Two particles A and B move anticlockwise with the same speed v in a circle of radius R and are diametrically opposite to each other. At `t=0`, A is given a constant accelerartion (tangential) `a_(t)=(72v^(2))/(25 pi R)`. Calculate the time in which a collides with B, the angle traced by A, its angular velocity and radial acceleration at the time of collision.

Explanation

To solve the problem step by step, we will break down the calculations for each part of the question:

### Step 1: Calculate the Time of Collision
1. **Understanding the Setup**: Particles A and B are moving in a circle of radius R and are initially diametrically opposite. Both have the same speed \( v \), but particle A is given a tangential acceleration \( a_t = \frac{72v^2}{25\pi R} \).

2. **Distance to be Traveled**: The distance A needs to travel to collide with B is half the circumference of the circle:
\[ s = \frac{1}{2} \times 2\pi R = \pi R \]

3. **Using the Kinematic Equation**: The kinematic equation relating distance, initial velocity, acceleration, and time is:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( u = 0 \) (relative velocity in A's frame), \( a = a_t = \frac{72v^2}{25\pi R} \), and \( s = \pi R \).

4. **Substituting Values**:
\[ \pi R = 0 + \frac{1}{2} \left(\frac{72v^2}{25\pi R}\right) t^2 \]
Simplifying gives:
\[ \pi R = \frac{36v^2}{25\pi R} t^2 \]
Rearranging:
\[ t^2 = \frac{25\pi^2 R^2}{36v^2} \]
Taking the square root:
\[ t = \frac{5\pi R}{6v} \]

### Step 2: Calculate the Angle Traced by A
1. **Initial Angular Velocity**: The initial angular velocity \( \omega_0 \) of A is:
\[ \omega_0 = \frac{v}{R} \]

2. **Angular Acceleration**: The angular acceleration \( \alpha \) is given by:
\[ \alpha = \frac{a_t}{R} = \frac{72v^2}{25\pi R^2} \]

3. **Using the Angular Displacement Equation**:
\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]
Substituting the values:
\[ \theta = \left(\frac{v}{R}\right) \left(\frac{5\pi R}{6v}\right) + \frac{1}{2} \left(\frac{72v^2}{25\pi R^2}\right) \left(\frac{5\pi R}{6v}\right)^2 \]
Simplifying:
\[ \theta = \frac{5\pi}{6} + \frac{1}{2} \cdot \frac{72v^2}{25\pi R^2} \cdot \frac{25\pi^2 R^2}{36v^2} \]
After canceling terms:
\[ \theta = \frac{5\pi}{6} + \pi = \frac{11\pi}{6} \]

### Step 3: Calculate the Final Angular Velocity
1. **Using the Angular Velocity Equation**:
\[ \omega_f = \omega_0 + \alpha t \]
Substituting the known values:
\[ \omega_f = \frac{v}{R} + \left(\frac{72v^2}{25\pi R^2}\right) \left(\frac{5\pi R}{6v}\right) \]
Simplifying:
\[ \omega_f = \frac{v}{R} + \frac{12v}{5R} \]
Combining terms:
\[ \omega_f = \frac{17v}{5R} \]

### Step 4: Calculate the Radial Acceleration
1. **Using the Radial Acceleration Formula**:
\[ a_r = \omega_f^2 R \]
Substituting \( \omega_f \):
\[ a_r = \left(\frac{17v}{5R}\right)^2 R = \frac{289v^2}{25R} \]

### Final Results
- **Time of Collision**: \( t = \frac{5\pi R}{6v} \)
- **Angle Traced by A**: \( \theta = \frac{11\pi}{6} \)
- **Final Angular Velocity**: \( \omega_f = \frac{17v}{5R} \)
- **Radial Acceleration**: \( a_r = \frac{289v^2}{25R} \)