A bird is flying with a speed of 40 km/hr. in the north direction. A train is moving with a speed of 40km/hr. in the west direction. A passenger sitting in the train will see the bird moving with velocity:-

Options

A. 40 km/hr in NE direction
B. `40sqrt(2)` km/hr in NE direction
C. 40 km/hr in. NW direction
D. `40 sqrt(2) km//hr` in NW direction

Answer

To solve the problem of how a passenger on a train perceives the motion of a bird flying in the north direction while the train is moving in the west direction, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Directions and Velocities**:
- Let the north direction be represented by the positive y-axis and the west direction by the negative x-axis.
- The velocity of the bird (V_bird) is 40 km/hr in the north direction, which can be represented as:
\[ V_{\text{bird}} = 40 \hat{j} \text{ km/hr} \]
- The velocity of the train (V_train) is 40 km/hr in the west direction, represented as:
\[ V_{\text{train}} = -40 \hat{i} \text{ km/hr} \]

2. **Determine the Velocity of the Bird Relative to the Train**:
- The passenger on the train will see the bird moving with respect to the train's velocity. This is calculated using the relative velocity formula:
\[ V_{\text{bird, relative}} = V_{\text{bird}} - V_{\text{train}} \]
- Substituting the values:
\[ V_{\text{bird, relative}} = 40 \hat{j} - (-40 \hat{i}) = 40 \hat{j} + 40 \hat{i} \]

3. **Calculate the Magnitude of the Relative Velocity**:
- The magnitude of the relative velocity can be found using the Pythagorean theorem:
\[ |V_{\text{bird, relative}}| = \sqrt{(40)^2 + (40)^2} = \sqrt{1600 + 1600} = \sqrt{3200} = 40\sqrt{2} \text{ km/hr} \]

4. **Determine the Direction of the Relative Velocity**:
- The direction of the relative velocity can be found using the tangent function:
\[ \tan(\theta) = \frac{V_y}{V_x} = \frac{40}{40} = 1 \]
- Therefore, the angle θ is:
\[ \theta = 45^\circ \]
- Since the bird's relative velocity has components in the positive x (east) and positive y (north) directions, the direction is northeast.

5. **Conclusion**:
- The passenger will see the bird moving with a velocity of \( 40\sqrt{2} \) km/hr at an angle of 45 degrees northeast.

### Final Answer:
The bird is moving with a velocity of \( 40\sqrt{2} \) km/hr at 45 degrees northeast relative to the passenger on the train.

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