**Relation between Distance and Time for a Particle with Uniform Acceleration**

When a particle starts from rest and undergoes uniform acceleration, the relationship between the distance traveled and the time taken can be determined using kinematic equations. Let's analyze the situation and derive the relation between the distances traveled in the first 2 seconds (x) and the next 2 seconds (y).

**Assumptions:**
1. The particle starts from rest, implying an initial velocity of zero.
2. The acceleration remains constant throughout the motion.

**Kinematic Equations:**
To solve this problem, we can utilize the following kinematic equation that relates distance, initial velocity, time, and acceleration:

\[ s = ut + \frac{1}{2}at^2 \]

where:
- \( s \) is the distance traveled
- \( u \) is the initial velocity
- \( t \) is the time
- \( a \) is the acceleration

**Analysis:**
In the first 2 seconds, the particle travels a distance \( x \). Therefore, we can use the kinematic equation to express this distance as:

\[ x = 0 \cdot 2 + \frac{1}{2}a(2)^2 \]
\[ x = 2a \]

In the next 2 seconds, the particle travels a distance \( y \). We can again use the kinematic equation to express this distance as:

\[ y = 0 \cdot 2 + \frac{1}{2}a(2)^2 \]
\[ y = 2a \]

**Relation between x and y:**
Comparing the expressions for \( x \) and \( y \), we observe that the distances traveled in the first 2 seconds (x) and the next 2 seconds (y) are equal. Therefore, the relation between x and y is:

\[ x = y \]

Hence, the distances traveled in the first 2 seconds and the next 2 seconds are equal when a particle starts from rest with uniform acceleration.

This relation can also be explained intuitively. Since the particle starts from rest, the initial velocity is zero. As a result, the distance traveled in any given time period will be directly proportional to the time squared. Consequently, the distances traveled in the first 2 seconds and the next 2 seconds will be the same, regardless of the magnitude of acceleration.

y=2x