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. (a)` (n( 3n+ 4))/(8) at^(2) `
B. (b)` (n(n+1))/(2) at^(2) `
C. (c)` ((n^(2) +n+1))/( 4) at ^(2) `
D. (d)`nat^(2) `

Explanation

To solve the problem, we need to analyze the motion of the particle during one complete cycle of acceleration and retardation. Let's break it down step by step.

### Step 1: Analyze the motion during one cycle
1. **Acceleration phase**: The particle accelerates with an acceleration 'a' for a time 't'.
- Initial velocity (u) = 0
- Final velocity (v) after time 't' can be calculated using the equation:
\[ v = u + at = 0 + at = at \]

2. **Retardation phase**: The particle then decelerates (retardation) with a magnitude 'a' for a time of \( \frac{t}{2} \).
- Initial velocity for this phase = \( at \)
- Final velocity (v') after \( \frac{t}{2} \) can be calculated as:
\[ v' = v - at' = at - a\left(\frac{t}{2}\right) = at - \frac{at}{2} = \frac{at}{2} \]

### Step 2: Calculate the displacement during one cycle
1. **Displacement during the acceleration phase**:
- Using the formula for displacement (s) during uniform acceleration:
\[ s_1 = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} a t^2 = \frac{1}{2} a t^2 \]

2. **Displacement during the retardation phase**:
- Using the same formula for displacement during uniform retardation:
\[ s_2 = v t' - \frac{1}{2} a (t')^2 = at\left(\frac{t}{2}\right) - \frac{1}{2} a \left(\frac{t}{2}\right)^2 \]
- Substituting \( t' = \frac{t}{2} \):
\[ s_2 = \frac{at^2}{2} - \frac{1}{2} a \left(\frac{t^2}{4}\right) = \frac{at^2}{2} - \frac{at^2}{8} = \frac{4at^2}{8} - \frac{at^2}{8} = \frac{3at^2}{8} \]

3. **Total displacement in one cycle**:
- The total displacement \( S \) in one complete cycle is:
\[ S = s_1 + s_2 = \frac{1}{2} a t^2 + \frac{3}{8} a t^2 = \frac{4}{8} a t^2 + \frac{3}{8} a t^2 = \frac{7}{8} a t^2 \]

### Step 3: Calculate total displacement after n cycles
- The total displacement after \( n \) cycles is:
\[ S_n = n \cdot S = n \cdot \frac{7}{8} a t^2 = \frac{7n}{8} a t^2 \]

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