A body starts from rest with uniform acceleration. The velocity of the body after t seconds is v. The displacement of the body in last three seconds is

Options

A. `(3v)/2(t-3)`
B. `(3v)/2(t+3)`
C. `3v[1-3/(2t)]`
D. `3v[1+3/(2t)]`

Explanation

To solve the problem step by step, we will calculate the displacement of the body in the last three seconds of its motion, given that it starts from rest with uniform acceleration.

### Step 1: Understand the problem
The body starts from rest, which means its initial velocity (u) is 0. It accelerates uniformly, and after time \( t \), its velocity is \( v \). We need to find the displacement of the body during the last three seconds of its motion.

### Step 2: Define the time intervals
The last three seconds of motion can be defined as the time interval from \( t - 3 \) seconds to \( t \) seconds.

### Step 3: Use the equations of motion
We will use the second equation of motion to find the displacement during the last three seconds. The displacement \( s \) from time \( t_1 \) to \( t_2 \) is given by:
\[ s = u(t_2 - t_1) + \frac{1}{2} a (t_2^2 - t_1^2) \]
In our case:
- \( u = 0 \) (initial velocity)
- \( t_1 = t - 3 \)
- \( t_2 = t \)

### Step 4: Calculate the displacement from \( t - 3 \) to \( t \)
Substituting into the equation:
\[ s = 0 \cdot 3 + \frac{1}{2} a \left(t^2 - (t - 3)^2\right) \]
Now, simplify \( (t - 3)^2 \):
\[ (t - 3)^2 = t^2 - 6t + 9 \]
Thus,
\[ t^2 - (t - 3)^2 = t^2 - (t^2 - 6t + 9) = 6t - 9 \]
Now substituting back into the displacement equation:
\[ s = \frac{1}{2} a (6t - 9) \]

### Step 5: Relate acceleration to velocity
From the first equation of motion:
\[ v = u + at \implies v = 0 + at \implies a = \frac{v}{t} \]
Substituting \( a \) into the displacement equation:
\[ s = \frac{1}{2} \left(\frac{v}{t}\right)(6t - 9) \]
This simplifies to:
\[ s = \frac{v}{2t}(6t - 9) = \frac{v}{2t}(3(2t - 3)) = \frac{3v(2t - 3)}{2} \]

### Final Answer
Thus, the displacement of the body in the last three seconds is:
\[ s = \frac{3v(2t - 3)}{2} \]