Does a certainty in the position implies infinite variation in speed?

[Giuseppino32]

TL;DR Summary

As position uncertainty decreases, the uncertainty in momentum and velocity increases, potentially approaching the speed of light. While Δv can be large, it doesn’t imply any arbitrary velocity from 0 to 𝑐; it's probabilistic and constrained.

I would like to know if this thought makes any sence or if i'm missing something

Heisenberg principle states that: ΔxΔρ ≥ ħ/2 ⇒ Δρ ≥ ħ /2Δx
If we consider a scenario where we increase the precision of our measurement of position, we have Δx ⇒ 0 the principle implies:
Δρ ≥ ħ/2Δx → Δρ ⇒ ∞

Assuming that any of that makes any sense, If we consider the maximum possible speed in the universe, c (the speed of light, approximately 3×10⁸m/s) the minimum possible speed as 0, and assume the mass of the particle remains constant, then the uncertainty in velocity Δv is given by: Δρ = 3.10⁸ - 0 = 3.10⁸ m/s

This illustrates that as the uncertainty in position decreases, the uncertainty in momentum—and thus velocity—can increase significantly, potentially approaching the speed of light.
Does that makes physical sense? Lets say we know where an electron e⁻ is in space so if its mass is constant 9.1093837 × 10-31 kilograms. If we find out its position does it mean it could have any value from 0 to the speed of light, in any direction, without the necessity of a force?

 

[hutchphd]

How doe we "know" this very defined location?? Likely there are some nontrivial "forces" involved.. The Heisenberg uncertainty limit is difficult to circumvent.