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 finding the retardation of a train that is brought to rest from a speed of 72 km/h over a distance of 200 m, we can follow these steps:

### Step-by-Step Solution

1. **Convert the Speed from km/h to m/s**:
The initial speed \( u \) is given as 72 km/h. We need to convert this to meters per second (m/s).
\[ u = 72 \, \text{km/h} \times \frac{5}{18} = 20 \, \text{m/s} \]

2. **Identify the Final Velocity**:
Since the train is brought to rest, the final velocity \( v \) is:
\[ v = 0 \, \text{m/s} \]

3. **Use the Kinematic Equation**:
We can use the kinematic equation that relates initial velocity, final velocity, acceleration (or retardation), and distance:
\[ v^2 = u^2 + 2as \]
Here, \( s \) is the distance (200 m), and \( a \) is the retardation we want to find.

4. **Substituting the Known Values**:
Substitute \( v = 0 \, \text{m/s} \), \( u = 20 \, \text{m/s} \), and \( s = 200 \, \text{m} \) into the equation:
\[ 0 = (20)^2 + 2a(200) \]
This simplifies to:
\[ 0 = 400 + 400a \]

5. **Solving for Acceleration (Retardation)**:
Rearranging the equation gives:
\[ 400a = -400 \]
Dividing both sides by 400:
\[ a = -1 \, \text{m/s}^2 \]

6. **Interpret the Result**:
The negative sign indicates that this is a retardation (deceleration). Therefore, the magnitude of the retardation is:
\[ \text{Retardation} = 1 \, \text{m/s}^2 \]

### Final Answer:
The retardation required to bring the train to rest in a distance of 200 m is \( 1 \, \text{m/s}^2 \).

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