1. We are given the velocity as a function of position:

$v = \alpha \sqrt{x}$

Also,

$v = \frac{dx}{dt}$

So,

$\frac{dx}{dt} = \alpha \sqrt{x}$

2. Rearrange to separate variables:

$\frac{dx}{\sqrt{x}} = \alpha \, dt$

3. Integrate both sides:

$\int \frac{dx}{\sqrt{x}} = \int \alpha \, dt$

Since

$\int x^{-1/2} dx = 2\sqrt{x}$

we get

$2\sqrt{x} = \alpha t + C$

4. Use the initial condition: at $t=0$, $x=0$.

So,

$2\sqrt{0} = \alpha(0) + C \Rightarrow C=0$

Hence,

$2\sqrt{x} = \alpha t$

5. Solve for $x$:

$\sqrt{x} = \frac{\alpha t}{2}$

Squaring both sides,

$x = \frac{\alpha^2 t^2}{4}$

Thus displacement is proportional to $t^2$:

$x \propto t^2$

6. Evaluate options:

• A: $t^2$ ✅
• B: $t$ ❌
• C: $t^{1/2}$ ❌
• D: $t^3$ ❌

Therefore, the correct option is A.