If displacement of a particle is directly proportional to the square of time. Then particle is moving with

The correct answer is Uniform acceleration.

Let us understand the relationship between displacement, velocity, and acceleration step-by-step.

We are given that the displacement of the particle, let's denote it as $s$, is directly proportional to the square of time $t$.

Mathematically, we can write this relationship as:
$s\propto t^2$
To convert this proportionality into an equation, we introduce a constant of proportionality, $k$:
$s=k{t}^2$
where $k$ is a constant ($k\ne 0$).

To find the velocity $v$ of the particle, we take the first derivative of displacement with respect to time $t$:
$v=\frac{ds}{dt}=\frac{d}{dt}\left(k{t}^2\right)$
Using the power rule of differentiation ($\frac{d}{dt}\left(t^n\right)=n{t}^{n-1}$), we get:
$v=2kt$
Since velocity $v$ depends on time $t$, the particle is moving with a variable velocity.

Next, to find the acceleration $a$ of the particle, we take the derivative of velocity with respect to time $t$:
$a=\frac{dv}{dt}=\frac{d}{dt}\left(2kt\right)$
Differentiating, we get:
$a=2k$
Since $k$ is a constant, the acceleration $a$ is also a constant and does not change with time. Therefore, the particle is moving with a constant or uniform acceleration.