If a train travelling at 72 km/h is to be brought to rest in a distance of 200m, then its retardation should be:-

Options

A. `20 m//s^(2)`
B. `2m//s^(2)`
C. `10 m//s^(2)`
D. `1 m//s^(2)`

Answer

To solve the problem of determining the retardation required to bring a train traveling at 72 km/h to rest over a distance of 200 meters, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Initial velocity (u) = 72 km/h
- Final velocity (v) = 0 m/s (since the train is brought to rest)
- Distance (s) = 200 m

2. **Convert Initial Velocity to m/s**:
- To convert km/h to m/s, we use the conversion factor \( \frac{5}{18} \).
- \( u = 72 \times \frac{5}{18} = 20 \) m/s.

3. **Use the Third Equation of Motion**:
- The third equation of motion is given by:
\[ v^2 = u^2 + 2as \]
- Here, \( a \) is the acceleration (which will be negative for retardation).

4. **Substitute the Known Values**:
- Substitute \( v = 0 \), \( u = 20 \) m/s, and \( s = 200 \) m into the equation:
\[ 0 = (20)^2 + 2a(200) \]

5. **Simplify the Equation**:
- This simplifies to:
\[ 0 = 400 + 400a \]

6. **Solve for Acceleration (a)**:
- Rearranging gives:
\[ 400a = -400 \]
- Therefore:
\[ a = -1 \text{ m/s}^2 \]

7. **Interpret the Result**:
- The negative sign indicates that this is a retardation (deceleration).
- Thus, the required retardation is \( 1 \text{ m/s}^2 \).

### Final Answer:
The retardation required to bring the train to rest is \( 1 \text{ m/s}^2 \).

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