(A):The displacement of a particle starting from rest is directly proportional to the cube of the time of travel then the particle is moving with non uniform acceleration.
(R ):Acceleration `veca=(d^(2)vecs)/(dt^(2))`

Answer

To solve the problem, we need to analyze the relationship between displacement, time, and acceleration of a particle starting from rest.

### Step-by-Step Solution:

1. **Understanding the Problem Statement:**
The displacement \( s \) of a particle starting from rest is directly proportional to the cube of the time \( t \). This can be mathematically expressed as:
\[ s \propto t^3 \]

2. **Expressing the Proportionality:**
To remove the proportionality, we introduce a constant \( c \):
\[ s = c t^3 \]

3. **Finding Velocity:**
The velocity \( v \) of the particle is the first derivative of displacement with respect to time:
\[ v = \frac{ds}{dt} = \frac{d}{dt}(c t^3) = 3c t^2 \]

4. **Finding Acceleration:**
The acceleration \( a \) of the particle is the second derivative of displacement with respect to time, or the first derivative of velocity:
\[ a = \frac{dv}{dt} = \frac{d}{dt}(3c t^2) = 6c t \]

5. **Analyzing the Acceleration:**
From the expression \( a = 6c t \), we can see that the acceleration is directly proportional to time \( t \). This indicates that the acceleration is not constant; rather, it increases linearly with time.

6. **Conclusion:**
Since the acceleration is a function of time (specifically, it varies linearly with time), we conclude that the particle is moving with non-uniform acceleration.

### Final Statement:
Thus, the statement (A) is true, and the reasoning (R) is also correct. The particle is indeed moving with non-uniform acceleration.

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