If numerically, the momentum and position uncertainties are equal, what is the uncertainty in the velocity?

Do note that this question doesn't make physical sense. Position and momentum have different units...

We refer to the Heisenberg Uncertainty Principle:

#DeltaxDeltap_x >= ℏ//2#

where #Deltax# and #Deltap_x# are the uncertainties in position and momentum, respectively. #ℏ = h//2pi# is the reduced Planck's constant.

And so, if #Deltax = Deltap_x# NUMERICALLY (which makes no physical sense), then:

#(Deltap_x)^2 >= ℏ//2#

Assuming the relativistic mass is equal to the rest mass, and knowing consequently that if #p = mv#, then #Deltap = mDeltav#:

#(mDeltav_x)^2 >= ℏ//2#

#m^2(Deltav_x)^2 >= ℏ//2#

And so, the uncertainty in the speed is:

#color(blue)(Deltav_x >= (ℏ/(2m^2))^"1/2" "m"/"s")#

#ℏ# is very small, so it is very difficult to have a large uncertainty in speed unless the object is very light. For an electron:

#Deltav_x >= ((6.626 xx 10^(-34))/(4pi(9.109 xx 10^(-31))^2))^"1/2" "m"/"s"#

#>= ul(7.97 xx 10^12 " m/s")#

As expected, if we know the mass and momentum of an ELECTRON to the same uncertainty (whatever that means... they are different observables), the speed is not known well.