A train starts from a station P and travels some distance with a uniform acceleration `a_1` then it goes with uniform retardation `a_2` for some more distance and come to rest at the station Q. If the distance between the stations P and Q is 4 km and the train takes 4 minutes to complete this journey, then `1/(a_1) + 1/(a_2)` =

Options

A. `2 m^(-1) s^2`
B. `4 m^(-1) s^2`
C. `7.2 m^(-1) s^2`
D. `72 m^(-1) s^2`

Explanation

To solve the problem, we need to analyze the motion of the train in two phases: the acceleration phase and the deceleration phase. Let's break it down step by step.

### Step 1: Understand the problem
The train starts from rest (initial velocity \( u = 0 \)) and travels a total distance of 4 km (4000 m) in 4 minutes (240 seconds). It accelerates with uniform acceleration \( a_1 \) for some distance, then decelerates with uniform retardation \( a_2 \) until it comes to rest at station Q.

### Step 2: Equations of motion
1. **Acceleration Phase**:
- The distance covered during acceleration (\( S_1 \)) can be given by the equation:
\[ S_1 = \frac{1}{2} a_1 t_1^2 \]
- The final velocity at the end of this phase (\( V \)) is:
\[ V = u + a_1 t_1 = a_1 t_1 \]

2. **Deceleration Phase**:
- The distance covered during deceleration (\( S_2 \)) can be given by the equation:
\[ S_2 = V t_2 - \frac{1}{2} a_2 t_2^2 \]
- Since the train comes to rest, the final velocity is 0.

### Step 3: Total distance and time
The total distance \( S \) is the sum of \( S_1 \) and \( S_2 \):
\[ S_1 + S_2 = 4000 \text{ m} \]
The total time \( T \) is the sum of \( t_1 \) and \( t_2 \):
\[ t_1 + t_2 = 240 \text{ s} \]

### Step 4: Substitute and simplify
From the equations derived, we can express \( S_2 \) in terms of \( V \) and \( a_2 \):
\[ S_2 = V t_2 - \frac{1}{2} a_2 t_2^2 \]
Substituting \( V = a_1 t_1 \) into \( S_2 \):
\[ S_2 = a_1 t_1 t_2 - \frac{1}{2} a_2 t_2^2 \]

### Step 5: Combine equations
Now, we can write:
\[ \frac{1}{2} a_1 t_1^2 + a_1 t_1 t_2 - \frac{1}{2} a_2 t_2^2 = 4000 \]
And from the time equation:
\[ t_2 = 240 - t_1 \]

### Step 6: Substitute \( t_2 \) into the distance equation
Substituting \( t_2 \) into the distance equation gives us a single equation in terms of \( t_1 \) and \( a_1, a_2 \).

### Step 7: Solve for \( \frac{1}{a_1} + \frac{1}{a_2} \)
After manipulating the equations, we arrive at:
\[ \frac{V^2}{2a_1} + \frac{V^2}{2a_2} = 4000 \]
This can be rearranged to find:
\[ \frac{1}{a_1} + \frac{1}{a_2} = \frac{4000}{\frac{V^2}{2}} = \frac{8000}{V^2} \]

### Step 8: Calculate \( V \)
Using the time equation, we can find \( V \) and substitute it back to find \( \frac{1}{a_1} + \frac{1}{a_2} \).

### Final Answer
After performing the calculations, we find that:
\[ \frac{1}{a_1} + \frac{1}{a_2} = 7.2 \text{ s}^2/\text{km} \]