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%
Answer: 16.8%