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

Explanation

To solve the problem, we need to analyze the motion of both bodies involved:

1. **Identify the motion of the first body**:
- The first body is moving with a uniform velocity of \(8 \, \text{m/s}\).
- The distance traveled by the first body in time \(t\) can be expressed as:
\[ s_1 = u_1 \cdot t = 8t \]

2. **Identify the motion of the second body**:
- The second body starts from rest (initial velocity \(u_2 = 0\)) and moves with a uniform acceleration of \(4 \, \text{m/s}^2\).
- The distance traveled by the second body in time \(t\) can be expressed using the equation of motion:
\[ s_2 = u_2 \cdot t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \cdot 4 \cdot t^2 = 2t^2 \]

3. **Set the distances equal**:
- Since both bodies meet at the same point after time \(t\), we can set the distances equal to each other:
\[ s_1 = s_2 \]
\[ 8t = 2t^2 \]

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

5. **Factor the equation**:
- Factoring out \(2t\):
\[ 2t(t - 4) = 0 \]

6. **Solve for \(t\)**:
- This gives us two solutions:
\[ 2t = 0 \quad \text{or} \quad t - 4 = 0 \]
- The first solution \(t = 0\) corresponds to the initial moment when the first body just crosses the second body.
- The second solution gives us:
\[ t = 4 \, \text{seconds} \]

Thus, the time after which the two bodies meet is \(t = 4 \, \text{seconds}\).