A swimmer is swimming with constant velocity 2root2 m/s due north east in a calm lake. He observes his image his image fitted at the rear of a boat moving with constant velocity 1m/s due east. Velocity of his image as observed by him in the mirror will be?

To determine the velocity of the swimmer's image as observed in the mirror of the boat, we need to analyze the situation using vector addition. The swimmer is moving at a constant velocity of 2√2 m/s towards the northeast, while the boat is moving at 1 m/s towards the east. Let's break this down step by step.

Understanding the Velocities

First, we need to express the velocities of both the swimmer and the boat in vector form. The northeast direction can be represented as a 45-degree angle from the north or east axis. Therefore, we can decompose the swimmer's velocity into its north and east components.

Swimmer's Velocity Components

The swimmer's velocity of 2√2 m/s can be broken down as follows:

• North component: \( V_{swimmer, north} = 2\sqrt{2} \cdot \cos(45^\circ) = 2\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 2 \, \text{m/s} \)
• East component: \( V_{swimmer, east} = 2\sqrt{2} \cdot \sin(45^\circ) = 2\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 2 \, \text{m/s} \)

Thus, the swimmer's velocity vector can be represented as:

V_swimmer = (2, 2) m/s (where the first component is east and the second is north).

Boat's Velocity

The boat is moving due east at 1 m/s, which can be represented as:

V_boat = (1, 0) m/s.

Relative Velocity of the Swimmer's Image

To find the velocity of the swimmer's image as observed in the mirror of the boat, we need to consider the relative velocity. The image in the mirror will appear to move in the opposite direction of the swimmer's motion relative to the boat. Therefore, we can calculate the relative velocity of the swimmer with respect to the boat.

Calculating Relative Velocity

The relative velocity of the swimmer with respect to the boat is given by:

V_relative = V_swimmer - V_boat

Substituting the values:

V_relative = (2, 2) - (1, 0) = (2 - 1, 2 - 0) = (1, 2) m/s.

Velocity of the Image

Since the image appears to move in the opposite direction to the swimmer's relative velocity, we take the negative of the relative velocity:

V_image = -V_relative = (-1, -2) m/s.

Magnitude and Direction

To find the magnitude of the velocity of the image, we can use the Pythagorean theorem:

|V_image| = √((-1)² + (-2)²) = √(1 + 4) = √5 m/s.

The direction can be found using the arctangent function:

θ = tan^-1(2/1) = tan^-1(2), which gives us an angle in the southwest direction since both components are negative.

Final Result

In summary, the swimmer observes his image moving with a velocity of approximately √5 m/s towards the southwest. This analysis illustrates how relative motion can affect the perception of velocity in different frames of reference.