A train takes f s to perform a journey. It travels for `(t)/(n)` s with uniform acceleration, then for `(n -3) (t)/(n)` with uniform spced v and finally it comes to rest with unifom retardation. The average of the speed of the train is

Options

A. `(3n-2) (nu)/(2n)`
B. `(2n-3) (nu)/(2n)`
C. `(3n-2)(nu)/(3n)`
D. `(2n-3)(nu)/(3n)`

Explanation

To solve the problem step by step, we will analyze the journey of the train in three segments: uniform acceleration, uniform speed, and uniform retardation.

### Step 1: Define the time intervals
The total time taken for the journey is \( t \) seconds. The train travels for:
1. \( \frac{t}{n} \) seconds with uniform acceleration.
2. \( (n - 3) \frac{t}{n} \) seconds with uniform speed \( v \).
3. The remaining time with uniform retardation.

### Step 2: Calculate the time for uniform retardation
To find the time for uniform retardation, we first calculate the time spent in the first two segments:
- Time for uniform acceleration: \( \frac{t}{n} \)
- Time for uniform speed: \( (n - 3) \frac{t}{n} \)

Total time for acceleration and uniform speed:
\[ \text{Total time} = \frac{t}{n} + (n - 3) \frac{t}{n} = \frac{t + (n - 3)t}{n} = \frac{nt - 2t}{n} = \frac{(n-2)t}{n} \]
Thus, the time for uniform retardation is:
\[ t - \frac{(n-2)t}{n} = \frac{nt - (n-2)t}{n} = \frac{2t}{n} \]

### Step 3: Calculate the distances for each segment
1. **Distance during uniform acceleration**:
Using the formula for distance under uniform acceleration:
\[ d_1 = \frac{1}{2} a t^2 \]
where \( t = \frac{t}{n} \). The final velocity \( v \) at the end of this segment can be calculated using:
\[ v = a \cdot \frac{t}{n} \]
However, we will express \( d_1 \) in terms of \( v \) later.

2. **Distance during uniform speed**:
The distance traveled at uniform speed \( v \) for time \( (n - 3) \frac{t}{n} \) is:
\[ d_2 = v \cdot (n - 3) \frac{t}{n} \]

3. **Distance during uniform retardation**:
For the time \( \frac{2t}{n} \), if the initial speed at the start of this segment is \( v \) and it comes to rest, we can use:
\[ d_3 = v \cdot \frac{2t}{n} - \frac{1}{2} a' \left(\frac{2t}{n}\right)^2 \]
where \( a' \) is the retardation.

### Step 4: Total distance and average speed
The total distance \( D \) is the sum of the distances:
\[ D = d_1 + d_2 + d_3 \]

The average speed \( V_{avg} \) is given by:
\[ V_{avg} = \frac{D}{t} \]

### Step 5: Substitute and simplify
After substituting the expressions for \( d_1, d_2, \) and \( d_3 \) and simplifying, we arrive at:
\[ V_{avg} = \frac{V}{2n} (2n - 3) \]

### Final Answer
The average speed of the train is:
\[ V_{avg} = \frac{V}{2n} (2n - 3) \]

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