A particle starts from rest with uniform acceleration `a`. Its velocity after 'n' second is 'v'. The displacement of the body in the last two second is

Options

A. `(2v(n-1))/n`
B. `(v(n-1))/n`
C. `(v(n+1))/n`
D. `(2v(2n+1))/n`

Explanation

To solve the problem, we need to find the displacement of the particle during the last two seconds of its motion, given that it starts from rest with uniform acceleration \( a \) and its velocity after \( n \) seconds is \( v \).

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The particle starts from rest, which means its initial velocity \( u = 0 \).
- The acceleration is uniform and denoted by \( a \).

2. **Finding the Velocity after \( n \) Seconds**:
- The formula for velocity under uniform acceleration is given by:
\[ v = u + at \]
- Substituting \( u = 0 \) and \( t = n \):
\[ v = 0 + an \implies v = an \]
- From this, we can express the acceleration \( a \) as:
\[ a = \frac{v}{n} \tag{1} \]

3. **Calculating the Displacement after \( n \) Seconds**:
- The displacement \( s_n \) after \( n \) seconds can be calculated using the formula:
\[ s_n = ut + \frac{1}{2} a t^2 \]
- Again substituting \( u = 0 \) and \( t = n \):
\[ s_n = 0 + \frac{1}{2} a n^2 = \frac{1}{2} a n^2 \]

4. **Calculating the Displacement at \( n-2 \) Seconds**:
- Now, we need to find the displacement \( s_{n-2} \) at \( n-2 \) seconds:
\[ s_{n-2} = u(n-2) + \frac{1}{2} a (n-2)^2 \]
- Substituting \( u = 0 \):
\[ s_{n-2} = 0 + \frac{1}{2} a (n-2)^2 = \frac{1}{2} a (n^2 - 4n + 4) = \frac{1}{2} a n^2 - 2an + 2a \]

5. **Finding the Displacement in the Last 2 Seconds**:
- The displacement during the last 2 seconds is given by:
\[ s_{n} - s_{n-2} \]
- Substituting the expressions for \( s_n \) and \( s_{n-2} \):
\[ s_{n} - s_{n-2} = \left( \frac{1}{2} a n^2 \right) - \left( \frac{1}{2} a n^2 - 2an + 2a \right) \]
- Simplifying this:
\[ s_{n} - s_{n-2} = 2an - 2a = 2a(n - 1) \]

6. **Substituting for \( a \)**:
- From equation (1), we can substitute \( a = \frac{v}{n} \):
\[ s_{n} - s_{n-2} = 2 \left(\frac{v}{n}\right)(n - 1) = \frac{2v(n - 1)}{n} \]

### Final Answer:
The displacement of the body in the last two seconds is:
\[ s_{n} - s_{n-2} = \frac{2v(n - 1)}{n} \]