A body is moving with uniform velocity of `8 ms^-1`. When the body just crossed another body, the second one starts and moves with uniform acceleration of `4 ms^-2`. The time after which two bodies meet will be :

Options

A. 2 s
B. 4 s
C. 6 s
D. 8 s

Explanation

To solve the problem step by step, we need to analyze the motion of both bodies and find the time at which they meet.

### Step 1: Understand the motion of both bodies
- **Body A** is moving with a uniform velocity of \(8 \, \text{m/s}\).
- **Body B** starts from rest (initial velocity \(u = 0\)) and accelerates uniformly with an acceleration of \(4 \, \text{m/s}^2\).

### Step 2: Write the equations of motion
- The distance covered by Body A after time \(t\) is given by:
\[ d_A = v_A \cdot t = 8t \]

- The distance covered by Body B after time \(t\) is given by the equation of motion for uniformly accelerated motion:
\[ d_B = u_B \cdot t + \frac{1}{2} a_B t^2 = 0 \cdot t + \frac{1}{2} \cdot 4 \cdot t^2 = 2t^2 \]

### Step 3: Set the distances equal
Since both bodies meet at the same distance after time \(t\), we can set the distances equal to each other:
\[ d_A = d_B \]
Substituting the expressions we derived:
\[ 8t = 2t^2 \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ 2t^2 - 8t = 0 \]

### Step 5: Factor the equation
Factoring out \(2t\) from the equation:
\[ 2t(t - 4) = 0 \]

### Step 6: Solve for \(t\)
Setting each factor to zero gives us:
1. \(2t = 0 \Rightarrow t = 0\) (This is the initial time when they just cross each other)
2. \(t - 4 = 0 \Rightarrow t = 4\) seconds

### Conclusion
The time after which the two bodies meet is \(t = 4\) seconds.

### Final Answer
The time after which the two bodies meet is **4 seconds**.

---