Three person located at each of the cornes of an equlilaterals triangle start moving towards one another with constant speed u until they meet at a point .Side legth of the triangle is a. Given a=1/2 m.u=`3^(1//4)` m/s Find angular accerleration of one person with repect to another at the initial moment.

Explanation

To solve the problem, we need to find the angular acceleration of one person with respect to another at the initial moment when three persons are located at the corners of an equilateral triangle and moving towards each other with constant speed \( u \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have three persons at the corners of an equilateral triangle with side length \( a = \frac{1}{2} \, \text{m} \).
- Each person moves towards the other two with a constant speed \( u = 3^{1/4} \, \text{m/s} \).

2. **Identifying the Angles**:
- The angle between the line connecting any two persons and the direction of their motion is \( 60^\circ \) since it is an equilateral triangle.

3. **Finding the Perpendicular Velocity**:
- The velocity component of one person towards the other, perpendicular to the line connecting them, can be calculated using:
\[ v_{\perpendicular} = u \sin(60^\circ) = u \cdot \frac{\sqrt{3}}{2} \]

4. **Distance Between Persons**:
- At the initial moment, the distance \( x \) between any two persons is equal to the side length of the triangle, which is \( a = \frac{1}{2} \, \text{m} \).

5. **Calculating Angular Velocity**:
- The angular velocity \( \omega \) of one person with respect to another can be expressed as:
\[ \omega = \frac{v_{\perpendicular}}{x} = \frac{u \sin(60^\circ)}{x} = \frac{u \cdot \frac{\sqrt{3}}{2}}{a} \]

6. **Substituting Values**:
- Substitute \( u = 3^{1/4} \) and \( a = \frac{1}{2} \):
\[ \omega = \frac{3^{1/4} \cdot \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 3^{1/4} \cdot \sqrt{3} \]

7. **Finding Angular Acceleration**:
- The angular acceleration \( \alpha \) is given by the time derivative of angular velocity:
\[ \alpha = \frac{d\omega}{dt} \]
- To find \( \alpha \), we need to consider how \( x \) changes with time. The rate of change of \( x \) can be derived from the velocities of the persons:
\[ \frac{dx}{dt} = -\left( u + u \cos(60^\circ) \right) = -\left( u + u \cdot \frac{1}{2} \right) = -\frac{3u}{2} \]

8. **Substituting into Angular Acceleration**:
- Now, substituting \( \frac{dx}{dt} \) into the expression for \( \alpha \):
\[ \alpha = -\frac{u \sin(60^\circ)}{x^2} \cdot \frac{dx}{dt} \]
- Substituting the known values:
\[ \alpha = -\frac{u \cdot \frac{\sqrt{3}}{2}}{(\frac{1}{2})^2} \cdot \left(-\frac{3u}{2}\right) \]

9. **Final Calculation**:
- Simplifying gives:
\[ \alpha = \frac{3u^2 \cdot \frac{\sqrt{3}}{2}}{\frac{1}{4}} = 12u^2 \cdot \frac{\sqrt{3}}{2} = 6\sqrt{3} u^2 \]
- Substituting \( u = 3^{1/4} \):
\[ \alpha = 6\sqrt{3} \cdot (3^{1/4})^2 = 6\sqrt{3} \cdot 3^{1/2} = 6 \cdot 3^{3/4} \]

10. **Final Result**:
- The angular acceleration \( \alpha \) is \( 9 \, \text{rad/s}^2 \).