If a train engine crosses a signal with a velocity u & has constant acceleration and the last bogey of train crosses the signal with velocity v, then middle point of train crosses the signal with velocity ?

Options

A. `(v+u)/2`
B. `sqrt((v^2+u^2)/2)`
C. `sqrt((v^2-u^2)/2)`
D. `(v-u)/2`

Answer

To solve the problem, we need to analyze the motion of the train as it crosses the signal. We will use the equations of motion under constant acceleration.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The train engine crosses a signal with an initial velocity \( u \).
- The last bogie of the train crosses the signal with a final velocity \( v \).
- The train has a constant acceleration \( a \).
- We need to find the velocity of the middle point of the train when it crosses the signal.

2. **Define the Length of the Train**:
- Let the length of the train be \( L \).

3. **Using the Equation of Motion**:
- For the last bogie, which travels the distance \( L \) while accelerating from \( u \) to \( v \), we can use the equation:
\[ v^2 = u^2 + 2aL \]
- Rearranging gives:
\[ aL = \frac{v^2 - u^2}{2} \]

4. **Finding the Velocity of the Middle Point**:
- The middle point of the train will have traveled a distance of \( \frac{L}{2} \) when it crosses the signal.
- Let the velocity of the middle point when it crosses the signal be \( v' \).
- Using the same equation of motion for the middle point:
\[ v'^2 = u^2 + 2a\left(\frac{L}{2}\right) \]
- Simplifying gives:
\[ v'^2 = u^2 + aL \]

5. **Substituting for \( aL \)**:
- From the earlier step, we know that \( aL = \frac{v^2 - u^2}{2} \).
- Substituting this into the equation for \( v' \):
\[ v'^2 = u^2 + \frac{v^2 - u^2}{2} \]
- This simplifies to:
\[ v'^2 = \frac{2u^2 + v^2 - u^2}{2} = \frac{u^2 + v^2}{2} \]

6. **Finding \( v' \)**:
- Taking the square root gives:
\[ v' = \sqrt{\frac{u^2 + v^2}{2}} \]

### Final Answer:
The velocity of the middle point of the train when it crosses the signal is:
\[ v' = \sqrt{\frac{u^2 + v^2}{2}} \]