A bota has a velocity `4m//s` towards east with respect to river is flowing north with velocity `2m//s` Wind is blowing towards north with velocity `6m//s`. The direction of the flag blown over by the wind hoisted on the boat is `:`

Options

A. north`-`west
B. south`-`east
C. `tan^(-1)``(1)/(2)` with east
D. north

Explanation

To solve the problem, we need to find the direction of the flag blown over by the wind hoisted on the boat. We will follow these steps:

### Step 1: Define the velocities
1. **Velocity of the boat with respect to the river (V_BR)**: The boat has a velocity of 4 m/s towards the east. In vector form, this is represented as:
\[ \vec{V}_{BR} = 4 \hat{i} \, \text{m/s} \]

2. **Velocity of the river (V_R)**: The river flows north with a velocity of 2 m/s. In vector form, this is:
\[ \vec{V}_R = 2 \hat{j} \, \text{m/s} \]

3. **Velocity of the wind (V_W)**: The wind blows north with a velocity of 6 m/s. In vector form, this is:
\[ \vec{V}_W = 6 \hat{j} \, \text{m/s} \]

### Step 2: Find the velocity of the boat with respect to the ground
To find the velocity of the boat with respect to the ground (V_B), we add the velocity of the boat with respect to the river and the velocity of the river:
\[ \vec{V}_B = \vec{V}_{BR} + \vec{V}_R = 4 \hat{i} + 2 \hat{j} \, \text{m/s} \]

### Step 3: Find the velocity of the wind with respect to the boat
Now, we need to find the velocity of the wind with respect to the boat (V_WB). This is given by:
\[ \vec{V}_{WB} = \vec{V}_W - \vec{V}_B \]
Substituting the values we have:
\[ \vec{V}_{WB} = 6 \hat{j} - (4 \hat{i} + 2 \hat{j}) = 6 \hat{j} - 4 \hat{i} - 2 \hat{j} = -4 \hat{i} + 4 \hat{j} \]

### Step 4: Determine the direction of the flag
The velocity of the wind with respect to the boat is:
\[ \vec{V}_{WB} = -4 \hat{i} + 4 \hat{j} \]
This indicates that the wind is blowing towards the northwest (since it has a negative x-component and a positive y-component).

### Step 5: Calculate the angle of the flag's direction
To find the angle θ that the wind makes with the east direction (positive x-axis), we can use the tangent function:
\[ \tan(\theta) = \frac{|\text{vertical component}|}{|\text{horizontal component}|} = \frac{4}{4} = 1 \]
Thus,
\[ \theta = 45^\circ \]

### Conclusion
The direction of the flag blown over by the wind hoisted on the boat is towards the northwest, making an angle of 45 degrees with the east direction.

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