A particle starts from rest with uniform acceleration and its velocity after `n` seconds is `v`. The displacement of the body in last two seconds is.

Options

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

Explanation

To solve the problem step by step, we will use the equations of motion for a particle under uniform acceleration.

### Step 1: Identify the given information
- The particle starts from rest, so the initial velocity \( U = 0 \).
- The velocity after \( n \) seconds is \( v \).
- We need to find the displacement of the particle in the last 2 seconds.

### Step 2: Find the acceleration
Using the first equation of motion:
\[ v = U + A \cdot n \]
Since \( U = 0 \), we have:
\[ v = A \cdot n \]
From this, we can express the acceleration \( A \):
\[ A = \frac{v}{n} \]

### Step 3: Calculate the total displacement \( S_n \) after \( n \) seconds
Using the second equation of motion:
\[ S_n = U \cdot n + \frac{1}{2} A \cdot n^2 \]
Again, since \( U = 0 \):
\[ S_n = \frac{1}{2} A \cdot n^2 \]
Substituting the value of \( A \):
\[ S_n = \frac{1}{2} \cdot \frac{v}{n} \cdot n^2 = \frac{1}{2} v n \]

### Step 4: Calculate the displacement \( S_{n-2} \) after \( n-2 \) seconds
Using the same formula for displacement:
\[ S_{n-2} = U \cdot (n-2) + \frac{1}{2} A \cdot (n-2)^2 \]
Again, since \( U = 0 \):
\[ S_{n-2} = \frac{1}{2} A \cdot (n-2)^2 \]
Substituting the value of \( A \):
\[ S_{n-2} = \frac{1}{2} \cdot \frac{v}{n} \cdot (n-2)^2 \]

### Step 5: Calculate the displacement in the last 2 seconds
The displacement in the last 2 seconds \( S_{last\ 2} \) is given by:
\[ S_{last\ 2} = S_n - S_{n-2} \]
Substituting the values we found:
\[ S_{last\ 2} = \frac{1}{2} v n - \frac{1}{2} \cdot \frac{v}{n} \cdot (n-2)^2 \]
Factoring out \( \frac{1}{2} v \):
\[ S_{last\ 2} = \frac{1}{2} v \left( n - \frac{(n-2)^2}{n} \right) \]

### Step 6: Simplify the expression
Now simplify \( n - \frac{(n-2)^2}{n} \):
\[ = n - \left( \frac{n^2 - 4n + 4}{n} \right) = n - (n - 4 + \frac{4}{n}) = 4 - \frac{4}{n} \]
Thus,
\[ S_{last\ 2} = \frac{1}{2} v \left( 4 - \frac{4}{n} \right) = 2v - \frac{2v}{n} \]

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