A body is moving with uniform velocity of `8 m s^(-1)`. When the body just crosses another body, the second one starts and moves with uniform acceleration of `4 m s^(-2)`.
The distance comered by the second body when they meet is .

Options

A. `8 m`
B. `16 m`
C. `24 m`
D. `32 m`

Explanation

To solve the problem, we need to find the distance covered by the second body when it meets the first body. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify the motion of the first body:**
The first body is moving with a uniform velocity of \(8 \, \text{m/s}\). This means it covers a distance \(d_1\) given by the formula:
\[ d_1 = v \cdot t \]
where \(v = 8 \, \text{m/s}\) and \(t\) is the time in seconds.

2. **Identify the motion of the second body:**
The second body starts moving with an initial velocity \(u = 0\) and accelerates uniformly with an acceleration \(a = 4 \, \text{m/s}^2\). The distance \(d_2\) covered by the second body is given by the formula:
\[ d_2 = ut + \frac{1}{2} a t^2 \]
Substituting the values, we have:
\[ d_2 = 0 \cdot t + \frac{1}{2} \cdot 4 \cdot t^2 = 2t^2 \]

3. **Set the distances equal:**
Since both bodies meet at the same point, the distances covered by both bodies will be equal:
\[ d_1 = d_2 \]
Therefore, we can write:
\[ 8t = 2t^2 \]

4. **Rearranging the equation:**
Rearranging gives us:
\[ 2t^2 - 8t = 0 \]
Factoring out \(2t\):
\[ 2t(t - 4) = 0 \]
This gives us two solutions: \(t = 0\) (when they just start) and \(t = 4\) seconds.

5. **Calculate the distance covered by the second body:**
Now, we can find the distance covered by the second body when \(t = 4\) seconds:
\[ d_2 = 2t^2 = 2(4^2) = 2 \cdot 16 = 32 \, \text{meters} \]

### Final Answer:
The distance covered by the second body when they meet is \(32 \, \text{meters}\).

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