A trian starts from rest from a station with acceleration `0.2 m/s^(2)` on a straight track and after attaining maximum speed it comes to rest on another station due to retardation of `0.4 m/s^(2)`. If total time spent is half an hour, then distance between two stations is (Neglect length of train) :-

Options

A. 216 km
B. 512 km
C. 728 km
D. 1296 km

Answer

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

### Step 1: Define Variables
- Let \( a_1 = 0.2 \, \text{m/s}^2 \) (acceleration)
- Let \( a_2 = -0.4 \, \text{m/s}^2 \) (retardation)
- Let \( D_1 \) be the distance covered during acceleration.
- Let \( D_2 \) be the distance covered during deceleration.
- Let \( T_1 \) be the time taken to accelerate.
- Let \( T_2 \) be the time taken to decelerate.

### Step 2: Total Time
The total time for the journey is given as half an hour, which is:
\[ T_1 + T_2 = 0.5 \, \text{hours} = 1800 \, \text{seconds} \]

### Step 3: Equations of Motion for Acceleration
Using the equations of motion for the first half of the journey (acceleration):
1. From \( v^2 = u^2 + 2a s \):
\[ v^2 = 0 + 2 \cdot 0.2 \cdot D_1 \implies v^2 = 0.4 D_1 \quad \text{(Equation 1)} \]
2. From \( v = u + at \):
\[ v = 0 + 0.2 T_1 \implies v = 0.2 T_1 \quad \text{(Equation 2)} \]

### Step 4: Equations of Motion for Deceleration
Using the equations of motion for the second half of the journey (deceleration):
1. From \( v^2 = u^2 + 2a s \):
\[ 0 = v^2 - 2 \cdot 0.4 \cdot D_2 \implies v^2 = 0.8 D_2 \quad \text{(Equation 3)} \]
2. From \( v = u + at \):
\[ 0 = v - 0.4 T_2 \implies v = 0.4 T_2 \quad \text{(Equation 4)} \]

### Step 5: Equate Velocities
From Equations 2 and 4, we can equate the expressions for \( v \):
\[ 0.2 T_1 = 0.4 T_2 \implies T_1 = 2 T_2 \quad \text{(Equation 5)} \]

### Step 6: Substitute into Total Time Equation
Substituting Equation 5 into the total time equation:
\[ 2 T_2 + T_2 = 1800 \implies 3 T_2 = 1800 \implies T_2 = 600 \, \text{seconds} \]
Now, substituting back to find \( T_1 \):
\[ T_1 = 2 \cdot 600 = 1200 \, \text{seconds} \]

### Step 7: Calculate Maximum Speed \( v \)
Using \( T_2 \) in Equation 4:
\[ v = 0.4 \cdot 600 = 240 \, \text{m/s} \]

### Step 8: Calculate Distances \( D_1 \) and \( D_2 \)
Using \( v \) in Equation 1:
\[ v^2 = 0.4 D_1 \implies (240)^2 = 0.4 D_1 \implies 57600 = 0.4 D_1 \implies D_1 = \frac{57600}{0.4} = 144000 \, \text{meters} \]
Using \( v \) in Equation 3:
\[ v^2 = 0.8 D_2 \implies (240)^2 = 0.8 D_2 \implies 57600 = 0.8 D_2 \implies D_2 = \frac{57600}{0.8} = 72000 \, \text{meters} \]

### Step 9: Calculate Total Distance
The total distance \( D \) between the two stations is:
\[ D = D_1 + D_2 = 144000 + 72000 = 216000 \, \text{meters} \]
Converting to kilometers:
\[ D = \frac{216000}{1000} = 216 \, \text{km} \]

### Final Answer
The distance between the two stations is **216 kilometers**.