A train has to negotiate a curve of 40 m radius. What is the super elevation of the outer rail required for a speed of `54 kmh^(-1)`. The distance between the rails is one metre ?

Explanation

To find the super elevation of the outer rail required for a train negotiating a curve, we can use the formula:

\[ h = \frac{V^2 \cdot L}{g \cdot R} \]

Where:
- \( h \) = super elevation (height of the outer rail above the inner rail)
- \( V \) = speed of the train in meters per second (m/s)
- \( L \) = distance between the rails (in meters)
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, m/s^2 \))
- \( R \) = radius of the curve (in meters)

### Step-by-Step Solution:

1. **Convert Speed from km/h to m/s**:
The speed of the train is given as \( 54 \, km/h \). To convert this to meters per second:
\[ V = 54 \, km/h \times \frac{1000 \, m}{1 \, km} \times \frac{1 \, h}{3600 \, s} = \frac{54000}{3600} = 15 \, m/s \]

2. **Identify Given Values**:
- Radius \( R = 40 \, m \)
- Distance between rails \( L = 1 \, m \)
- Acceleration due to gravity \( g = 9.8 \, m/s^2 \)

3. **Substitute Values into the Formula**:
Now, substitute the values into the super elevation formula:
\[ h = \frac{V^2 \cdot L}{g \cdot R} \]
\[ h = \frac{(15 \, m/s)^2 \cdot 1 \, m}{9.8 \, m/s^2 \cdot 40 \, m} \]

4. **Calculate \( V^2 \)**:
\[ V^2 = 15^2 = 225 \, m^2/s^2 \]

5. **Calculate the Denominator**:
\[ g \cdot R = 9.8 \, m/s^2 \cdot 40 \, m = 392 \, m^2/s^2 \]

6. **Complete the Calculation**:
Now, substitute these values back into the equation:
\[ h = \frac{225 \, m^2/s^2 \cdot 1 \, m}{392 \, m^2/s^2} = \frac{225}{392} \]
\[ h \approx 0.574 \, m \]

### Final Answer:
The super elevation of the outer rail required for the train to negotiate the curve is approximately \( 0.574 \, m \).