A train is moving with uniform acceleration. The two ends of the train pass through a point on the track with velocity `v_(1)` and `v_(2)`. With what velocity the middle point of the train would pass through the same point ?

Options

A. `[(v_(1)^(2)+v_(2)^(2))/(2)]^(1/2)`
B. `((v_(1)^(2)-v_(2)^(2)))/(2)`
C. `(v_(1)+v_(2))/(2)`
D. `(v_(1)-v_(2))/(2)`

Answer

To solve the problem of finding the velocity with which the middle point of the train passes through the same point, we will use the equations of motion under uniform acceleration.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a train moving with uniform acceleration.
- The front end of the train passes a point with velocity \( v_1 \).
- The back end of the train passes the same point with velocity \( v_2 \).
- We need to find the velocity \( v_m \) of the middle point of the train as it passes through the same point.

2. **Using the Third Equation of Motion**:
- The third equation of motion states:
\[ v^2 = u^2 + 2as \]
- Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the distance.

3. **Applying the Equation for the Back End of the Train**:
- For the back end of the train (velocity \( v_2 \)):
\[ v_2^2 = v_1^2 + 2aL \]
- This equation relates the velocities and acceleration over the length \( L \) of the train.

4. **Applying the Equation for the Middle Point of the Train**:
- For the middle point of the train (velocity \( v_m \)):
\[ v_m^2 = v_1^2 + 2a\left(\frac{L}{2}\right) \]
- This equation accounts for the distance covered by the middle point, which is half the length of the train.

5. **Rearranging the Equations**:
- From the first equation:
\[ v_2^2 = v_1^2 + 2aL \quad \text{(Equation 1)} \]
- From the second equation:
\[ v_m^2 = v_1^2 + aL \quad \text{(Equation 2)} \]

6. **Subtracting Equation 1 from Equation 2**:
- We can express \( v_m^2 \) in terms of \( v_1 \) and \( v_2 \):
\[ v_m^2 = \frac{v_1^2 + v_2^2}{2} \]

7. **Final Expression for the Velocity of the Middle Point**:
- Taking the square root gives:
\[ v_m = \sqrt{\frac{v_1^2 + v_2^2}{2}} \]

### Final Answer:
The velocity with which the middle point of the train passes through the same point is:
\[ v_m = \sqrt{\frac{v_1^2 + v_2^2}{2}} \]