A particle starts from rest with uniform acceleration `a`. Its velocity after `n` seconds is `v`. The displacement of the particle in the two seconds 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 step by step, we will use the equations of motion for uniformly accelerated motion.

### Step 1: Understand the initial conditions
The particle starts from rest, which means:
- Initial velocity \( u = 0 \)
- Uniform acceleration \( a \)

### Step 2: Relate acceleration to final velocity
The final velocity \( v \) after \( n \) seconds can be expressed using the equation:
\[ v = u + at \]
Since \( u = 0 \), we have:
\[ v = 0 + a \cdot n \implies a = \frac{v}{n} \]

### Step 3: Calculate the displacement after \( n \) seconds
The displacement \( S_n \) after \( n \) seconds can be calculated using the equation:
\[ S_n = ut + \frac{1}{2} a t^2 \]
Substituting \( u = 0 \) and \( a = \frac{v}{n} \):
\[ S_n = 0 + \frac{1}{2} \left(\frac{v}{n}\right) n^2 = \frac{v n}{2} \]

### Step 4: Calculate the displacement after \( n - 2 \) seconds
Now, we calculate the displacement \( S_{n-2} \) after \( n - 2 \) seconds:
\[ S_{n-2} = u(n-2) + \frac{1}{2} a (n-2)^2 \]
Again substituting \( u = 0 \) and \( a = \frac{v}{n} \):
\[ S_{n-2} = 0 + \frac{1}{2} \left(\frac{v}{n}\right) (n-2)^2 \]
Expanding this:
\[ S_{n-2} = \frac{v}{2n} (n^2 - 4n + 4) = \frac{v}{2n} (n^2 - 4n + 4) \]

### Step 5: Calculate the displacement in the last 2 seconds
The displacement during the last 2 seconds \( S_2 \) can be found by subtracting the two displacements:
\[ S_2 = S_n - S_{n-2} \]
Substituting the values we found:
\[ S_2 = \frac{v n}{2} - \frac{v}{2n} (n^2 - 4n + 4) \]
Factoring out \( \frac{v}{2n} \):
\[ S_2 = \frac{v}{2n} \left(n^2 - (n^2 - 4n + 4)\right) \]
Simplifying the expression inside the parentheses:
\[ S_2 = \frac{v}{2n} \left(4n - 4\right) = \frac{v}{2n} \cdot 4(n - 1) \]
Thus:
\[ S_2 = \frac{2v(n - 1)}{n} \]

### Final Answer
The displacement of the particle in the last 2 seconds is:
\[ S_2 = \frac{2v(n - 1)}{n} \]