in each.

The mean downstream current, weighted by these equal times, is

\[ u_{\text{mean}}=\frac{5.0+0}{2} =2.50\,\mathrm{m\,s^{-1}} \]

The heading that cancels the total displacement satisfies

\[ \begin{aligned} 4.0\sin\theta&=2.50\\ \sin\theta&=0.625\\ \theta&=38.7^\circ \end{aligned} \]

This works for the final landing position in both arrangements.

The across-river ground velocity is

\[ \begin{aligned} v_y&=\sqrt{4.0^2-2.50^2}\\ &=3.12\,\mathrm{m\,s^{-1}} \end{aligned} \]

Each 50 m band takes approximately

\[ t=\frac{50}{3.12} =16.01\,\mathrm{s} \]

to cross.

In the strong-current band, the downstream ground velocity is

\[ v_x=5.0-2.50 =2.50\,\mathrm{m\,s^{-1}} \]

Therefore, the downstream displacement accumulated in that band is

\[ \begin{aligned} x&=2.50(16.01)\\ &=40.0\,\mathrm{m} \end{aligned} \]

In still water, the boat has an upstream component of 2.50 m/s and accumulates an equal displacement in the opposite direction.

The order matters:

• In River A, the boat moves 40.0 m downstream, then returns to B.
• In River B, it moves 40.0 m upstream of the A-B line, then returns to B.

Both land at B, but neither satisfies the corridor restriction using this constant heading.

b) Can changing heading save either crossing?

River A can succeed. River B cannot.

Start with River A.

The skipper must first cross 50 m of water flowing downstream at 5.0 m/s, which is faster than the boat’s through-water speed of 4.0 m/s.

The boat must drift downstream in this band, no matter how it points.

The objective is to minimise that unavoidable displacement so it stays within 38.0 m.

A tempting choice is to point almost directly upstream. But this leaves almost no across-river velocity, giving the strong current a very long time to carry the boat downstream.

The best heading in this band is actually

\[ \sin\theta_1=\frac{4.0}{5.0} \]

giving

\[ \theta_1=53.1^\circ \]

upstream.

At this heading,

\[ \begin{aligned} v_{x1}&=5.0-4.0(0.800)\\ &=1.80\,\mathrm{m\,s^{-1}} \end{aligned} \]

and

\[ \begin{aligned} v_{y1}&=4.0(0.600)\\ &=2.40\,\mathrm{m\,s^{-1}} \end{aligned} \]

The crossing time for the first band is

\[ t_1=\frac{50}{2.40} =20.83\,\mathrm{s} \]

The downstream displacement is

\[ \begin{aligned} x_1&=1.80(20.83)\\ &=37.5\,\mathrm{m} \end{aligned} \]

That fits inside the 38.0 m corridor.

Now the boat enters still water, 37.5 m downstream of the A-B line.

To reach B, it must travel 50 m across and 37.5 m upstream. Its through-water velocity must therefore have an upstream-to-across component ratio of

\[ \frac{37.5}{50}=0.750 \]

Thus,

\[ \begin{aligned} \tan\theta_2&=0.750\\ \theta_2&=36.9^\circ \end{aligned} \]

upstream.

At this angle, the boat has components

\[ v_{x2}=-2.40\,\mathrm{m\,s^{-1}} \]

and

\[ v_{y2}=3.20\,\mathrm{m\,s^{-1}} \]

in still water.

It crosses the final 50 m in

\[ t_2=\frac{50}{3.20} =15.625\,\mathrm{s} \]

and travels upstream by

\[ x_2=-2.40(15.625) =-37.5\,\mathrm{m} \]

The total displacement parallel to the banks is

\[ x_{\text{total}}=37.5-37.5=0 \]

The path remains within the corridor throughout: it first increases from \(x=0\) to \(x=37.5\) m, then decreases back to \(x=0\).

So River A is possible.

Now consider River B.

The boat crosses still water first and the stronger current last.

In the final band, the current is 5.0 m/s downstream and the boat can supply at most 4.0 m/s upstream.

Its downstream ground velocity must therefore be positive throughout the final band.

To finish at \(x=0\), the boat would have to enter that band at a negative \(x\)-position, upstream of the A-B line.

But the corridor forbids \(x<0\).

No steering change can fix this.

The important difference between the rivers is not their average current. It’s the order of the regions, combined with the requirement to remain on one side of the A-B line.

c) What happens when the corridor narrows to 37.0 m?

Neither river is possible.

River B is already impossible because the final strong-current region forces downstream movement.

For River A, we need to show that even the best steering cannot keep the displacement within 37.0 m while crossing the strong-current band.

Take any heading \(\theta\) that gives positive across-river velocity.

In the strong-current band, the downstream displacement per metre travelled across the river is

\[ \frac{\Delta x}{\Delta y} = \frac{5-4\sin\theta} {4\cos\theta} \]

Here, \(\Delta x\) and \(\Delta y\) refer to small downstream and across-river displacements during a segment with that heading.

We can establish a minimum value using

\[ 4\sin\theta+3\cos\theta\le5 \]

This inequality follows because the maximum magnitude of a sum of sine and cosine terms is

\[ \sqrt{4^2+3^2}=5 \]

Rearranging,

\[ 5-4\sin\theta\ge3\cos\theta \]

Since \(\cos\theta>0\) for forward progress across the river,

\[ \frac{5-4\sin\theta} {4\cos\theta} \ge\frac34 \]

So every metre travelled across this strong-current region requires at least 0.75 m of downstream movement.

Over 50 m, the minimum downstream displacement is

\[ \begin{aligned} x_{\min}&=\frac34(50)\\ &=37.5\,\mathrm{m} \end{aligned} \]

Equality occurs at the \(53.1^\circ\) heading used in part b.

Because the inequality applies to every forward-moving segment, changing the heading repeatedly cannot reduce the total displacement below 37.5 m. Every small segment still contributes at least 0.75 m downstream for each metre across.

River A must therefore reach at least \(x=37.5\) m before leaving the strong-current band. A corridor ending at \(x=37.0\) m cannot contain that movement.

This is why the problem is more demanding than simply comparing \(u\) with \(v\).

The average current correctly predicts whether a suitable constant heading can produce zero final displacement when the path is unrestricted. But the corridor introduces a second requirement: every intermediate position must also be allowed.

Even an exact final landing can be impossible once the path itself matters.

The next decision in a river-crossing problem is whether you need to control the boat’s final position, predict its complete ground track, or minimise the time taken. Those are different conditions, even though they use the same velocity vectors.

For more practice calculating ground tracks and crossing times, use Boats Crossing a Current: Velocity, Track and Crossing Time. If the harder part is identifying which velocity an observer measures, Can Two Objects Have the Same Speed but Different Velocities? develops the reference-frame reasoning that sits behind the method.