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 will analyze the motion of both bodies and set up equations based on their respective motions.

### 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 and accelerates uniformly with an acceleration of \(4 \, \text{m/s}^2\) after Body A crosses it.

### Step 2: Define the Displacement Equations
- The displacement of Body A after time \(t\) can be expressed as:
\[ s_A = v \cdot t = 8t \]
- The displacement of Body B after time \(t\) can be expressed using the equation of motion for uniformly accelerated motion:
\[ s_B = ut + \frac{1}{2} a t^2 \]
Since Body B starts from rest, \(u = 0\), so:
\[ s_B = \frac{1}{2} a t^2 = \frac{1}{2} \cdot 4 \cdot t^2 = 2t^2 \]

### Step 3: Set the Displacements Equal
Since both bodies meet at the same point, their displacements will be equal:
\[ s_A = s_B \]
Thus, we have:
\[ 8t = 2t^2 \]

### Step 4: Rearrange the Equation
Rearranging the equation gives us:
\[ 2t^2 - 8t = 0 \]
Factoring out \(2t\):
\[ 2t(t - 4) = 0 \]

### Step 5: Solve for \(t\)
Setting each factor to zero gives:
1. \(2t = 0 \Rightarrow t = 0\) (This is the initial time when both bodies are at the same position)
2. \(t - 4 = 0 \Rightarrow t = 4\)

Thus, the time after which the two bodies meet is:
\[ t = 4 \, \text{seconds} \]

### Conclusion
The time after which the two bodies meet is \(4 \, \text{seconds}\).

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