A swimmer wants to cross a river from point A to point B. Line AB makes an angle of 30 degrees with

Q: Why must the actual path be exactly along the line AB?

Because the swimmer's goal is to reach point B starting from point A. To achieve this, their net displacement, and consequently their net absolute velocity (as seen from the ground), must be directed perfectly along the line connecting A and B.

Q: What is the difference between $\vec{v}_M$ and $\vec{v}_{MR}$?

$\vec{v}_M$ is the actual velocity of the swimmer as observed by someone standing still on the riverbank. $\vec{v}_{MR}$ is the velocity of the swimmer relative to the water—it represents the physical effort and direction the swimmer is aiming their body.

Q: Why did we form a rhombus in the vector diagram?

The vector addition $\vec{v}_M = \vec{v}_{MR} + \vec{v}_R$ naturally forms a parallelogram. Since the problem explicitly states that the magnitudes $|\vec{v}_{MR}|$ and $|\vec{v}_R|$ are equal, the parallelogram has equal adjacent sides, which is the exact definition of a rhombus.

Q: Can I solve this using components instead of geometry?

Absolutely! You can equate the ratio of the y and x components of the net velocity to $\tan(30^\circ)$ and solve the resulting trigonometric equation. However, recognizing the geometric symmetry of the rhombus is much faster and less prone to calculation errors.

Q: What if the river was flowing faster than the swimmer?

If $v_R > v_{MR}$, the vector diagram would form a general parallelogram, not a rhombus. Depending on the required angle, the swimmer might be swept too far downstream and could find it physically impossible to reach point B directly.

Question

A swimmer wants to cross a river from point $A$ to point $B$. Line $AB$ makes an angle of $30^\circ$ with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle $	heta$ with the line $AB$ should be ......$^\circ$, so that the swimmer reaches point $B$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the bank of a river that is flowing steadily. A swimmer at point $A$ wants to reach a specific destination, point $B$, on the opposite bank. The line connecting $A$ and $B$ makes a $30^\circ$ angle with the direction of the river's flow.

To successfully reach point $B$, the swimmer's **actual path** (as seen by an observer on the ground) must lie exactly along the line $AB$. This actual path is represented by the resultant velocity vector, $\vec{v}_M$.

### The Master Equation

In relative velocity problems, the absolute velocity of an object is the vector sum of its relative velocity and the velocity of the frame of reference. For our swimmer, this relationship is given by:

$$\vec{v}_M = \vec{v}_{MR} + \vec{v}_R$$

Here, $\vec{v}_{MR}$ is the velocity of the swimmer relative to the river (the effort the swimmer puts in), and $\vec{v}_R$ is the velocity of the river itself. Geometrically, this vector addition forms a parallelogram where $\vec{v}_{MR}$ and $\vec{v}_R$ are adjacent sides, and $\vec{v}_M$ is the diagonal starting from the same origin.

### The Geometric Trick

The problem provides a beautiful constraint that turns a potentially tedious calculation into a geometric masterpiece: the magnitude of the swimmer's velocity relative to the river is exactly equal to the magnitude of the river's velocity.

$$|\vec{v}_{MR}| = |\vec{v}_R|$$

When two adjacent sides of a parallelogram are equal in length, the shape is no longer just a parallelogram—it is a **rhombus**.

### The Power of the Rhombus

A rhombus possesses a magical geometric property: its diagonal perfectly bisects the angle between its adjacent sides.

Since the resultant vector $\vec{v}_M$ is the diagonal of our rhombus, it must bisect the total angle between $\vec{v}_{MR}$ and $\vec{v}_R$. We already know that the angle between the resultant $\vec{v}_M$ (which lies along $AB$) and the river flow $\vec{v}_R$ is $30^\circ$.

Because the diagonal bisects the total angle, the angle *above* the diagonal must equal the angle *below* it. Therefore, the angle $	heta$ between the swimmer's heading $\vec{v}_{MR}$ and the line $AB$ must also be $30^\circ$.

$$	heta = 30^\circ$$

### The Algebraic Alternative

What if you didn't spot the rhombus? You could still solve this using standard vector components. Let's align the x-axis with the river flow.

The y-component of the resultant velocity must match the y-component of the swimmer's effort:
$$v_M \sin(30^\circ) = v_{MR} \sin(30^\circ + 	heta)$$

The x-component of the resultant velocity is the sum of the river's flow and the swimmer's x-effort:
$$v_M \cos(30^\circ) = v_R + v_{MR} \cos(30^\circ + 	heta)$$

Dividing the two equations gives:
$$	an(30^\circ) = \frac{v_{MR} \sin(30^\circ + 	heta)}{v_R + v_{MR} \cos(30^\circ + 	heta)}$$

Since $v_{MR} = v_R$, we can cancel them out:
$$	an(30^\circ) = \frac{\sin(30^\circ + 	heta)}{1 + \cos(30^\circ + 	heta)}$$

Using the half-angle trigonometric identity $\frac{\sin(x)}{1 + \cos(x)} = 	an(\frac{x}{2})$, we get:
$$	an(30^\circ) = 	an\left(\frac{30^\circ + 	heta}{2}ight)$$

$$30^\circ = \frac{30^\circ + 	heta}{2} \implies 60^\circ = 30^\circ + 	heta \implies 	heta = 30^\circ$$

While the algebra confirms our result, the geometric intuition of the rhombus gets us to the answer in seconds. Always look for symmetry in physics problems!

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Relative Velocity Equation$

$Applying the Constraint$

$The Rhombus Property$

$Calculating the Angle$

$Final Answer$

$The Way Forward$

The Sigma Insight: Relative Velocity

Analyzing the Setup

The Master Equation

The Geometric Trick

The Power of the Rhombus

The Algebraic Alternative

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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; } Visualizing the Setup

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Constraint

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Rhombus Property

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Angle

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward