A segment $AB$ is marked on a line $t$. The segment is moved on the plane so that it remains parallel to $t$ and that the traces of points $A$ and $B$ do not intersect. The segment finally returns onto $t$. How far can point $A$ now be from its initial position?
1. **Understanding the Problem:**
   - We have a segment \( AB \) on a line \( t \).
   - The segment \( AB \) is moved in the plane while remaining parallel to \( t \).
   - The traces of points \( A \) and \( B \) do not intersect.
   - The segment \( AB \) finally returns to the line \( t \).

2. **Analyzing the Movement:**
   - Since the segment \( AB \) remains parallel to \( t \), the movement can be visualized as a series of parallel translations.
   - The condition that the traces of points \( A \) and \( B \) do not intersect implies that the segment \( AB \) cannot rotate or cross over itself.

3. **Returning to the Line \( t \):**
   - The segment \( AB \) must return to the line \( t \) after some movements.
   - This means that the final position of \( A \) and \( B \) on \( t \) must be such that the segment \( AB \) is still parallel to its initial position.

4. **Calculating the Maximum Displacement:**
   - Consider moving the segment \( AB \) vertically by a small height \( h \) and then translating it horizontally by a distance \( d \).
   - If we repeat this process, the segment can trace out a path that forms a series of small steps.
   - By making the height \( h \) arbitrarily small, the horizontal displacement \( d \) can be made arbitrarily large without the traces of \( A \) and \( B \) intersecting.

5. **Conclusion:**
   - Since the height \( h \) can be made arbitrarily small, the horizontal displacement \( d \) can be made arbitrarily large.
   - Therefore, the point \( A \) can be displaced arbitrarily far from its initial position while still satisfying the given conditions.

\[
\boxed{\text{The displacement can be arbitrarily large.}}
\]