A body starts from rest and moves for n seconds with uniform acceleration a, its velocity after n seconds is v. The displacement of the body in last 3 seconds is :

Options

A. `(v(6n-9))/(2n)`
B. `(2v(6n-9))/(n)`
C. `(2v(2n+1))/(n)`
D. `(2v(n-1))/(n)`

Answer

To solve the problem, we need to find the displacement of a body in the last 3 seconds of its motion, given that it starts from rest and moves with uniform acceleration. Let's break down the solution step by step.

### Step 1: Understand the Given Information
- The body starts from rest, so the initial velocity \( u = 0 \).
- The body accelerates uniformly with acceleration \( a \).
- The total time of motion is \( n \) seconds.
- The final velocity after \( n \) seconds is \( v \).

### Step 2: Relate Final Velocity to Acceleration
Using the equation of motion:
\[ v = u + at \]
Substituting \( u = 0 \) and \( t = n \):
\[ v = 0 + a \cdot n \implies a = \frac{v}{n} \]

### Step 3: Calculate Total Displacement in \( n \) Seconds
The displacement \( S_n \) in \( n \) seconds can be calculated using the equation:
\[ S_n = ut + \frac{1}{2} a t^2 \]
Substituting \( u = 0 \), \( a = \frac{v}{n} \), and \( t = n \):
\[ S_n = 0 + \frac{1}{2} \cdot \frac{v}{n} \cdot n^2 = \frac{vn}{2} \]

### Step 4: Calculate Displacement in \( n - 3 \) Seconds
Now, we find the displacement \( S_{n-3} \) in the first \( n - 3 \) seconds:
\[ S_{n-3} = u(n - 3) + \frac{1}{2} a (n - 3)^2 \]
Again substituting \( u = 0 \) and \( a = \frac{v}{n} \):
\[ S_{n-3} = 0 + \frac{1}{2} \cdot \frac{v}{n} \cdot (n - 3)^2 \]
Expanding \( (n - 3)^2 \):
\[ (n - 3)^2 = n^2 - 6n + 9 \]
Thus,
\[ S_{n-3} = \frac{1}{2} \cdot \frac{v}{n} \cdot (n^2 - 6n + 9) = \frac{v}{2n} (n^2 - 6n + 9) \]

### Step 5: Calculate Displacement in the Last 3 Seconds
The displacement in the last 3 seconds \( S_{last} \) can be found by subtracting \( S_{n-3} \) from \( S_n \):
\[ S_{last} = S_n - S_{n-3} \]
Substituting the values:
\[ S_{last} = \frac{vn}{2} - \frac{v}{2n} (n^2 - 6n + 9) \]
Factoring out \( \frac{v}{2} \):
\[ S_{last} = \frac{v}{2} \left( n - \frac{1}{n} (n^2 - 6n + 9) \right) \]
Simplifying the expression inside the parentheses:
\[ n - \frac{n^2 - 6n + 9}{n} = n - (n - 6 + \frac{9}{n}) = 6 - \frac{9}{n} \]
Thus, we have:
\[ S_{last} = \frac{v}{2} \left( 6 - \frac{9}{n} \right) \]

### Final Result
The displacement of the body in the last 3 seconds is:
\[ S_{last} = \frac{v}{2} \left( 6 - \frac{9}{n} \right) \]