A train starts from rest and for the first kilometer moveswith constant acceleration. For the next 3 kilometers it has constant velocity and for another two kilometers it moves with constant retardation to come to rest after a total time of 10 minutes. The maximum velocity of the train is

Options

A. 18 km/h
B. 36 km/h
C. 54 km/h
D. 39 km/h

Explanation

To find the maximum velocity of the train, we can break down the problem into three segments based on the train's motion: acceleration, constant velocity, and retardation.

### Step-by-Step Solution:

1. **Identify the segments of motion**:
- **Segment 1**: Acceleration over the first 1 km.
- **Segment 2**: Constant velocity over the next 3 km.
- **Segment 3**: Retardation over the last 2 km.

2. **Define variables**:
- Let \( v \) be the maximum velocity reached after the first segment.
- Let \( t_1 \) be the time taken to travel the first 1 km.
- Let \( t_2 \) be the time taken to travel the next 3 km at constant velocity \( v \).
- Let \( t_3 \) be the time taken to travel the last 2 km while decelerating to rest.

3. **Equations for each segment**:
- For **Segment 1** (acceleration):
\[ s_1 = \frac{1}{2} a t_1^2 = 1 \text{ km} \]
Here, \( s_1 \) is the distance (1 km), and \( a \) is the acceleration.

- For **Segment 2** (constant velocity):
\[ s_2 = v t_2 = 3 \text{ km} \]

- For **Segment 3** (retardation):
\[ s_3 = v t_3 - \frac{1}{2} a t_3^2 = 2 \text{ km} \]
Here, \( a \) is the deceleration.

4. **Total time**:
The total time for the journey is given as 10 minutes, which we convert to hours:
\[ t_{\text{total}} = 10 \text{ minutes} = \frac{10}{60} \text{ hours} = \frac{1}{6} \text{ hours} \]
Therefore:
\[ t_1 + t_2 + t_3 = \frac{1}{6} \text{ hours} \]

5. **Relate distances and times**:
From the equations for the distances:
- From Segment 1, we can express \( t_1 \) in terms of \( v \):
\[ v = a t_1 \quad \text{(from } v = u + at \text{, where } u = 0\text{)} \]
Using \( s_1 = 1 \text{ km} \):
\[ 1 = \frac{1}{2} a t_1^2 \implies a = \frac{2}{t_1^2} \]
Substitute \( a \) into the equation for \( v \):
\[ v = \frac{2}{t_1^2} t_1 = \frac{2}{t_1} \]

6. **Substituting into total time**:
Now we can express \( t_2 \) and \( t_3 \) in terms of \( v \):
- From Segment 2:
\[ t_2 = \frac{3}{v} \]
- From Segment 3:
Using \( s_3 = 2 \text{ km} \):
\[ 2 = v t_3 - \frac{1}{2} a t_3^2 \]
Substitute \( a \):
\[ 2 = v t_3 - \frac{1}{2} \left(\frac{2}{t_1^2}\right) t_3^2 \]
Rearranging gives:
\[ 2 = v t_3 - \frac{t_3^2}{t_1^2} \]

7. **Combine equations**:
Substitute \( t_1, t_2, t_3 \) into the total time equation:
\[ t_1 + \frac{3}{v} + t_3 = \frac{1}{6} \]

8. **Solve for \( v \)**:
After substituting and simplifying, we find that:
\[ v = 54 \text{ km/h} \]

### Final Answer:
The maximum velocity of the train is **54 km/h**.