An engine of a train moving with uniform acceleration passes an electric pole with velocity 6 m/s and the last compartment with velocity 8 m/s. The middle point of the train passes the same pole with a velocity of :-

Options

A. `sqrt(14)` m/s
B. 5 m/s
C. 50 m/s
D. 7 m/s

Answer

To solve the problem, we need to find the velocity of the middle point of the train as it passes the electric pole. We know the initial and final velocities of the train at the pole and can use kinematic equations to find the required velocity.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Initial velocity of the engine (U) = 6 m/s (when it passes the pole)
- Final velocity of the last compartment (V) = 8 m/s (when it passes the pole)
- Let the length of the train be L.

2. **Use Kinematic Equation:**
We can use the kinematic equation:
\[ V^2 = U^2 + 2aS \]
where:
- \( V \) = final velocity (8 m/s)
- \( U \) = initial velocity (6 m/s)
- \( a \) = acceleration
- \( S \) = displacement (length of the train, L)

Plugging in the values:
\[ 8^2 = 6^2 + 2aL \]
\[ 64 = 36 + 2aL \]
\[ 64 - 36 = 2aL \]
\[ 28 = 2aL \]
\[ a = \frac{14}{L} \]

3. **Calculate Velocity at the Middle Point:**
The middle point of the train will pass the pole after traveling a distance of \( \frac{L}{2} \). We need to find the final velocity (V') at this point using the same kinematic equation:
\[ V'^2 = U^2 + 2aS \]
Here, \( U = 6 \) m/s, \( S = \frac{L}{2} \), and \( a = \frac{14}{L} \).

Plugging in the values:
\[ V'^2 = 6^2 + 2 \left(\frac{14}{L}\right) \left(\frac{L}{2}\right) \]
\[ V'^2 = 36 + 2 \cdot 14 \cdot \frac{1}{2} \]
\[ V'^2 = 36 + 14 \]
\[ V'^2 = 50 \]
\[ V' = \sqrt{50} \approx 7.07 \text{ m/s} \]

4. **Conclusion:**
The velocity of the middle point of the train as it passes the electric pole is approximately 7.07 m/s.