**Understanding the Problem**

We are given that an object is dropped from rest and falls a distance of x during the first second. After that, it falls an additional distance of y in the next two seconds. We need to find the ratio of x to y.

**Analysis**

To analyze this problem, we can break it down into two parts:

1. The object's motion during the first second.
2. The object's motion during the next two seconds.

Let's analyze these two parts separately.

**1. Motion during the First Second**

When an object is dropped from rest, it is subjected to a constant acceleration due to gravity (g) in the downward direction. The distance covered by the object during this time can be calculated using the equation:

\[x = \frac{1}{2}gt^2\]

Where:
- x is the distance covered
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time (1 second in this case)

**2. Motion during the Next Two Seconds**

After the first second, the object continues to fall with a constant acceleration due to gravity. However, the total time for this part is 2 seconds. Using the same equation:

\[y = \frac{1}{2}g(2t)^2\]

Where:
- y is the additional distance covered in the next two seconds
- g is the acceleration due to gravity
- t is the time (2 seconds in this case)

**Finding the Ratio**

To find the ratio x/y, we need to substitute the values of x and y into the equation:

\[ \frac{x}{y} = \frac{\frac{1}{2}gt^2}{\frac{1}{2}g(2t)^2} \]

Simplifying the equation:

\[ \frac{x}{y} = \frac{t^2}{(2t)^2} \]

\[ \frac{x}{y} = \frac{t^2}{4t^2} \]

\[ \frac{x}{y} = \frac{1}{4} \]

Therefore, the ratio of x to y is 1:4.

**Conclusion**

In conclusion, when an object is dropped from rest and falls a distance x during the first second and an additional distance y in the next two seconds, the ratio of x to y is 1:4. This can be derived by analyzing the object's motion using the equations of motion and simplifying the ratio.