Animated Solution for Physics - Kinematics: A boat which has a speed of $5\text{ km/h}$ in still water crosses a river of width $1\text{ km}$ along the shortest possible path in $15\text{ min}$. The velocity of the river water in km/h is

Imagine you are standing on the bank of a mighty river. The river is exactly $1\text{ km}$ wide. You have a motorboat, and your mission is to cross this river along the shortest possible path. What does that mean geometrically? The shortest distance between two parallel lines (the river banks) is a straight line perpendicular to them.

So, your goal is to travel straight across. Your boat's engine is capable of pushing it at a speed of $5\text{ km/h}$ in perfectly still water. We are also given a crucial piece of information: it takes you exactly $15\text{ min}$ to complete this crossing. The mystery we need to solve is: how fast is the river flowing?

Here is where many students fall into a classic kinematics trap. If you simply point the nose of your boat straight across the river, will you travel along the shortest path? Absolutely not!

As your boat moves forward, the river's current, with velocity ${\vec{v}}_r$, will relentlessly sweep you downstream. Your actual path relative to the ground will be a diagonal line, which is significantly longer than the straight width of the river.

To achieve the true shortest path, you must fight the river. You must steer your boat diagonally upstream. By doing this, a component of your boat's velocity cancels out the river's flow, allowing your net movement to be perfectly straight across.

Let's visualize this struggle using vectors. Your boat's velocity in still water is ${\vec{v}}_{br}$. The river's velocity is ${\vec{v}}_r$. The resultant velocity, which is your actual velocity relative to the ground, is ${\vec{v}}_b$.

By the law of vector addition, your actual velocity is the sum of your boat's effort and the river's push:

Since we demand that ${\vec{v}}_b$ points straight across the river, it must be exactly perpendicular to the river's flow ${\vec{v}}_r$. When we draw these vectors head-to-tail, they form a beautiful right-angled triangle!

In this right-angled velocity triangle, the hypotenuse is the boat's speed in still water, $v_{br}$, because it is the longest vector. The other two perpendicular sides are the river's speed $v_r$ and your resultant speed straight across, $v_b$.

We can now summon the Pythagorean theorem:

We want to find our effective crossing speed, $v_b$. Rearranging the equation gives:

Now, we bridge the gap between geometry and kinematics. The time $t$ taken to cross the river is simply the distance $d$ divided by the velocity along that specific distance, which is $v_b$.

We are armed with our master equation and our given values: $d=1\text{ km}$, $v_{br}=5\text{ km/h}$, and $t=15\text{ min}$.

Warning: This is where the examiner tests your alertness. Do not mix minutes and hours! We must convert the time into hours to maintain consistent units across the board.

Now, we substitute these pristine values into our master equation:

To liberate our unknown variable $v_r$ from the square root, we square both sides of the equation. This is a standard algebraic maneuver to simplify radical expressions.

Since the numerators on both sides are identical (both are $1$), their denominators must be perfectly equal. This gives us a beautifully simple linear equation:

Rearranging the terms to isolate $v_r^2$:

Taking the square root of both sides reveals our final answer:

The river is flowing at a steady speed of $3\text{ km/h}$. The true elegance of this problem lies in how a physical constraint (the shortest path) translates perfectly into a geometric condition (a right-angled vector triangle), which then seamlessly yields a clean algebraic solution.