A train is moving at a constant speed `V` when its driverobserves another train in front of him on the same track and voing in the same direction with constant speed `v`. If the distance berween the trains is `x`. Trains is `x then what should be the minimum retardation of the train so as to avoed collision?.

Options

A. `(V_+v)^(2)/(x)`
B. `(V_+v)^(2)/(x)`
C. c. `(V_+v)^(2)/(2x)`
D. `(V_+v)^(2)/(2x)`

Explanation

To solve the problem of determining the minimum retardation required for the first train to avoid a collision with the second train, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let the speed of the first train be \( V \).
- Let the speed of the second train be \( v \).
- Let the distance between the two trains be \( x \).

2. **Relative Motion Concept**:
- Since both trains are moving in the same direction, we can analyze the situation using relative motion.
- The relative speed of the first train with respect to the second train is given by:
\[ u_{\text{relative}} = V - v \]

3. **Final Relative Velocity**:
- To avoid collision, the final relative velocity when the first train comes to a stop should be zero:
\[ v_{\text{relative, final}} = 0 \]

4. **Using the Equation of Motion**:
- We can use the third equation of motion, which states:
\[ v^2 = u^2 + 2as \]
- In our case, substituting for the relative motion:
\[ 0^2 = (V - v)^2 + 2(-a)x \]
- Here, \( a \) is the retardation (deceleration), and \( s \) is the distance \( x \).

5. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ 0 = (V - v)^2 - 2ax \]
- This can be rewritten as:
\[ 2ax = (V - v)^2 \]

6. **Solving for Retardation**:
- Now, solving for \( a \) (retardation):
\[ a = \frac{(V - v)^2}{2x} \]

### Final Answer:
The minimum retardation required for the first train to avoid a collision with the second train is given by:
\[ a = \frac{(V - v)^2}{2x} \]