Method 1: Algebraic Solution

Given Parameters:

• Velocity of swimmer relative to water: \( |\vec{v}_{s/w}| = \sqrt{3} \, \text{m/s} \)
• Angle of path with downstream: \( 30^\circ \)
• River velocity: \( \vec{u} \) (unknown)
• Condition: The path cuts the relative velocity locus at two points such that the absolute speeds satisfy \( v_2 = 2v_1 \).

Step 1: The Vector Equation

The absolute velocity of the swimmer \( \vec{v} \) is the vector sum of the river velocity \( \vec{u} \) and the swimmer’s relative velocity \( \vec{v}_{s/w} \).

$$ \vec{v} = \vec{u} + \vec{v}_{s/w} $$

Using the Cosine Rule in the vector triangle formed by these velocities, where the angle between \( \vec{u} \) and \( \vec{v} \) is \( 30^\circ \):

$$ |\vec{v}_{s/w}|^2 = |\vec{v}|^2 + |\vec{u}|^2 – 2|\vec{u}||\vec{v}| \cos(30^\circ) $$

Substituting the known values (\( |\vec{v}_{s/w}| = \sqrt{3} \)):

$$ (\sqrt{3})^2 = v^2 + u^2 – 2uv \left( \frac{\sqrt{3}}{2} \right) $$ $$ 3 = v^2 + u^2 – \sqrt{3}uv $$

Step 2: Quadratic Formation

Rearranging this into a quadratic equation in terms of the speed \( v \):

$$ v^2 – (\sqrt{3}u)v + (u^2 – 3) = 0 $$

Let the roots of this equation be \( v_1 \) and \( v_2 \). These represent the two possible speeds at which the swimmer can travel along the \( 30^\circ \) path. We are given that \( v_2 = 2v_1 \).

Step 3: Solving Using Roots

Using the properties of quadratic roots:

1. Sum of roots:

$$ v_1 + v_2 = \sqrt{3}u $$

Substitute \( v_2 = 2v_1 \):

$$ 3v_1 = \sqrt{3}u \implies v_1 = \frac{u}{\sqrt{3}} $$

2. Product of roots:

$$ v_1 v_2 = u^2 – 3 $$

Substitute \( v_2 = 2v_1 \):

$$ 2v_1^2 = u^2 – 3 $$

Now, substitute the expression for \( v_1 \) derived from the sum of roots:

$$ 2\left( \frac{u}{\sqrt{3}} \right)^2 = u^2 – 3 $$ $$ \frac{2u^2}{3} = u^2 – 3 $$

Multiply by 3 to clear the denominator:

$$ 2u^2 = 3u^2 – 9 $$ $$ u^2 = 9 $$

Method 2: Geometric Approach (Alternative)

Step 1: Sine Rule Application

Consider the vector triangle formed by \( \vec{u} \) (river velocity), \( \vec{v}_{s/w} \) (relative velocity), and the resultant velocity \( \vec{v} \). Let \( \theta \) be the angle between the relative velocity vector and the resultant path.

Applying the Sine Rule to this triangle:

$$ \frac{u}{\sin \theta} = \frac{v_{s/w}}{\sin 30^\circ} $$

Substituting \( v_{s/w} = \sqrt{3} \) and \( \sin 30^\circ = 1/2 \):

$$ u = 2\sqrt{3} \sin \theta \quad \dots(1) $$

Step 2: Geometric Condition for Velocity Ratio

The problem implies that the secant drawn from the origin intersects the circle at distances \( v_1 \) and \( v_2 \) such that \( v_2 = 2v_1 \).

Consider the projection of the center of the circle (tip of \( \vec{u} \)) onto the path line:

• The projection distance along the line is \( d_{proj} = u \cos 30^\circ \).
• This projection point is the midpoint of the chord formed by the two velocity vectors.
• Therefore, \( \frac{v_1 + v_2}{2} = u \cos 30^\circ \).

Using \( v_2 = 2v_1 \):

$$ \frac{3v_1}{2} = u \frac{\sqrt{3}}{2} \implies v_1 = \frac{u}{\sqrt{3}} $$

Now, relate this to the Sine Rule logic. In the vector triangle, if we drop a perpendicular from the tip of \( \vec{u} \) to the path line, we form a right-angled triangle with hypotenuse \( v_{s/w} \). The geometry required to satisfy the \( 1:2 \) ratio along with the \( 30^\circ \) path angle dictates that:

$$ \theta = 60^\circ $$

Step 3: Final Calculation

Substituting \( \theta = 60^\circ \) back into equation (1):

$$ u = 2\sqrt{3} \times \sin(60^\circ) $$ $$ u = 2\sqrt{3} \times \frac{\sqrt{3}}{2} $$ $$ u = 3 \, \text{m/s} $$