Road marks are 2.00 m apart. A vehicle crosses five gaps in 0.32 s on calibrated video. Calculate average speed and decide whether it exceeds a 30 m/s limit if timing uncertainty is ±0.02 s.

Question

Accepted Answer

Distance = 10.0 m and speed = 31.25 m/s; the uncertainty range is about 29.4–33.3 m/s, so the data alone do not establish the exceedance confidently. Five gaps cover 5 × 2.00 = 10.0 m. Best speed = 10.0/0.32 = 31.25 m/s. Using 0.34 s gives 29.4 m/s; using 0.30 s gives 33.3 m/s. Because the plausible range crosses 30 m/s, qualify the claim.

Speed-camera evidence

Road marks are 2.00 m apart. A vehicle crosses five gaps in 0.32 s on calibrated video. Calculate average speed and decide whether it exceeds a 30 m/s limit if timing uncertainty is ±0.02 s.

What a complete response should do

How did this one feel?