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. `dlt2.5`m
B. `dlt125`m
C. `dgt25`m
D. `dgt125`m

Answer

To solve the problem, we need to analyze the motion of both cars A and B, particularly focusing on the conditions under which they will not collide.

### Step-by-Step Solution:

1. **Identify the initial velocities:**
- Let the velocity of car A, \( v_A = 30 \, \text{m/s} \)
- Let the velocity of car B, \( v_B = 20 \, \text{m/s} \)

2. **Determine the relative velocity:**
- The relative velocity of car A with respect to car B is:
\[ v_{AB} = v_A - v_B = 30 \, \text{m/s} - 20 \, \text{m/s} = 10 \, \text{m/s} \]

3. **Identify the retardation of car A:**
- The retardation (deceleration) of car A is given as:
\[ a_A = -2 \, \text{m/s}^2 \]

4. **Set up the equation for no collision:**
- For car A to not collide with car B, the distance \( d \) must be such that the time taken by car A to stop is greater than the time taken by car B to cover the distance \( d \).
- We can use the equation of motion:
\[ v^2 = u^2 + 2as \]
- Here, for car A, the final velocity \( v = 0 \) (when it stops), initial velocity \( u = 30 \, \text{m/s} \), and acceleration \( a = -2 \, \text{m/s}^2 \). The distance \( s \) will be \( d \).

5. **Apply the equation:**
- Rearranging the equation gives:
\[ 0 = (30)^2 + 2(-2)d \]
\[ 0 = 900 - 4d \]
\[ 4d = 900 \]
\[ d = \frac{900}{4} = 225 \, \text{m} \]

6. **Conclusion:**
- The distance \( d \) must be at least \( 225 \, \text{m} \) for car A to not collide with car B.

### Final Answer:
The distance \( d \) behind car B at which car A must be to avoid collision is \( 225 \, \text{m} \).

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