According to the Heisenberg Uncertainty Principle, the product of uncertainties in position ($\Delta x$) and momentum ($\Delta p$) is given by: $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$

Given that $\Delta x = \Delta p$, we substitute into the equation: $(\Delta p)^2 \ge \frac{h}{4\pi}$ $\Delta p \ge \sqrt{\frac{h}{4\pi}} = \frac{1}{2}\sqrt{\frac{h}{\pi}}$

We know that uncertainty in momentum is related to uncertainty in velocity by $\Delta p = m \cdot \Delta v$. Substituting this: $m \cdot \Delta v \ge \frac{1}{2}\sqrt{\frac{h}{\pi}}$ $\Delta v \ge \frac{1}{2m}\sqrt{\frac{h}{\pi}}$

Thus, the minimum uncertainty in velocity is $\frac{1}{2m}\sqrt{\frac{h}{\pi}}$.