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: Understand the motion of the mass
The mass starts from rest and accelerates with a constant acceleration \( a \) for the first 2 seconds. After 2 seconds, the acceleration changes direction (becomes \(-a\)), meaning the mass will decelerate.

### Step 2: Calculate the displacement during the first 2 seconds
Using the equation of motion, the displacement \( s \) during the first 2 seconds can be calculated as:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the initial velocity \( u = 0 \) (the mass starts from rest), we have:
\[ s = 0 + \frac{1}{2} a (2)^2 = 2a \]

### Step 3: Determine the velocity at \( t = 2 \) seconds
The velocity \( v \) at \( t = 2 \) seconds can be calculated using:
\[ v = u + at \]
Again, since \( u = 0 \):
\[ v = 0 + a \cdot 2 = 2a \]

### Step 4: Analyze the motion after \( t = 2 \) seconds
After 2 seconds, the mass starts decelerating with an acceleration of \(-a\). Let \( t_1 \) be the time taken to return to the initial position after \( t = 2 \) seconds.

### Step 5: Write the equation for displacement after \( t = 2 \) seconds
The displacement during the time \( t_1 \) can be expressed as:
\[ \text{Displacement} = \text{Initial Displacement} + \text{Displacement during } t_1 \]
Since the mass returns to the initial position, the total displacement must equal zero:
\[ -2a = (2a)t_1 - \frac{1}{2} a t_1^2 \]
This simplifies to:
\[ -2a = 2at_1 - \frac{1}{2} a t_1^2 \]

### Step 6: Simplify the equation
Dividing through by \( a \) (assuming \( a \neq 0 \)):
\[ -2 = 2t_1 - \frac{1}{2} t_1^2 \]
Rearranging gives:
\[ \frac{1}{2} t_1^2 - 2t_1 - 2 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ t_1^2 - 4t_1 - 4 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t_1 = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 + 16}}{2} = \frac{4 \pm \sqrt{32}}{2} = \frac{4 \pm 4\sqrt{2}}{2} = 2 \pm 2\sqrt{2} \]
Since time cannot be negative, we take:
\[ t_1 = 2 + 2\sqrt{2} \]

### Step 8: Calculate the total time \( t_0 \)
The total time \( t_0 \) from the start until the mass returns to the initial position is:
\[ t_0 = 2 + t_1 = 2 + (2 + 2\sqrt{2}) = 4 + 2\sqrt{2} \]

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