A train starts, from station A with uniform acceleration `f_(1)` for some distance and then goes with uniform retardation `f_(2)` for some more distance to come to rest at station B. The distance between the stations A and B is 4 km and the train takes 4 minutes to complete this journey. If `f_(1)` and `f_(2)` are in km-minute units, then `(1)/(f_(1))+(1)/(f_(2))=`

Options

A. 1
B. 2
C. 3
D. 4

Answer

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

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The train accelerates from rest with acceleration \( f_1 \) for a distance \( S_1 \).
- It then decelerates with retardation \( f_2 \) for a distance \( S_2 \) until it comes to rest at station B.
- The total distance \( S_1 + S_2 = 4 \) km.
- The total time taken for the journey is 4 minutes.

2. **Using the Average Velocity**:
- The average velocity during the entire journey can be calculated as:
\[ \text{Average Velocity} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{4 \text{ km}}{4 \text{ min}} = 1 \text{ km/min} \]

3. **Finding Distances in Terms of Time**:
- For the first part (acceleration), the average velocity is:
\[ S_1 = \text{Average Velocity} \times T_1 = \frac{V}{2} \times T_1 \]
- For the second part (retardation), the average velocity is:
\[ S_2 = \text{Average Velocity} \times T_2 = \frac{V}{2} \times T_2 \]

4. **Setting Up the Equations**:
- We know \( S_1 + S_2 = 4 \) km, thus:
\[ \frac{V}{2} T_1 + \frac{V}{2} T_2 = 4 \]
- Since \( T_1 + T_2 = 4 \) min, we can substitute \( T_1 + T_2 \) into the equation:
\[ \frac{V}{2} \cdot 4 = 4 \implies V = 2 \text{ km/min} \]

5. **Finding \( S_1 \) and \( S_2 \)**:
- Now, we can express \( S_1 \) and \( S_2 \) in terms of \( T_1 \) and \( T_2 \):
\[ S_1 = \frac{V}{2} T_1 = \frac{2}{2} T_1 = T_1 \text{ km} \]
\[ S_2 = \frac{V}{2} T_2 = \frac{2}{2} T_2 = T_2 \text{ km} \]
- Thus, we have:
\[ T_1 + T_2 = 4 \text{ km} \]

6. **Using Kinematic Equations**:
- For the acceleration phase:
\[ V^2 = 0 + 2 f_1 S_1 \implies 2^2 = 2 f_1 S_1 \implies 4 = 2 f_1 T_1 \implies f_1 = \frac{2}{T_1} \]
- For the retardation phase:
\[ 0 = V^2 - 2 f_2 S_2 \implies 0 = 2^2 - 2 f_2 T_2 \implies 4 = 2 f_2 T_2 \implies f_2 = \frac{2}{T_2} \]

7. **Finding \( \frac{1}{f_1} + \frac{1}{f_2} \)**:
- Now substituting \( f_1 \) and \( f_2 \):
\[ \frac{1}{f_1} = \frac{T_1}{2}, \quad \frac{1}{f_2} = \frac{T_2}{2} \]
- Therefore:
\[ \frac{1}{f_1} + \frac{1}{f_2} = \frac{T_1}{2} + \frac{T_2}{2} = \frac{T_1 + T_2}{2} = \frac{4}{2} = 2 \]

### Final Answer:
\[ \frac{1}{f_1} + \frac{1}{f_2} = 2 \]