Question

A man who can swim at the rate of 2 km/h (in still river) crosses a river to a point exactly opposite on the other bank by swimming in a direction of 120° to the flow of river water. The velocity of water current in km/h is

• A. 1
• B. 2
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

Answer: (A)

1

Answer 2

The problem involves relative motion, specifically a man swimming across a river. We are given the man's speed in still water and the direction he swims relative to the river flow. The condition that he reaches a point exactly opposite on the other bank implies that his resultant velocity is perpendicular to the river flow.

Let's define the velocities:

• $\vec{v_{m\_w}}$: Velocity of the man relative to the water (swimming velocity).
• $\vec{v_{w\_g}}$: Velocity of the water relative to the ground (river current).
• $\vec{v_{m\_g}}$: Velocity of the man relative to the ground (resultant velocity).

According to the principle of relative motion, the resultant velocity of the man is the vector sum of his velocity relative to water and the velocity of the water relative to the ground:

$\vec{v_{m\_g}} = \vec{v_{m\_w}} + \vec{v_{w\_g}}$

Given information:

1. Magnitude of the man's speed in still water: $|\vec{v_{m\_w}}| = 2 \text{ km/h}$.
2. The man swims in a direction of 120° to the flow of river water. Let's assume the river flows along the positive x-axis. Therefore, $\vec{v_{w\_g}}$ is along the positive x-axis. The angle of $\vec{v_{m\_w}}$ with the positive x-axis is 120°.
3. The man reaches a point exactly opposite on the other bank. This means his resultant velocity $\vec{v_{m\_g}}$ has no component along the direction of the river flow (i.e., its x-component is zero). It is directed purely across the river (along the y-axis).

Let's resolve the velocities into their x and y components:

• Velocity of man relative to water ($\vec{v_{m\_w}}$):

  • x-component: $v_{m\_w\_x} = |\vec{v_{m\_w}}| \cdot \cos(120°) = 2 \cdot (-\frac{1}{2}) = -1 \text{ km/h}$.
  • y-component: $v_{m\_w\_y} = |\vec{v_{m\_w}}| \cdot \sin(120°) = 2 \cdot (\frac{\sqrt{3}}{2}) = \sqrt{3} \text{ km/h}$.

• Velocity of water relative to ground ($\vec{v_{w\_g}}$):

  • Let $v_r$ be the magnitude of the river current.
  • x-component: $v_{w\_g\_x} = v_r$.
  • y-component: $v_{w\_g\_y} = 0$ (since the river flows horizontally).

• Resultant velocity of man relative to ground ($\vec{v_{m\_g}}$):

  • x-component: $v_{m\_g\_x} = v_{m\_w\_x} + v_{w\_g\_x} = -1 + v_r$.
  • y-component: $v_{m\_g\_y} = v_{m\_w\_y} + v_{w\_g\_y} = \sqrt{3} + 0 = \sqrt{3} \text{ km/h}$.

Since the man reaches a point exactly opposite, his resultant velocity has no x-component:

$v_{m\_g\_x} = 0$

Therefore, $-1 + v_r = 0$

$v_r = 1 \text{ km/h}$

The velocity of the water current is 1 km/h.