A train accelerates from rest at a constant rate `alpha` for some time and then it retards to rest at the constant rate `beta`. If the total distance covered by the train is s, then the velocity of the train is

Options

A. `[(alpha + beta)/(2alphabeta) xx s]^(1//2)`
B. `[(alpha - beta)/(2alphabeta) xx s]^(1//2)`
C. `[(alpha beta)/(alpha + beta) xx s]^(1//2)`
D. `[(alpha beta)/(alpha - beta) xx s]^(1//2)`

Explanation

To solve the problem, we need to analyze the motion of the train in two phases: acceleration and deceleration.

### Step-by-Step Solution:

1. **Define the Variables**:
- Let \( \alpha \) be the constant acceleration of the train.
- Let \( \beta \) be the constant deceleration (retardation) of the train.
- Let \( s \) be the total distance covered by the train.
- Let \( S_1 \) be the distance covered during acceleration.
- Let \( S_2 \) be the distance covered during deceleration.
- Let \( V \) be the final velocity of the train after the acceleration phase.

2. **Acceleration Phase (A to C)**:
- The train starts from rest, so the initial velocity \( u = 0 \).
- Using the equation of motion:
\[ V^2 = u^2 + 2 \alpha S_1 \]
Substituting \( u = 0 \):
\[ V^2 = 0 + 2 \alpha S_1 \implies V^2 = 2 \alpha S_1 \quad \text{(Equation 1)} \]

3. **Deceleration Phase (C to B)**:
- The initial velocity for this phase is \( V \) and the final velocity is \( 0 \).
- Using the equation of motion:
\[ 0 = V^2 + 2(-\beta) S_2 \]
Rearranging gives:
\[ V^2 = 2 \beta S_2 \quad \text{(Equation 2)} \]

4. **Total Distance**:
- The total distance covered by the train is the sum of the distances during acceleration and deceleration:
\[ s = S_1 + S_2 \]

5. **Expressing \( S_1 \) and \( S_2 \)**:
- From Equation 1:
\[ S_1 = \frac{V^2}{2\alpha} \]
- From Equation 2:
\[ S_2 = \frac{V^2}{2\beta} \]

6. **Substituting \( S_1 \) and \( S_2 \) into the Total Distance Equation**:
- Substitute \( S_1 \) and \( S_2 \) into the total distance equation:
\[ s = \frac{V^2}{2\alpha} + \frac{V^2}{2\beta} \]
- Taking a common denominator:
\[ s = V^2 \left( \frac{1}{2\alpha} + \frac{1}{2\beta} \right) = V^2 \left( \frac{\beta + \alpha}{2\alpha\beta} \right) \]

7. **Solving for \( V^2 \)**:
- Rearranging gives:
\[ V^2 = \frac{2s \alpha \beta}{\alpha + \beta} \]

8. **Finding \( V \)**:
- Taking the square root:
\[ V = \sqrt{\frac{2\alpha\beta s}{\alpha + \beta}} \]

### Final Answer:
\[ V = \sqrt{\frac{2\alpha\beta s}{\alpha + \beta}} \]