A time varying force `F=6t-2t^(2)N`, at t=0 starts acting on a body of mass 2kg initially at rest, where t is in second. The force is withdrawn just at the instant when the body comes to rest again. We can see that at `t=0` the force `F=0`. Now answer the following:
Find the duration for which the force acts on the body.

Options

A. 2s
B. 3s
C. 3.5s
D. 4.5s

Explanation

To solve the problem, we need to determine the duration for which the force \( F = 6t - 2t^2 \) acts on the body. The force is applied from the moment it starts acting until it becomes zero again.

### Step-by-Step Solution:

1. **Identify the Force Equation:**
The force acting on the body is given by:
\[ F(t) = 6t - 2t^2 \]

2. **Find When the Force is Zero:**
To find the duration for which the force acts, we need to determine when the force \( F(t) \) becomes zero. We set the equation to zero:
\[ 6t - 2t^2 = 0 \]

3. **Factor the Equation:**
We can factor out \( 2t \) from the equation:
\[ 2t(3 - t) = 0 \]

4. **Solve for t:**
Setting each factor to zero gives us:
\[ 2t = 0 \quad \Rightarrow \quad t = 0 \]
\[ 3 - t = 0 \quad \Rightarrow \quad t = 3 \]

5. **Determine the Duration:**
The force starts acting at \( t = 0 \) and stops acting when \( t = 3 \). Therefore, the duration for which the force acts on the body is:
\[ \text{Duration} = t_{\text{end}} - t_{\text{start}} = 3 - 0 = 3 \text{ seconds} \]

### Final Answer:
The duration for which the force acts on the body is **3 seconds**.

---