Question

Consider a 2850-lb automobile clocked by law-enforcement radar at a speed of 85.5 mph (miles/hour). If the position of the car is known to within 5.0 feet at the time of the measurement, what is the uncertainty in the velocity of the car?

Answer 1

rom H.U.P.
.. ∆x * ∆p ≥ ħ / 2

where
.. ∆x = uncertainty in the position
.. ∆p = uncertainty in the momemtum
.. ħ = reduced planks constant = 1.054x10^-34 J x sec

and we know that momentum (p) = m x v... (m = mass, v = velocity)
and we assume the mass is exact and all the variance in p is due to variance in v
and let's make it an equality for now..

so that
.. ∆v = ħ / (2m∆x)

and finally..
∆v = (1.054x10^-34 J x s) / (2 x 2330lb x 0.45359 kg/lb x 5.0ft x 1m/3.281ft) x (1kgm²/s² / 1J) = 3.27x10^-38 m/s

converting that to miles per hour
(3.27x10^-38 m/s) x (3.281ft / m) x (1mi / 5280ft) x (3600 sec / hr) = 7.32x10^-38 mph

so... your velocity would be 85.5 ± 0.0 mph via that analysis.