The speed of a train increases at a constant rate `alpha` from zero to v and then remains constant for an interval and finally decreases to zero at a constant rate `beta`. The total distance travelled by the train is l. The time taken to complete the journey is t. Then,

Options

A. `t= (l(alpha + beta))/(alpha beta)`
B. `t= l/v +v/2 (1/alpha + 1/beta)`
C. t is minimum when `v= sqrt((2 l alpha beta)/(alpha - beta))`
D. t is minimum when `v= sqrt((2 l alpha beta)/(alpha + beta))`

Explanation

To solve the problem step-by-step, we will analyze the motion of the train in three phases: acceleration, constant speed, and deceleration.

### Step 1: Understanding the Phases of Motion
The train's journey can be divided into three distinct phases:
1. **Acceleration Phase**: The train accelerates from rest (0) to speed \( v \) at a constant rate \( \alpha \).
2. **Constant Speed Phase**: The train moves at a constant speed \( v \) for a certain time \( t_2 \).
3. **Deceleration Phase**: The train decelerates from speed \( v \) to rest (0) at a constant rate \( \beta \).

### Step 2: Calculate Time for Each Phase
1. **Acceleration Phase**:
- The time taken to accelerate from 0 to \( v \) is given by:
\[ t_1 = \frac{v}{\alpha} \]

2. **Constant Speed Phase**:
- The time taken at constant speed \( v \) is \( t_2 \).

3. **Deceleration Phase**:
- The time taken to decelerate from \( v \) to 0 is given by:
\[ t_3 = \frac{v}{\beta} \]

### Step 3: Total Time
The total time \( t \) for the journey is the sum of the times for each phase:
\[ t = t_1 + t_2 + t_3 = \frac{v}{\alpha} + t_2 + \frac{v}{\beta} \]

### Step 4: Calculate Distances for Each Phase
1. **Distance during Acceleration Phase**:
- The distance covered while accelerating is:
\[ d_1 = \frac{1}{2} v t_1 = \frac{1}{2} v \left(\frac{v}{\alpha}\right) = \frac{v^2}{2\alpha} \]

2. **Distance during Constant Speed Phase**:
- The distance covered at constant speed is:
\[ d_2 = v \cdot t_2 \]

3. **Distance during Deceleration Phase**:
- The distance covered while decelerating is:
\[ d_3 = \frac{1}{2} v t_3 = \frac{1}{2} v \left(\frac{v}{\beta}\right) = \frac{v^2}{2\beta} \]

### Step 5: Total Distance
The total distance \( L \) covered by the train is the sum of the distances for each phase:
\[ L = d_1 + d_2 + d_3 = \frac{v^2}{2\alpha} + v t_2 + \frac{v^2}{2\beta} \]

### Step 6: Rearranging the Equations
From the total distance equation, we can express \( t_2 \):
\[ L = \frac{v^2}{2\alpha} + v t_2 + \frac{v^2}{2\beta} \]
Rearranging gives:
\[ v t_2 = L - \left(\frac{v^2}{2\alpha} + \frac{v^2}{2\beta}\right) \]
Thus,
\[ t_2 = \frac{L - \left(\frac{v^2}{2\alpha} + \frac{v^2}{2\beta}\right)}{v} \]

### Step 7: Substitute \( t_2 \) into Total Time Equation
Substituting \( t_2 \) back into the total time equation:
\[ t = \frac{v}{\alpha} + \frac{L - \left(\frac{v^2}{2\alpha} + \frac{v^2}{2\beta}\right)}{v} + \frac{v}{\beta} \]

### Step 8: Final Relationships
After simplification, we can derive the relationships between \( t \), \( L \), \( \alpha \), \( \beta \), and \( v \).