A cat, on seeing a rat at a distance `d=5 m`, starts velocity `u=5 m s^(-1)` and moves with acceleration `alpha =2.5 m s^(-2)` in order to catch it, while the rate with acceleration `beta` starts from rest. For what minimum value of `beta` will the overtake the rat?. (in `m s^(-2)`).

Explanation

To solve the problem, we need to determine the minimum acceleration \( \beta \) of the rat so that the cat can catch it. We will analyze the motion of both the cat and the rat.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Distance \( d = 5 \, \text{m} \)
- Initial velocity of the cat \( u = 5 \, \text{m/s} \)
- Acceleration of the cat \( \alpha = 2.5 \, \text{m/s}^2 \)
- Initial velocity of the rat \( v_r = 0 \) (starts from rest)
- Acceleration of the rat \( \beta \) (unknown)

2. **Relative Motion Analysis:**
- We will analyze the situation from the frame of reference of the rat. In this frame, the rat is stationary, and the cat is moving towards it with an initial velocity \( u \) and acceleration \( \alpha \) while the rat has an acceleration \( \beta \) in the opposite direction.

3. **Using the Equations of Motion:**
- The equation of motion for the cat can be expressed as:
\[ s_c = ut + \frac{1}{2} \alpha t^2 \]
- The equation of motion for the rat can be expressed as:
\[ s_r = \frac{1}{2} \beta t^2 \]

4. **Setting Up the Equation:**
- For the cat to catch the rat, the distance covered by the cat must equal the initial distance plus the distance covered by the rat:
\[ s_c = d + s_r \]
- Substituting the equations of motion:
\[ ut + \frac{1}{2} \alpha t^2 = d + \frac{1}{2} \beta t^2 \]

5. **Rearranging the Equation:**
- Rearranging gives:
\[ ut + \frac{1}{2} \alpha t^2 - \frac{1}{2} \beta t^2 = d \]
- This simplifies to:
\[ ut + \frac{1}{2} (\alpha - \beta) t^2 = d \]

6. **Finding the Time of Catching:**
- For the cat to just catch the rat, we can consider the condition when the relative velocity becomes zero. This occurs when:
\[ u + \alpha t = \beta t \]
- Rearranging gives:
\[ t(\beta - \alpha) = u \]
- Thus, we can express \( t \) as:
\[ t = \frac{u}{\beta - \alpha} \]

7. **Substituting Back to Find \( \beta \):**
- Substitute \( t \) back into the distance equation:
\[ d = u \left(\frac{u}{\beta - \alpha}\right) + \frac{1}{2} (\alpha - \beta) \left(\frac{u}{\beta - \alpha}\right)^2 \]
- After simplification, we find:
\[ d = \frac{u^2}{\beta - \alpha} + \frac{1}{2} (\alpha - \beta) \frac{u^2}{(\beta - \alpha)^2} \]

8. **Solving for \( \beta \):**
- After rearranging and solving for \( \beta \), we find:
\[ \beta = \alpha + \frac{u^2}{2d} \]

9. **Substituting Known Values:**
- Now substitute \( \alpha = 2.5 \, \text{m/s}^2 \), \( u = 5 \, \text{m/s} \), and \( d = 5 \, \text{m} \):
\[ \beta = 2.5 + \frac{5^2}{2 \times 5} = 2.5 + \frac{25}{10} = 2.5 + 2.5 = 5 \, \text{m/s}^2 \]

### Final Answer:
The minimum value of \( \beta \) for the rat to be overtaken by the cat is \( \beta = 5 \, \text{m/s}^2 \).