Between two stations a train starting from rest first accelerates uniformly, then moves with constant velocity and finally retarts uniformly to come to rest. If the ratio of the time taken be `1 : 8 : 1` and the maximum speed attained be `60 km//h`, then what is the average speed over the whole journey ?

Options

A. `48 km//h`
B. `52 km//h`
C. `54 km//h`
D. `56 km//h`

Explanation

To solve the problem step by step, we will break down the journey of the train into three parts as described in the question.

### Step 1: Understand the Journey
The train travels in three phases:
1. **Acceleration Phase**: The train accelerates from rest to its maximum speed.
2. **Constant Velocity Phase**: The train moves at a constant speed.
3. **Retardation Phase**: The train decelerates uniformly to come to rest.

### Step 2: Define Variables
- Let \( t_1 \) be the time taken during the acceleration phase.
- Let \( t_2 \) be the time taken during the constant velocity phase.
- Let \( t_3 \) be the time taken during the retardation phase.

According to the problem, the ratio of the times is given as:
\[ t_1 : t_2 : t_3 = 1 : 8 : 1 \]

Let \( t_1 = t \), \( t_2 = 8t \), and \( t_3 = t \).

### Step 3: Total Time
The total time \( T \) for the journey is:
\[ T = t_1 + t_2 + t_3 = t + 8t + t = 10t \]

### Step 4: Maximum Speed
The maximum speed attained by the train is given as \( 60 \text{ km/h} \).

### Step 5: Calculate Distances
1. **Distance during Acceleration Phase**:
- The train starts from rest and accelerates to \( 60 \text{ km/h} \).
- Using the formula for distance during uniform acceleration:
\[ d_1 = \frac{1}{2} a t_1^2 \]
where \( a \) is the acceleration.

Since \( v = u + at \) and \( u = 0 \):
\[ 60 = 0 + a t_1 \implies a = \frac{60}{t} \]
Substituting \( a \) into the distance formula:
\[ d_1 = \frac{1}{2} \left(\frac{60}{t}\right) t^2 = 30t \]

2. **Distance during Constant Velocity Phase**:
\[ d_2 = v \cdot t_2 = 60 \cdot 8t = 480t \]

3. **Distance during Retardation Phase**:
- The distance during retardation is the same as during acceleration, so:
\[ d_3 = 30t \]

### Step 6: Total Distance
The total distance \( D \) covered by the train is:
\[ D = d_1 + d_2 + d_3 = 30t + 480t + 30t = 540t \]

### Step 7: Average Speed
The average speed \( V_{avg} \) is given by:
\[ V_{avg} = \frac{D}{T} = \frac{540t}{10t} = 54 \text{ km/h} \]

### Final Answer
The average speed over the whole journey is \( 54 \text{ km/h} \).

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