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^(-1)`
B. `52 km h^(-1)`
C. `45 kmh^(-1)`
D. `56 km h^(-1)`

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Understand the Motion
The train undergoes three phases:
1. Accelerates uniformly (time ratio = 1)
2. Moves with constant velocity (time ratio = 8)
3. Retards uniformly to come to rest (time ratio = 1)

### Step 2: Define Time Variables
Let the time for the first phase (acceleration) be \( T_0 \). Then:
- Time for acceleration = \( T_0 \)
- Time for constant velocity = \( 8T_0 \)
- Time for retardation = \( T_0 \)

### Step 3: Calculate Total Time
The total time \( T \) for the journey is:
\[ T = T_0 + 8T_0 + T_0 = 10T_0 \]

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

### Step 5: Calculate Distances for Each Phase
1. **Distance during acceleration**:
- Using the formula \( d = \frac{1}{2} a t^2 \) and knowing that the final velocity \( v = a T_0 \) (where \( a \) is acceleration):
- We can express \( a \) as \( a = \frac{60 \, \text{km/h}}{T_0} \) (converting \( 60 \, \text{km/h} \) to \( \text{m/s} \) if needed).
- Distance during acceleration \( d_1 = \frac{1}{2} a T_0^2 = \frac{1}{2} \left(\frac{60}{T_0}\right) T_0^2 = 30 T_0 \).

2. **Distance during constant velocity**:
- The distance during this phase is given by \( d_2 = v \cdot t = 60 \cdot 8T_0 = 480T_0 \).

3. **Distance during retardation**:
- Similar to the acceleration phase, the distance during retardation is also \( d_3 = 30 T_0 \).

### Step 6: Calculate Total Distance
The total distance \( D \) is:
\[ D = d_1 + d_2 + d_3 = 30T_0 + 480T_0 + 30T_0 = 540T_0 \]

### Step 7: Calculate Average Speed
The average speed \( V_{avg} \) is given by:
\[ V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{540T_0}{10T_0} = 54 \, \text{km/h} \]

### Final Answer
Thus, the average speed over the whole journey is:
\[ \boxed{54 \, \text{km/h}} \]