Two cars A and B are moving in same direction with velocities 30 m/s and 20 m/s. When car A is at a distance d behind the car B, the driver of the car A applies brakes producing uniform retardation of `2m//s^(2)`. There will be no collision when :-

Options

A. `d lt 2.5m`
B. `d gt 125m`
C. `d gt 25m`
D. `d lt 125m`

Answer

To solve the problem, we need to analyze the motion of both cars A and B. Car A is initially moving faster than car B but begins to decelerate. We want to determine the minimum distance \( d \) that car A must be behind car B to avoid a collision.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Velocity of car A, \( u_A = 30 \, \text{m/s} \)
- Velocity of car B, \( u_B = 20 \, \text{m/s} \)
- Retardation of car A, \( a = -2 \, \text{m/s}^2 \) (negative because it is deceleration)

2. **Determine the final velocity of car A:**
- Car A will decelerate until its velocity equals that of car B. Therefore, we set the final velocity of car A, \( v_A = u_B = 20 \, \text{m/s} \).

3. **Use the kinematic equation to find the distance \( S \) that car A travels while decelerating:**
\[ v^2 = u^2 + 2aS \]
Substituting the known values:
\[ (20)^2 = (30)^2 + 2(-2)S \]
This simplifies to:
\[ 400 = 900 - 4S \]

4. **Rearranging the equation to solve for \( S \):**
\[ 4S = 900 - 400 \]
\[ 4S = 500 \]
\[ S = \frac{500}{4} = 125 \, \text{m} \]

5. **Conclusion:**
- For car A to avoid a collision with car B, the distance \( d \) between them must be greater than \( S \). Therefore, we conclude that:
\[ d > 125 \, \text{m} \]

### Final Answer:
The correct condition for there to be no collision is that the distance \( d \) must be greater than 125 m.