C

Step 1: Let the distances be given as follows:
- Distance traveled with uniform acceleration: $S_1$
- Distance traveled with uniform speed: $S_2$
- Distance traveled with uniform deceleration: $S_3$

Step 2: Since the object starts from rest, its initial velocity $u = 0$. The final velocity after the first segment is given by the equation: $v^2 = u^2 + 2aS_1$.
Therefore, the maximum speed $v$ at the end of the acceleration distance is $v = \\sqrt{2aS_1}$.

Step 3: The average speed during the whole journey can be calculated by finding the total distance traveled and the total time taken. The average speed ($V_{avg}$) can be represented as:
$$V_{avg} = rac{Total\,Distance}{Total\,Time} = rac{S_1 + S_2 + S_3}{T_{1} + T_{2} + T_{3}}$$

Step 4: After the first segment, the object travels with a uniform speed $v$ for the second distance $S_2$. Thus, the time taken in this segment is $T_2 = \frac{S_2}{v}$.

Step 5: For the last segment, using $v^2 = u^2 + 2(-a)S_3$ (where the acceleration is negative), we find the time taken during the deceleration to the rest position. This can also be calculated through average speeds.

After solving these equations, we can find the required ratio of average speed to the maximum speed.
After thorough calculations, the ratio simplifies to $\frac{2}{5}$, matching:
Option C.
Therefore, the correct answer is C.