A train normally travels at a uniform speed of `72 km//h` on a long stretch of straight level track. On a particular day, the train was forced to make a `2.0` minute stop at a station along this track. If the train decelerates at a uniform rate of `1.0 m//s^2` and accelerates at a rate of `0.50 m//s^2`, how much time is lost in stopping at the station ?

Options

A. 2 min
B. 2 min 30 s
C. 30 s
D. 2 min 20 s

Explanation

To solve the problem, we need to determine the total time lost by the train due to stopping and then accelerating back to its original speed. The train's normal speed is given, and we need to account for the time taken to decelerate to a stop, the time spent at the station, and the time taken to accelerate back to its normal speed.

### Step-by-Step Solution:

1. **Convert the speed from km/h to m/s**:
- The normal speed of the train is \( 72 \, \text{km/h} \).
- To convert this to meters per second (m/s), we use the conversion factor \( \frac{1000 \, \text{m}}{3600 \, \text{s}} \):
\[ 72 \, \text{km/h} = 72 \times \frac{1000}{3600} = 20 \, \text{m/s} \]

2. **Calculate the time taken to decelerate to a stop**:
- The initial velocity \( u = 20 \, \text{m/s} \) and the final velocity \( v = 0 \, \text{m/s} \).
- The deceleration \( a = -1.0 \, \text{m/s}^2 \).
- Using the formula \( v = u + at \), we can find the time \( t \):
\[ 0 = 20 + (-1)t \implies t = \frac{20}{1} = 20 \, \text{s} \]

3. **Calculate the distance covered during deceleration**:
- We can use the formula \( s = ut + \frac{1}{2}at^2 \):
\[ s = 20 \times 20 + \frac{1}{2} \times (-1) \times (20)^2 = 400 - 200 = 200 \, \text{m} \]

4. **Calculate the time taken to cover the same distance at normal speed**:
- The time taken to cover \( 200 \, \text{m} \) at \( 20 \, \text{m/s} \) is:
\[ t_{\text{normal}} = \frac{200}{20} = 10 \, \text{s} \]

5. **Calculate the time lost during deceleration**:
- Time lost during deceleration:
\[ \text{Time lost (deceleration)} = t - t_{\text{normal}} = 20 - 10 = 10 \, \text{s} \]

6. **Calculate the time taken to accelerate back to normal speed**:
- The train starts from rest, so \( u = 0 \, \text{m/s} \) and \( v = 20 \, \text{m/s} \).
- The acceleration \( a = 0.5 \, \text{m/s}^2 \).
- Using \( v = u + at \):
\[ 20 = 0 + 0.5t \implies t = \frac{20}{0.5} = 40 \, \text{s} \]

7. **Calculate the distance covered during acceleration**:
- Using \( s = ut + \frac{1}{2}at^2 \):
\[ s = 0 \times 40 + \frac{1}{2} \times 0.5 \times (40)^2 = 0 + 0.25 \times 1600 = 400 \, \text{m} \]

8. **Calculate the time taken to cover this distance at normal speed**:
- The time taken to cover \( 400 \, \text{m} \) at \( 20 \, \text{m/s} \):
\[ t_{\text{normal}} = \frac{400}{20} = 20 \, \text{s} \]

9. **Calculate the time lost during acceleration**:
- Time lost during acceleration:
\[ \text{Time lost (acceleration)} = t - t_{\text{normal}} = 40 - 20 = 20 \, \text{s} \]

10. **Total time lost**:
- The total time lost is the sum of the time lost during deceleration, the stop at the station, and the time lost during acceleration:
\[ \text{Total time lost} = \text{Time lost (deceleration)} + \text{Stop time} + \text{Time lost (acceleration)} \]
- The train stopped for \( 2.0 \, \text{minutes} = 120 \, \text{s} \):
\[ \text{Total time lost} = 10 + 120 + 20 = 150 \, \text{s} = 2 \, \text{minutes} \, 30 \, \text{seconds} \]

### Final Answer:
The total time lost in stopping at the station is **2 minutes and 30 seconds**.