A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The distance covered by the boggy and distance covered by the train in the same time has relation

Options

A. Both will be equal
B. First will be half of second
C. First will be 1/4 of second
D. No definite ratio

Explanation

To solve the problem, we will analyze the motion of the bogey after it detaches from the train and compare the distances covered by both the bogey and the train in the same time interval.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( u \) be the initial velocity of the bogey at the moment it detaches from the train.
- Let \( a_b \) be the retardation (negative acceleration) of the bogey after detachment.
- Let \( S_b \) be the distance covered by the bogey before it stops.
- Let \( S_t \) be the distance covered by the train in the same time \( t \).

2. **Equation of Motion for the Bogey**:
- The final velocity \( V_b \) of the bogey when it stops is \( 0 \).
- Using the equation of motion:
\[ V_b^2 = u^2 + 2 a_b S_b \]
- Substituting \( V_b = 0 \):
\[ 0 = u^2 + 2 a_b S_b \]
- Rearranging gives:
\[ S_b = -\frac{u^2}{2 a_b} \]
- Since \( a_b \) is negative (retardation), we can write:
\[ S_b = \frac{u^2}{2 |a_b|} \]

3. **Time Taken for the Bogey to Stop**:
- Using the equation \( V = u + a t \):
\[ 0 = u + a_b t \]
- Rearranging gives:
\[ t = -\frac{u}{a_b} \]
- Since \( a_b \) is negative, we can express \( t \) as:
\[ t = \frac{u}{|a_b|} \]

4. **Distance Covered by the Train**:
- The train continues to move with the same initial velocity \( u \).
- The distance covered by the train in time \( t \) is:
\[ S_t = u t = u \left( \frac{u}{|a_b|} \right) = \frac{u^2}{|a_b|} \]

5. **Finding the Ratio of Distances**:
- Now we have:
\[ S_b = \frac{u^2}{2 |a_b|} \quad \text{and} \quad S_t = \frac{u^2}{|a_b|} \]
- The ratio of the distance covered by the bogey to the distance covered by the train is:
\[ \frac{S_b}{S_t} = \frac{\frac{u^2}{2 |a_b|}}{\frac{u^2}{|a_b|}} = \frac{1}{2} \]

6. **Conclusion**:
- Therefore, the distance covered by the bogey is half of the distance covered by the train in the same time.
- The correct option is \( \frac{1}{2} \) or option B.