Starting from rest a body moves with uniform acceleration and acquires a velocity v after n seconds. Then its displacement in the last two seconds is

Options

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

Explanation

To solve the problem step by step, we will follow the physics equations of motion under uniform acceleration.

### Step 1: Identify the given values
- Initial velocity, \( u = 0 \) (the body starts from rest)
- Final velocity after \( n \) seconds, \( v = v \)
- Time, \( t = n \)

### Step 2: Calculate the acceleration
Using the equation of motion:
\[ v = u + at \]
Substituting the known values:
\[ v = 0 + a \cdot n \implies a = \frac{v}{n} \]

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

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

### Step 5: Calculate the displacement in the last 2 seconds
The displacement in the last 2 seconds is given by:
\[ S_{last\ 2\ seconds} = S_n - S_{n-2} \]
Substituting the values we found:
\[ S_{last\ 2\ seconds} = \frac{v n}{2} - \frac{v(n^2 - 4n + 4)}{2n} \]
Finding a common denominator:
\[ S_{last\ 2\ seconds} = \frac{v n^2}{2n} - \frac{v(n^2 - 4n + 4)}{2n} \]
Combining the fractions:
\[ S_{last\ 2\ seconds} = \frac{v(n^2 - (n^2 - 4n + 4))}{2n} = \frac{v(4n - 4)}{2n} = \frac{2v(n - 1)}{n} \]

### Final Result
Thus, the displacement in the last 2 seconds is:
\[ \text{Displacement in last 2 seconds} = \frac{2v(n - 1)}{n} \]