Two identical boats are moving relative to the water current with equal speed v_{b/w} = 1.0 m/s. To

Q: Why does Boat 1 need to steer upstream?

To move straight across relative to the ground, Boat 1 must counteract the river's current pushing it downstream. By steering upstream, its horizontal velocity component exactly cancels the river's velocity.

Q: Why is Boat 2's vertical speed just $v$?

The boy on the raft moves with the river. If he sees the boat moving straight across, it means the boat is pointing its nose directly at the opposite shore. Thus, its entire speed relative to the water, $v$, is directed vertically.

Q: Can we assume the time interval $\Delta t$ is 1 second?

You could, for the sake of calculation, since the ratio $\Delta y_1 / \Delta y_2$ is independent of $\Delta t$. However, keeping it as a variable and eliminating it is mathematically more rigorous.

Q: What if Boat 1 was faster than the river current?

It must be! If $v < u$, the term $\sqrt{v^2 - u^2}$ would be imaginary, meaning the boat could never move straight across the river; it would always be swept downstream.

Q: Why do we use a raft for Boat 2's observer?

A freely floating raft is a perfect physical representation of a frame of reference moving at the river's velocity. It helps visualize 'velocity relative to water' intuitively.

Question

Two identical boats are moving relative to the water current with equal speed $v_{b/w} = 1.0 	ext{ m/s}$. To a boy standing on the ground, the first boat appears moving perpendicular to the river current and to another boy standing on a raft in the river, the second boat appears moving perpendicular to the shoreline. In a certain time interval, distances of the boats from the shoreline increase by $\Delta y_1 = 4.0 	ext{ m}$ and $\Delta y_2 = 5.0 	ext{ m}$ respectively. Calculate speed of the river current.

Sigma Percentile · Accepted Answer

### The Tale of Two Observers

Imagine standing on the bank of a steadily flowing river. You watch two identical boats, both capable of the same speed relative to the water, attempt to cross.

However, they employ entirely different strategies, and their motions are judged by two different observers. This classic relative velocity problem is a beautiful exercise in understanding frames of reference.

### Analyzing Boat 1: The Ground Perspective

For the first boat, a boy standing on the ground observes it moving straight across the river. This means the boat's absolute velocity, $\vec{v}_1$, is directed perfectly perpendicular to the shoreline.

To achieve this, the boat cannot simply point its nose straight across. If it did, the river would sweep it downstream. Instead, it must steer slightly upstream.

By doing so, the horizontal component of its velocity relative to the water exactly cancels out the river's current. We can visualize this with a right-angled vector triangle.

The hypotenuse is the boat's speed relative to the water, $v$. The base is the river's speed, $u$. Using the Pythagorean theorem, the vertical speed that actually carries the boat across the river is:

$$v_{1y} = \sqrt{v^2 - u^2}$$

### Analyzing Boat 2: The Raft Perspective

Now, let's shift our perspective to the second boat. This time, the observer is a boy on a raft drifting freely with the river current.

To this boy, the second boat appears to be moving straight across the river. Because the raft is moving at the exact speed of the river, it serves as a moving frame of reference.

If the boat appears to move straight across from this moving frame, it means the boat's velocity *relative to the water*, $\vec{v}_{2/w}$, is directed perfectly perpendicular to the shore.

Since the boat is pointing its nose straight across, its entire engine effort is dedicated to crossing the river. Therefore, its vertical speed is simply its full speed relative to the water:

$$v_{2y} = v$$

### The Master Equation

We are given a crucial piece of information: in a certain time interval $\Delta t$, the distances of the boats from the shoreline increase by $\Delta y_1$ and $\Delta y_2$ respectively.

Distance is simply vertical speed multiplied by time. We can write the displacement equations for both boats:

$$\Delta y_1 = \sqrt{v^2 - u^2} \Delta t$$

$$\Delta y_2 = v \Delta t$$

Since the time interval $\Delta t$ is the same for both boats, we can elegantly eliminate it by dividing the first equation by the second. This gives us a ratio of their vertical displacements:

$$\frac{\Delta y_1}{\Delta y_2} = \frac{\sqrt{v^2 - u^2}}{v}$$

### Final Calculation

Now, we just need to isolate the river's speed, $u$. Let's square both sides of our ratio equation to remove the square root:

$$\frac{\Delta y_1^2}{\Delta y_2^2} = 1 - \frac{u^2}{v^2}$$

Rearranging the terms to solve for $u^2/v^2$, we get:

$$\frac{u^2}{v^2} = \frac{\Delta y_2^2 - \Delta y_1^2}{\Delta y_2^2}$$

Taking the square root gives us the final algebraic expression for the river's speed:

$$u = v \frac{\sqrt{\Delta y_2^2 - \Delta y_1^2}}{\Delta y_2}$$

Finally, we substitute the given numerical values: $v = 1.0 	ext{ m/s}$, $\Delta y_1 = 4.0 	ext{ m}$, and $\Delta y_2 = 5.0 	ext{ m}$.

Notice the numbers under the square root: $5.0^2 - 4.0^2$. This is the classic 3-4-5 right triangle! The square root evaluates perfectly to $3.0$.

$$u = 1.0 	imes \frac{3.0}{5.0}$$

$$u = 0.6 	ext{ m/s}$$

**The speed of the river current is exactly $0.6 	ext{ m/s}$.**

Enter Numerical Value:

Visualized Solution

\text{Visualizing the River and Boats}

\text{Velocity of Boat 1}

\text{Y-component of Boat 1}

\text{Velocity of Boat 2}

\text{Y-component of Boat 2}

\text{Relating Displacements}

\text{Eliminating } \Delta t

\text{Solving for River Speed } u

\text{Final Calculation}

The Sigma Insight: Relative Velocity

The Tale of Two Observers

Analyzing Boat 1

Analyzing Boat 2

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Visualizing the River and Boats} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Velocity of Boat 1} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Y-component of Boat 1} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Velocity of Boat 2} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Y-component of Boat 2} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Relating Displacements} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Eliminating } \Delta t .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Solving for River Speed } u .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Final Calculation} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }