 Time and distance averages of speed

For forward motion with nonnegative speed $v(t)$ over duration $T>0$ and distance $L={\int }_0^{T}v(t)dt>0$, define the time-averaged speed $V_t=L/T$ and distance-averaged speed $V_x=L^{-1}\int vdx$. The change of variables $dx=vdt$ gives $V_x={\mathbb{E}}_t[v^2]/{\mathbb{E}}_t[v]$, where ${\mathbb{E}}_t$ averages uniformly over time. ThereforeEquality holds exactly for constant speed almost everywhere. This also follows directly from the Cauchy-Schwarz inequality applied to $v$ and one on the time interval. The reason for the difference is that distance averaging puts more weight on periods when the vehicle covers more distance per unit time. These formulas assume a finite second time moment; if that moment is infinite, the distance average is infinite and the inequality holds in the extended sense.