A body moves for a total of nine second starting from rest with uniform acceleration and then with uniform retardation, which is twice the value of acceleration and then stop. The duration of uniform acceleration is

Options

A. 3s
B. 4.5s
C. 5s
D. 6s

Explanation

To solve the problem step by step, we will analyze the motion of the body during the two phases: uniform acceleration and uniform retardation.

### Step 1: Define Variables
Let:
- \( a \) = acceleration (in m/s²)
- \( t_1 \) = time of uniform acceleration (in seconds)
- \( t_2 \) = time of uniform retardation (in seconds)
- The total time of motion is given as \( t_1 + t_2 = 9 \) seconds.

### Step 2: Express Velocity
Since the body starts from rest, the final velocity \( v \) at the end of the acceleration phase can be expressed as:
\[ v = a \cdot t_1 \]
This equation relates the final velocity to the acceleration and the time of acceleration.

### Step 3: Express Retardation
The retardation is given as twice the acceleration, so:
\[ \text{Retardation} = -2a \]
The time taken to come to rest from velocity \( v \) under retardation can be expressed as:
\[ t_2 = \frac{v}{2a} \]

### Step 4: Substitute for Total Time
From the total time equation, we have:
\[ t_1 + t_2 = 9 \]
Substituting \( t_2 \) from the previous step:
\[ t_1 + \frac{v}{2a} = 9 \]

### Step 5: Substitute for Velocity
Now substitute \( v = a \cdot t_1 \) into the equation:
\[ t_1 + \frac{a \cdot t_1}{2a} = 9 \]
This simplifies to:
\[ t_1 + \frac{t_1}{2} = 9 \]
Combining the terms gives:
\[ \frac{3t_1}{2} = 9 \]

### Step 6: Solve for \( t_1 \)
Now, solve for \( t_1 \):
\[ 3t_1 = 18 \quad \Rightarrow \quad t_1 = 6 \text{ seconds} \]

### Conclusion
The duration of uniform acceleration is \( t_1 = 6 \) seconds.