Animated Solution for Physics - Kinematics: To cross a river of width $b=320\text{ m}$ a boatman steers his boat always aiming toward a point that is directly opposite to the starting point. Velocity of the boat relative to the river current is $v_{b/w}=2.5\text{ m/s}$ and river current velocity is $v_w=1.5\text{ m/s}$ everywhere. Determine time, which the boat will take to cross the river.

The problem of a boat crossing a river is a classic staple of kinematics, but this particular variation introduces a fascinating twist. Instead of maintaining a constant heading angle, our boatman continuously adjusts his steering to always point directly at his destination. This creates a dynamic, ever-changing velocity triangle that makes standard formulas useless.

Let's embark on this mathematical journey and discover a breathtakingly elegant trick to solve it.

Imagine standing on the bank of a river. You are at the origin, $(0,0)$, and your destination is directly across from you at $(0,b)$, where $b=320\text{ m}$ is the width of the river.

As you start rowing, your boat has a speed $v_{b/w}=2.5\text{ m/s}$ relative to the water. Because you are always aiming at the destination, your relative velocity vector ${\vec{v}}_{b/w}$ always points along the line connecting your current position $P(x,y)$ to the destination $D(0,b)$.

However, the river isn't sitting still. It flows horizontally with a constant speed $v_w=1.5\text{ m/s}$. The river's current constantly sweeps you downstream (in the $+x$ direction). To compensate, you have to keep turning the boat upstream. Your actual path over the ground is not a straight line, but a curve known as a curve of pursuit.

To solve this, we need to translate our physical intuition into mathematics. Let $r$ be your instantaneous distance to the destination. Let $\theta$ be the angle your line of sight makes with the $y$-axis.

Your absolute velocity $\vec{v}$ is the vector sum of your steering velocity and the river's velocity:

Let's break this down into components. How fast are you moving horizontally? The river pushes you to the right with speed $v_w$, but your steering has a horizontal component pulling you to the left (upstream) with magnitude $v_{b/w}\sin \theta$. Therefore, the rate of change of your $x$-coordinate is:

Now, how fast are you approaching the destination? Your steering speed $v_{b/w}$ is directed entirely towards the target, reducing the distance $r$. However, the river's flow has a component $v_w\sin \theta$ that pushes you away along that same line of sight. Thus, the net rate at which your distance $r$ decreases is: Or, written in terms of the rate of change of $r$:

Eliminating the Angle

We now have a system of two coupled differential equations. The nightmare here is the angle $\theta$. It changes continuously as you move, making these equations incredibly difficult to integrate directly.

But look closely at the two equations. Do you see the symmetry? 1. $\frac{dx}{dt}=v_w-v_{b/w}\sin \theta$ 2. $\frac{dr}{dt}=-v_{b/w}+v_w\sin \theta$

We want to annihilate the $\sin \theta$ terms. What if we cross-multiply? Let's multiply the first equation by $v_w$ and the second equation by $v_{b/w}$:

Now, add them together! The $\sin \theta$ terms perfectly cancel each other out, leaving us with a magnificent constant:

This is a profound physical statement. It tells us that a specific linear combination of your horizontal drift rate and your approach rate is absolutely constant throughout the entire journey, regardless of the complex curved path you take!

Because the right side of our equation is a constant, we can easily integrate it over the total time $T$ it takes to cross the river:

This simplifies beautifully to:

Now we apply our boundary conditions. - Horizontal Displacement ($\Delta x$): You started at $x=0$. Because you successfully reached the destination directly opposite your starting point, your final $x$-coordinate is also $0$. Thus, $\Delta x=0$. - Change in Distance ($\Delta r$): You started at a distance $b$ from the target. When you arrive, your distance to the target is $0$. Thus, $\Delta r=0-b=-b$.

Substituting these into our integrated equation:

Rearranging for $T$, we get our master formula:

All the heavy lifting is done. Now we just plug in the numerical values given in the problem: - River width, $b=320\text{ m}$ - Boat speed, $v_{b/w}=2.5\text{ m/s}$ - River speed, $v_w=1.5\text{ m/s}$

The boat will take exactly 200 seconds to cross the river. By looking past the complex trajectory and focusing on the invariant rates of change, we turned a seemingly impossible calculus problem into a beautiful algebraic triumph.