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 break it down into parts: acceleration phase, constant speed phase, and deceleration phase.

### Step 1: Define Variables
Let:
- \( a = 0.2 \, \text{m/s}^2 \) (acceleration)
- \( b = 0.4 \, \text{m/s}^2 \) (retardation)
- \( T = 30 \, \text{minutes} = 1800 \, \text{seconds} \) (total time)
- \( v_{\text{max}} \) = maximum speed attained by the train
- \( T_1 \) = time taken to reach maximum speed
- \( T_2 \) = time taken to come to rest

### Step 2: Find Time Taken to Reach Maximum Speed
Using the formula for acceleration:
\[ v_{\text{max}} = a \cdot T_1 \implies T_1 = \frac{v_{\text{max}}}{0.2} \]

### Step 3: Find Time Taken to Come to Rest
Using the formula for retardation:
\[ 0 = v_{\text{max}} - b \cdot T_2 \implies T_2 = \frac{v_{\text{max}}}{0.4} \]

### Step 4: Total Time Equation
The total time spent is the sum of the time taken to accelerate and the time taken to decelerate:
\[ T_1 + T_2 = 1800 \implies \frac{v_{\text{max}}}{0.2} + \frac{v_{\text{max}}}{0.4} = 1800 \]

### Step 5: Solve for Maximum Speed
To solve for \( v_{\text{max}} \), we can find a common denominator:
\[ \frac{2v_{\text{max}}}{0.4} + \frac{v_{\text{max}}}{0.4} = 1800 \]
\[ \frac{3v_{\text{max}}}{0.4} = 1800 \]
\[ 3v_{\text{max}} = 1800 \times 0.4 \]
\[ 3v_{\text{max}} = 720 \implies v_{\text{max}} = \frac{720}{3} = 240 \, \text{m/s} \]

### Step 6: Calculate Distances
Now we can calculate the distances during the acceleration and deceleration phases.

#### Distance during Acceleration (\( S_1 \)):
Using the equation:
\[ S_1 = \frac{1}{2} a T_1^2 \]
Substituting \( T_1 \):
\[ S_1 = \frac{1}{2} \cdot 0.2 \cdot \left(\frac{240}{0.2}\right)^2 \]
\[ S_1 = 0.1 \cdot \left(1200\right)^2 = 0.1 \cdot 1440000 = 144000 \, \text{m} \]

#### Distance during Deceleration (\( S_2 \)):
Using the equation:
\[ S_2 = v_{\text{max}} T_2 - \frac{1}{2} b T_2^2 \]
Substituting \( T_2 \):
\[ S_2 = 240 \cdot \left(\frac{240}{0.4}\right) - \frac{1}{2} \cdot 0.4 \cdot \left(\frac{240}{0.4}\right)^2 \]
\[ S_2 = 240 \cdot 600 - 0.2 \cdot 1440000 = 144000 - 288000 = 144000 \, \text{m} \]

### Step 7: Total Distance
The total distance \( S \) between the two stations is:
\[ S = S_1 + S_2 = 144000 + 144000 = 288000 \, \text{m} = 288 \, \text{km} \]

### Final Answer
The distance between the two stations is **288 kilometers**.

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