If the uncertainty in position and momentum of a particle are equal, then the uncertainty in its velocity is:

According to Heisenberg's Uncertainty Principle:
$$\Delta x\cdot \Delta p\ge \frac{h}{4\pi }$$
Given that the uncertainty in position ($\Delta x$) is equal to the uncertainty in momentum ($\Delta p$):
$$\Delta x=\Delta p$$
Substituting this into the principle:
$$(\Delta p)\cdot (\Delta p)=\frac{h}{4\pi }$$
$$(\Delta p{)}^2=\frac{h}{4\pi }⟹\Delta p=\sqrt{\frac{h}{4\pi }}=\frac{1}{2}\sqrt{\frac{h}{\pi }}$$
We know that momentum $\Delta p=m\Delta v$ (where $m$ is mass and $\Delta v$ is uncertainty in velocity):
$$m\Delta v=\frac{1}{2}\sqrt{\frac{h}{\pi }}$$
$$\Delta v=\frac{1}{2m}\sqrt{\frac{h}{\pi }}$$