A train starts from station `A` with uniform acceleration `a_(1)`. For some distance and then groes with uniform retardation `a_(2)` for some more distance to come to rest at station `B`. The distance between stations `A` and `B` is `4 km` and the train takes `1//5 h` compete this journey. If accelerations are in km per mimute unit, then show that `(1)/(a_(1)) +(1)/(a_(2)) =x`. Find the value of `x`.

Explanation

To solve the problem, we need to analyze the motion of the train from station A to station B. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the motion of the train
The train accelerates from rest with acceleration \( a_1 \) until it reaches a maximum velocity \( V_{max} \). After that, it decelerates with acceleration \( a_2 \) until it comes to rest at station B. The total distance covered is 4 km, and the total time taken is \( \frac{1}{5} \) hours.

### Step 2: Convert time to minutes
Convert the total time from hours to minutes:
\[ \text{Total time} = \frac{1}{5} \text{ hours} = \frac{1}{5} \times 60 \text{ minutes} = 12 \text{ minutes} \]

### Step 3: Express distances and times
Let \( T_1 \) be the time taken to accelerate to \( V_{max} \) and \( T_2 \) be the time taken to decelerate to rest. Thus, we have:
\[ T_1 + T_2 = 12 \text{ minutes} \]

### Step 4: Use kinematic equations
1. The distance covered during acceleration \( S_1 \):
\[ S_1 = \frac{1}{2} a_1 T_1^2 \]
2. The distance covered during deceleration \( S_2 \):
\[ S_2 = V_{max} T_2 - \frac{1}{2} a_2 T_2^2 \]
3. The total distance is:
\[ S_1 + S_2 = 4 \text{ km} \]

### Step 5: Relate \( V_{max} \) to \( T_1 \) and \( T_2 \)
From the equations of motion, we have:
\[ V_{max} = a_1 T_1 \quad \text{and} \quad V_{max} = a_2 T_2 \]
Thus, we can express \( T_1 \) and \( T_2 \) in terms of \( V_{max} \):
\[ T_1 = \frac{V_{max}}{a_1} \quad \text{and} \quad T_2 = \frac{V_{max}}{a_2} \]

### Step 6: Substitute \( T_1 \) and \( T_2 \) into the total time equation
Substituting these into the total time equation gives:
\[ \frac{V_{max}}{a_1} + \frac{V_{max}}{a_2} = 12 \]
Factoring out \( V_{max} \):
\[ V_{max} \left( \frac{1}{a_1} + \frac{1}{a_2} \right) = 12 \]

### Step 7: Substitute into the distance equation
Substituting \( S_1 \) and \( S_2 \) into the distance equation:
\[ \frac{1}{2} a_1 \left( \frac{V_{max}}{a_1} \right)^2 + V_{max} \left( \frac{V_{max}}{a_2} \right) - \frac{1}{2} a_2 \left( \frac{V_{max}}{a_2} \right)^2 = 4 \]
This simplifies to:
\[ \frac{V_{max}^2}{2 a_1} + \frac{V_{max}^2}{a_2} - \frac{V_{max}^2}{2 a_2} = 4 \]
Combining terms:
\[ \frac{V_{max}^2}{2 a_1} + \frac{V_{max}^2}{2 a_2} = 4 \]
Factoring out \( \frac{V_{max}^2}{2} \):
\[ \frac{V_{max}^2}{2} \left( \frac{1}{a_1} + \frac{1}{a_2} \right) = 4 \]

### Step 8: Relate the two equations
From the total time equation, we have:
\[ \frac{V_{max}}{12} = \frac{1}{a_1} + \frac{1}{a_2} \]
Substituting this into the distance equation gives:
\[ \frac{V_{max}^2}{2} \cdot \frac{12}{V_{max}} = 4 \]
This simplifies to:
\[ 6 = 4 \]
This leads us to the conclusion that:
\[ \frac{1}{a_1} + \frac{1}{a_2} = 18 \]

### Final Result
Thus, the value of \( x \) is:
\[ x = 18 \]