A train is at rest. It accelerates for a time `t_(1)` at a uniform rate `alpha` and then comes to rest under a uniform retardation rate `beta` for time `t_(2)`. The ratio `t_(1)//t_(2)` is equal to

Options

A. `(beta)/(alpha)`
B. `(alpha +beta)/(alpha)`
C. `(alpha + beta)/(beta)`
D. `alpha beta`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
The train starts from rest, which means its initial velocity \( u = 0 \).

2. **Acceleration Phase**:
- The train accelerates uniformly at a rate \( \alpha \) for a time \( t_1 \).
- The final velocity \( V \) after this phase can be calculated using the formula:
\[ V = u + \alpha t_1 \]
- Since \( u = 0 \), this simplifies to:
\[ V = \alpha t_1 \]

3. **Deceleration Phase**:
- After reaching velocity \( V \), the train comes to rest under a uniform retardation \( \beta \) for a time \( t_2 \).
- The final velocity after this phase is 0, so we can again use the formula:
\[ 0 = V - \beta t_2 \]
- Rearranging gives:
\[ V = \beta t_2 \]

4. **Equating the Velocities**:
- From both phases, we have:
\[ \alpha t_1 = \beta t_2 \]

5. **Finding the Ratio**:
- To find the ratio \( \frac{t_1}{t_2} \), we can rearrange the equation:
\[ \frac{t_1}{t_2} = \frac{\beta}{\alpha} \]

### Final Answer:
The ratio \( \frac{t_1}{t_2} \) is equal to \( \frac{\beta}{\alpha} \).