A point mass starts moving in straight line with constant acceleration `a` from rest at `t=0`. At time `t=2s`, the acceleration changes the sign remaining the same in magnitude. The mass returns to the initial position at time `t=t_(0)` after start of motion. Here `t_(0)` is

Options

A. `4s`
B. `(4+2sqrt(2))s`
C. `(2+2sqrt(2))s`
D. `(4+4sqrt(2))s`

Answer

To solve the problem step by step, we will analyze the motion of the point mass under the given conditions.

### Step 1: Initial Setup
The point mass starts from rest at \( t = 0 \) with a constant acceleration \( a \). Therefore, the initial velocity \( u = 0 \).

### Step 2: Velocity After 2 Seconds
The velocity \( v \) of the mass at \( t = 2 \) seconds can be calculated using the formula:
\[ v = u + at \]
Substituting \( u = 0 \) and \( t = 2 \):
\[ v = 0 + a \cdot 2 = 2a \]

### Step 3: Change in Acceleration
At \( t = 2 \) seconds, the acceleration changes its sign, becoming \( -a \). The mass will now decelerate until it comes to rest.

### Step 4: Time to Come to Rest
Let \( t' \) be the time taken to come to rest after \( t = 2 \) seconds. The final velocity \( v_f = 0 \) and initial velocity \( v = 2a \). Using the equation:
\[ v_f = v - at' \]
Substituting the values:
\[ 0 = 2a - a t' \]
Rearranging gives:
\[ a t' = 2a \implies t' = 2 \text{ seconds} \]

### Step 5: Total Time Until Rest
The total time from the start until the mass comes to rest is:
\[ t_{\text{total}} = 2 + t' = 2 + 2 = 4 \text{ seconds} \]

### Step 6: Distance Covered Until Rest
Now, we calculate the distance covered while moving with acceleration \( a \) for the first 2 seconds and then with acceleration \( -a \) for the next 2 seconds.

1. **Distance during first 2 seconds**:
\[ s_1 = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} a (2^2) = 2a \]

2. **Distance during next 2 seconds (deceleration)**:
The initial velocity for this phase is \( 2a \) and the time is \( 2 \) seconds:
\[ s_2 = v t' - \frac{1}{2} a (t')^2 = (2a)(2) - \frac{1}{2} a (2^2) = 4a - 2a = 2a \]

### Step 7: Total Distance Covered
The total distance covered when the mass comes to rest is:
\[ s_{\text{total}} = s_1 + s_2 = 2a + 2a = 4a \]

### Step 8: Return to Initial Position
After coming to rest, the mass will move back to the initial position. The distance to cover is \( 4a \) with a constant acceleration \( a \) (now positive).

Using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( u = 0 \) and \( s = 4a \):
\[ 4a = 0 + \frac{1}{2} a t_{\text{return}}^2 \]
Solving for \( t_{\text{return}} \):
\[ 4a = \frac{1}{2} a t_{\text{return}}^2 \implies 8 = t_{\text{return}}^2 \implies t_{\text{return}} = 2\sqrt{2} \text{ seconds} \]

### Step 9: Total Time \( t_0 \)
The total time \( t_0 \) until the mass returns to the initial position is:
\[ t_0 = 4 + 2\sqrt{2} \]

### Final Answer
Thus, the total time \( t_0 \) is:
\[ t_0 = 4 + 2\sqrt{2} \text{ seconds} \]