Consider a 1260-kg automobile clocked by law-enforcement radar at a speed of 85.5 km/h. if the position of the car is known to within 2.0 m at the time of the measurement, what is the uncertainty in the velocity of the car?

The expected speed is v = 85.5 km/h
 v = 85.5 km/h = (85.5 km/h)*(0.2778 (m/s)/(km/h)) = 23.75 m/s

If there is an uncertainty of 2 meters in measuring the position, then within a 1-second time interval:
The lower measurement for the speed is v₁ = 21.75 m/s,
The upper measurement for the speed is v₂ = 25.75 m/s.
The range of variation is
Δv = v₂ - v₁ = 4 m/s

The uncertainty in measuring the speed is
Δv/v = 4/23.75 = 0.1684 = 16.84%

16.8%

What does that mean you dum bum