metro train starts from rest and in five seconds achieves a speed 108 km/h. After that it moves with constant velocity and comes to rest after travelling 45m with uniform retardation. If total distance travelled is 395 m, find total time of travelling.

Options

A. `12.2 s `
B. `15.3s`
C. 9s
D. `17.2s`

Answer

To solve the problem, we will break it down into several steps:

### Step 1: Convert the speed from km/h to m/s
The metro train achieves a speed of 108 km/h. We need to convert this speed into meters per second (m/s).

\[ \text{Speed in m/s} = \frac{108 \times 1000}{3600} = 30 \text{ m/s} \]

### Step 2: Calculate the acceleration during the first 5 seconds
The train starts from rest (initial velocity \( u = 0 \)) and reaches a speed of 30 m/s in 5 seconds. We can use the formula:

\[ v = u + at \]

Rearranging for acceleration \( a \):

\[ a = \frac{v - u}{t} = \frac{30 - 0}{5} = 6 \text{ m/s}^2 \]

### Step 3: Calculate the distance traveled during acceleration
Using the formula for distance traveled under constant acceleration:

\[ d_1 = ut + \frac{1}{2} a t^2 \]

Substituting the known values:

\[ d_1 = 0 \cdot 5 + \frac{1}{2} \cdot 6 \cdot (5^2) = \frac{1}{2} \cdot 6 \cdot 25 = 75 \text{ m} \]

### Step 4: Determine the distance traveled during retardation
The train comes to rest after traveling 45 m with uniform retardation. The initial velocity for this phase is \( u' = 30 \text{ m/s} \) and the final velocity \( v = 0 \text{ m/s} \).

Using the equation of motion:

\[ v^2 = u'^2 + 2a d \]

Rearranging for \( a \):

\[ 0 = (30)^2 + 2a(45) \]
\[ 0 = 900 + 90a \implies a = -10 \text{ m/s}^2 \]

### Step 5: Calculate the time taken to come to rest
Using the formula again:

\[ v = u' + at \]

Rearranging for time \( t \):

\[ 0 = 30 - 10t \implies t = \frac{30}{10} = 3 \text{ seconds} \]

### Step 6: Calculate the remaining distance traveled at constant speed
The total distance traveled is 395 m. The distances already accounted for are:

- Distance during acceleration: \( d_1 = 75 \text{ m} \)
- Distance during retardation: \( d_2 = 45 \text{ m} \)

Calculating the remaining distance:

\[ d_{\text{remaining}} = 395 - (75 + 45) = 275 \text{ m} \]

### Step 7: Calculate the time taken to travel the remaining distance
The train travels this remaining distance at a constant speed of 30 m/s. The time taken \( t_2 \) is given by:

\[ t_2 = \frac{d_{\text{remaining}}}{\text{speed}} = \frac{275}{30} \approx 9.17 \text{ seconds} \]

### Step 8: Calculate the total time of travel
The total time \( T \) is the sum of the time during acceleration, the time during constant speed, and the time during retardation:

\[ T = t_{\text{acceleration}} + t_{\text{constant speed}} + t_{\text{retardation}} = 5 + 9.17 + 3 = 17.17 \text{ seconds} \]

### Final Answer
The total time of traveling is approximately **17.17 seconds**.

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