A body starts from rest with a uniform acceleration of `8 m//"sec"^(2)`. Then the time it will take in traversing the second metre of its journey is

Options

A. `(1)/(2)` second
B. `(sqrt(2)-1)/(2)` second
C. `sqrt(2)` seconds
D. `(sqrt(2)+1)/(2)` second

Answer

To solve the problem, we need to find the time it takes for a body to traverse the second meter of its journey when it starts from rest with a uniform acceleration of \(8 \, \text{m/s}^2\).

### Step-by-Step Solution:

1. **Understanding the motion**: The body starts from rest, which means its initial velocity \(u = 0\). The uniform acceleration \(a = 8 \, \text{m/s}^2\).

2. **Using the equation of motion**: We can use the second equation of motion to find the distance covered in time \(t\):
\[ s = ut + \frac{1}{2} a t^2 \]
Since \(u = 0\), this simplifies to:
\[ s = \frac{1}{2} a t^2 \]
Substituting \(a = 8 \, \text{m/s}^2\):
\[ s = \frac{1}{2} \cdot 8 \cdot t^2 = 4t^2 \]

3. **Finding the time to cover the first meter**: To find the time taken to cover the first meter, we set \(s = 1 \, \text{m}\):
\[ 1 = 4t_1^2 \]
Solving for \(t_1\):
\[ t_1^2 = \frac{1}{4} \implies t_1 = \frac{1}{2} \, \text{s} \]

4. **Finding the time to cover the second meter**: Now, we need to find the total time \(t_2\) to cover 2 meters:
\[ 2 = 4t_2^2 \]
Solving for \(t_2\):
\[ t_2^2 = \frac{2}{4} = \frac{1}{2} \implies t_2 = \frac{1}{\sqrt{2}} \approx 0.707 \, \text{s} \]

5. **Calculating the time to traverse the second meter**: The time taken to traverse the second meter is given by:
\[ t_{\text{second meter}} = t_2 - t_1 \]
Substituting the values:
\[ t_{\text{second meter}} = \frac{1}{\sqrt{2}} - \frac{1}{2} \]

6. **Finding a common denominator**: To simplify:
\[ t_{\text{second meter}} = \frac{2 - \sqrt{2}}{2\sqrt{2}} \approx 0.207 \, \text{s} \]

### Final Answer:
The time it will take to traverse the second meter of its journey is approximately \(0.207 \, \text{s}\).

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