A point mass starts moving in a straight line with constant acceleration. After time `t_0` the acceleration changes its sign, remaining the same in magnitude. Determine the time T from the beginning of motion in which the point mass returns to the initial position.

Explanation

To solve the problem, we need to analyze the motion of the point mass in two phases:

1. The first phase where it moves with constant acceleration \( a \) for a time \( t_0 \).
2. The second phase where the acceleration changes to \( -a \) (the same magnitude but opposite direction).

### Step-by-Step Solution:

**Step 1: Calculate the distance traveled in the first phase.**

The distance \( s_1 \) traveled during the first phase can be calculated using the equation of motion:

\[ s_1 = ut + \frac{1}{2} a t^2 \]

Since the initial velocity \( u = 0 \) (the point mass starts from rest), we have:

\[ s_1 = 0 + \frac{1}{2} a t_0^2 = \frac{1}{2} a t_0^2 \]

**Step 2: Calculate the velocity at the end of the first phase.**

The final velocity \( v \) at the end of the first phase can be calculated using:

\[ v = u + at \]

Again, since \( u = 0 \):

\[ v = 0 + a t_0 = a t_0 \]

**Step 3: Analyze the second phase of motion.**

In the second phase, the point mass starts with an initial velocity \( v = a t_0 \) and moves with an acceleration of \( -a \). We need to find the time \( t \) it takes to return to the initial position.

The distance traveled in the second phase \( s_2 \) can be calculated as:

\[ s_2 = vt + \frac{1}{2} (-a) t^2 \]

Substituting \( v = a t_0 \):

\[ s_2 = (a t_0) t - \frac{1}{2} a t^2 \]

**Step 4: Set the total distance equal to zero.**

For the point mass to return to the initial position, the total distance traveled must equal the distance traveled in the first phase (but in the opposite direction):

\[ s_1 + s_2 = 0 \]

Substituting \( s_1 = \frac{1}{2} a t_0^2 \) and \( s_2 \):

\[ \frac{1}{2} a t_0^2 + (a t_0) t - \frac{1}{2} a t^2 = 0 \]

**Step 5: Simplify the equation.**

Factoring out \( a \) (assuming \( a \neq 0 \)):

\[ \frac{1}{2} t_0^2 + t t_0 - \frac{1}{2} t^2 = 0 \]

Multiplying through by 2 to eliminate the fraction:

\[ t_0^2 + 2tt_0 - t^2 = 0 \]

**Step 6: Rearranging the equation.**

Rearranging gives us:

\[ t^2 - 2tt_0 - t_0^2 = 0 \]

**Step 7: Solve the quadratic equation.**

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1 \), \( b = -2t_0 \), and \( c = -t_0^2 \):

\[ t = \frac{2t_0 \pm \sqrt{(-2t_0)^2 - 4(1)(-t_0^2)}}{2(1)} \]

\[ t = \frac{2t_0 \pm \sqrt{4t_0^2 + 4t_0^2}}{2} \]

\[ t = \frac{2t_0 \pm \sqrt{8t_0^2}}{2} \]

\[ t = \frac{2t_0 \pm 2\sqrt{2}t_0}{2} \]

\[ t = t_0(1 \pm \sqrt{2}) \]

Since time cannot be negative, we take:

\[ t = t_0(1 + \sqrt{2}) \]

**Step 8: Calculate the total time \( T \).**

The total time \( T \) from the beginning of the motion is:

\[ T = t_0 + t = t_0 + t_0(1 + \sqrt{2}) = t_0(2 + \sqrt{2}) \]

### Final Result:

The total time \( T \) from the beginning of motion in which the point mass returns to the initial position is:

\[ T = t_0(2 + \sqrt{2}) \]