An observer is travelling in east with a velocity of 2 m/s and observes that wind is blowing in north with a velocity of 2m/s. If observer doubles his velocity then find out the velocity of wind appears to observer ?

Answer

To solve the problem, we need to find the apparent velocity of the wind as observed by an observer who doubles his velocity. We will break this down step by step.

### Step 1: Understand the Initial Conditions
- The observer is moving east with a velocity of \( v_o = 2 \, \text{m/s} \).
- The wind is blowing north with a velocity of \( v_w = 2 \, \text{m/s} \).

### Step 2: Represent the Velocities as Vectors
- The observer's velocity can be represented as:
\[ \vec{v_o} = 2 \hat{i} \, \text{m/s} \]
- The wind's velocity can be represented as:
\[ \vec{v_w} = 2 \hat{j} \, \text{m/s} \]

### Step 3: Calculate the Relative Velocity of Wind with Respect to Observer
- The apparent velocity of wind with respect to the observer is given by:
\[ \vec{v_{w/o}} = \vec{v_w} - \vec{v_o} \]
- Substituting the values:
\[ \vec{v_{w/o}} = (2 \hat{j}) - (2 \hat{i}) = -2 \hat{i} + 2 \hat{j} \]

### Step 4: Find the Magnitude and Direction of the Relative Velocity
- The magnitude of the relative velocity can be calculated as:
\[ |\vec{v_{w/o}}| = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \, \text{m/s} \]
- The direction can be determined using the arctangent function:
\[ \theta = \tan^{-1}\left(\frac{2}{-2}\right) = \tan^{-1}(-1) \]
This indicates that the direction is towards the northwest.

### Step 5: Double the Observer's Velocity
- If the observer doubles his velocity, then:
\[ v_o' = 2 \times 2 = 4 \, \text{m/s} \]
- The new velocity vector of the observer becomes:
\[ \vec{v_o'} = 4 \hat{i} \, \text{m/s} \]

### Step 6: Calculate the New Apparent Velocity of Wind
- The new apparent velocity of wind with respect to the observer is:
\[ \vec{v_{w/o'}} = \vec{v_w} - \vec{v_o'} \]
- Substituting the values:
\[ \vec{v_{w/o'}} = (2 \hat{j}) - (4 \hat{i}) = -4 \hat{i} + 2 \hat{j} \]

### Step 7: Find the Magnitude and Direction of the New Relative Velocity
- The magnitude of the new relative velocity can be calculated as:
\[ |\vec{v_{w/o'}}| = \sqrt{(-4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \, \text{m/s} \]
- The direction can be determined using the arctangent function:
\[ \theta' = \tan^{-1}\left(\frac{2}{-4}\right) = \tan^{-1}\left(-\frac{1}{2}\right) \]
This indicates that the direction is towards the northwest.

### Final Answer
The velocity of the wind as observed by the observer after doubling his velocity is \( 2\sqrt{5} \, \text{m/s} \) towards the northwest.

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