Hypothetical claim: "If a car drives 60 mph for half the distance and 40 mph for the other half, its average speed for the trip is the arithmetic mean: (60+40)/2 = 50 mph."

Counterexample. Use 120 miles at each speed (total 240 miles).

• Leg 1: 120 mi ÷ 60 mph = 2 h
• Leg 2: 120 mi ÷ 40 mph = 3 h
• Total: 240 mi ÷ 5 h = 48 mph

48 ≠ 50, so the claim fails. The arithmetic mean of the two speeds overweights the faster leg because less time is spent there. The correct formula for equal-distance segments is the harmonic mean: 2·v₁·v₂/(v₁+v₂) = 2·60·40/100 = 48.

What this refutes: the shortcut of averaging speeds by arithmetic mean when segments are equal in distance. What it does not refute: if segments are equal in time, the arithmetic mean is correct (e.g., 1 h at 60 + 1 h at 40 = 100 mi in 2 h = 50 mph). The distinction is whether the weighting is by distance or by time.