A body starts moving from rest with constant acceleration covers displacement `S_(1)` in first `(p-1)` seconds and `S_(2)` in first "p" seconds.The displacement `S_(1)+S_(2)` will be made in time :

Options

A. `(2p-1)s`
B. `sqrt((2p^(2)-2p+1))s`
C. `(2p+1)s`
D. `(2p^(2)-2p+1)s`

Answer

To solve the problem, we need to find the total time taken to cover the displacement \( S_1 + S_2 \) when a body starts from rest with constant acceleration.

### Step-by-Step Solution:

1. **Understanding the Displacement Equations**:
- The body starts from rest, so the initial velocity \( u = 0 \).
- The displacement covered in time \( t \) with constant acceleration \( a \) is given by:
\[ S = ut + \frac{1}{2} a t^2 \]
- Since \( u = 0 \), the equation simplifies to:
\[ S = \frac{1}{2} a t^2 \]

2. **Calculating \( S_1 \)**:
- For the first \( (p-1) \) seconds, the displacement \( S_1 \) is:
\[ S_1 = \frac{1}{2} a (p-1)^2 \]

3. **Calculating \( S_2 \)**:
- For the first \( p \) seconds, the displacement \( S_2 \) is:
\[ S_2 = \frac{1}{2} a p^2 \]

4. **Finding Total Displacement \( S_1 + S_2 \)**:
- Now we add \( S_1 \) and \( S_2 \):
\[ S_1 + S_2 = \frac{1}{2} a (p-1)^2 + \frac{1}{2} a p^2 \]
- Factor out \( \frac{1}{2} a \):
\[ S_1 + S_2 = \frac{1}{2} a \left( (p-1)^2 + p^2 \right) \]

5. **Expanding the Expression**:
- Expanding \( (p-1)^2 \):
\[ (p-1)^2 = p^2 - 2p + 1 \]
- Therefore:
\[ S_1 + S_2 = \frac{1}{2} a \left( p^2 - 2p + 1 + p^2 \right) = \frac{1}{2} a (2p^2 - 2p + 1) \]

6. **Finding Time \( t \) for \( S_1 + S_2 \)**:
- We need to find the time \( t \) such that:
\[ S_1 + S_2 = \frac{1}{2} a t^2 \]
- Setting the two equations equal:
\[ \frac{1}{2} a (2p^2 - 2p + 1) = \frac{1}{2} a t^2 \]
- Canceling \( \frac{1}{2} a \) from both sides (assuming \( a \neq 0 \)):
\[ 2p^2 - 2p + 1 = t^2 \]

7. **Final Result**:
- Thus, the total time \( t \) taken to cover the displacement \( S_1 + S_2 \) is:
\[ t = \sqrt{2p^2 - 2p + 1} \]