A person travelling towards the north-east, finds that the wind appears to blow from the north , but when he doubles his speed it seems to come from a direction inclined at an angle ` cot^(-1)` 2 on the east of north. The true direction of the wind is towards

Options

A. North east
B. North
C. East
D. none of these

Answer

To solve the problem, we need to analyze the situation using vector components and the concept of relative velocity.

### Step 1: Define the Directions
Let’s denote:
- The velocity of the person (V_p) traveling towards the northeast as \( V_p = v \) at an angle of 45° to the north and east.
- The wind velocity (V_w) is unknown and we need to find its direction.

### Step 2: Analyze the First Condition
When the person is traveling at speed \( v \) towards the northeast, the wind appears to come from the north. This means that the relative velocity of the wind with respect to the person is directed towards the north.

Using vector components:
- The velocity of the person can be represented as:
\[ V_p = v \cos(45^\circ) \hat{i} + v \sin(45^\circ) \hat{j} = \frac{v}{\sqrt{2}} \hat{i} + \frac{v}{\sqrt{2}} \hat{j} \]
- The wind velocity can be represented as:
\[ V_w = V_{wx} \hat{i} + V_{wy} \hat{j} \]

The relative velocity of the wind with respect to the person is given by:
\[ V_{relative} = V_w - V_p \]
For the wind to appear to come from the north, the x-component of the relative velocity must be zero:
\[ V_{wx} - \frac{v}{\sqrt{2}} = 0 \implies V_{wx} = \frac{v}{\sqrt{2}} \]

### Step 3: Analyze the Second Condition
When the person doubles his speed (2v), the wind appears to come from a direction inclined at an angle \( \cot^{-1}(2) \) east of north. The angle \( \cot^{-1}(2) \) means that the tangent of the angle is \( \frac{1}{2} \), which gives us a slope of 1:2.

The new velocity of the person is:
\[ V_p' = 2v \cos(45^\circ) \hat{i} + 2v \sin(45^\circ) \hat{j} = \sqrt{2} v \hat{i} + \sqrt{2} v \hat{j} \]

The relative velocity now becomes:
\[ V_{relative}' = V_w - V_p' = (V_{wx} - \sqrt{2}v) \hat{i} + (V_{wy} - \sqrt{2}v) \hat{j} \]

### Step 4: Set up the Relationship for the Angle
From the angle \( \cot^{-1}(2) \):
\[ \tan(\theta) = \frac{V_{wy} - \sqrt{2}v}{V_{wx} - \sqrt{2}v} \]
Substituting \( V_{wx} = \frac{v}{\sqrt{2}} \):
\[ \tan(\cot^{-1}(2)) = \frac{V_{wy} - \sqrt{2}v}{\frac{v}{\sqrt{2}} - \sqrt{2}v} \]
Since \( \tan(\cot^{-1}(2)) = \frac{1}{2} \):
\[ \frac{1}{2} = \frac{V_{wy} - \sqrt{2}v}{\frac{v}{\sqrt{2}} - \sqrt{2}v} \]

### Step 5: Solve for Wind Velocity Components
Now we can solve for \( V_{wy} \):
\[ \frac{1}{2} = \frac{V_{wy} - \sqrt{2}v}{\frac{v}{\sqrt{2}} - 2v} \]
Simplifying gives us:
\[ \frac{1}{2} = \frac{V_{wy} - \sqrt{2}v}{\frac{v - 2\sqrt{2}v}{\sqrt{2}}} \]
Cross-multiplying and solving will give us the value of \( V_{wy} \).

### Step 6: Determine the True Direction of the Wind
Finally, we can find the angle of the wind direction using:
\[ \theta_w = \tan^{-1}\left(\frac{V_{wy}}{V_{wx}}\right) \]
This will give us the true direction of the wind.

### Conclusion
The true direction of the wind can be determined from the components calculated above.