The distance travelled by a particle is proportional to the squares of time, then the particle travels with:

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Accepted Answer

1.  **Mathematical Relationship:** Given that the distance ($s$) is proportional to the square of time ($t^2$), we can express this as $s = kt^2$, where $k$ is a constant.
2.  **Velocity:** The instantaneous velocity ($v$) is the first derivative of distance with respect to time ($v = \frac{ds}{dt}$). Differentiating $s = kt^2$, we get $v = 2kt$.
3.  **Acceleration:** The acceleration ($a$) is the rate of change of velocity ($a = \frac{dv}{dt}$). Differentiating $v = 2kt$, we get $a = 2k$. 
4.  **Conclusion:** Since $k$ is a constant, $2k$ is also a constant. This implies the acceleration is uniform (constant) throughout the motion. This aligns with the kinematic equation for a body starting from rest ($u=0$): $s = \frac{1}{2}at^2$, where $s \propto t^2$ .

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