Acceleration of a particle has a value 'a' for a time t. It is followed immediately by a retardation of magnitude 'a' for time t/2. Consider this as one cycle. Initial velocity of particle was zero. The displacement of the particle after n such cycles in succession is:

Options

A. ` (n( 3n+ 4))/(8) at^(2) `
B. ` (n(n+1))/(2) at^(2) `
C. ` ((n^(2) +n+1))/( 4) at ^(2) `
D. `nat^(2) `

Explanation

To solve the problem step by step, we will analyze the motion of the particle during the cycles of acceleration and retardation.

### Step 1: Understanding the Motion
The particle undergoes two phases in each cycle:
1. **Acceleration phase**: The particle accelerates with an acceleration \( a \) for a time \( t \).
2. **Retardation phase**: The particle then decelerates (retardation) with a magnitude of \( a \) for a time \( \frac{t}{2} \).

### Step 2: Finding the Velocity at the End of One Cycle
- **Initial velocity \( u = 0 \)**.
- After the acceleration phase for time \( t \):
\[ v_1 = u + at = 0 + at = at \]
- During the retardation phase for time \( \frac{t}{2} \):
\[ v_2 = v_1 - a \left(\frac{t}{2}\right) = at - a\left(\frac{t}{2}\right) = at - \frac{at}{2} = \frac{at}{2} \]
Thus, the velocity at the end of one complete cycle is \( v = \frac{at}{2} \).

### Step 3: Displacement During One Cycle
- **Displacement during the acceleration phase**:
\[ s_1 = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} a t^2 = \frac{1}{2} a t^2 \]
- **Displacement during the retardation phase**:
\[ s_2 = v_1 \left(\frac{t}{2}\right) - \frac{1}{2} a \left(\frac{t}{2}\right)^2 = at\left(\frac{t}{2}\right) - \frac{1}{2} a \left(\frac{t}{2}\right)^2 \]
\[ s_2 = \frac{at^2}{2} - \frac{1}{2} a \left(\frac{t^2}{4}\right) = \frac{at^2}{2} - \frac{at^2}{8} = \frac{4at^2}{8} - \frac{at^2}{8} = \frac{3at^2}{8} \]
- **Total displacement in one cycle**:
\[ s_{\text{cycle}} = s_1 + s_2 = \frac{1}{2} a t^2 + \frac{3}{8} a t^2 = \frac{4}{8} a t^2 + \frac{3}{8} a t^2 = \frac{7}{8} a t^2 \]

### Step 4: Total Displacement After \( n \) Cycles
Since the displacement in each cycle is the same, the total displacement after \( n \) cycles is:
\[ s_{\text{total}} = n \cdot s_{\text{cycle}} = n \cdot \frac{7}{8} a t^2 = \frac{7n}{8} a t^2 \]

### Final Answer
The displacement of the particle after \( n \) cycles is:
\[ \boxed{\frac{7n}{8} a t^2} \]