Part of the problem with the typical textbook introduction of "average velocity" is that it often doesn't clarify that
it's a
time-weighted average of velocities (which they usually have not yet defined).
So, often the student is often left incorrectly assuming that it's a straight-average of velocities.
For a trip with three piecewise-constant-velocity-legs
##
\begin{align*}
\vec v_{avg}
&\equiv \frac{\vec v_1\Delta t_1+\vec v_2\Delta t_2+\vec v_3\Delta t_3}{\Delta t_1+\Delta t_2+\Delta t_3} \\
&=
\frac{\Delta \vec x_1+\Delta\vec x_2+\Delta \vec x_3}{\Delta t_1+\Delta t_2+\Delta t_3}\\
&=
\frac{\Delta \vec x_{total}}{\Delta t_{total}}\\
\end{align*}
##
What is usually also missing is an interpretation of the "average velocity".
I use the following...
Can the trip from start position to end position over the same interval of time be done with a constant velocity
(rather than a varying one)? Yes, use the average-velocity over that interval.
Only when the velocity is constant over the interval
will the average-velocity equal the velocity...
that's the only time one can use "[constant] velocity=displacement/time".
For uniform acceleration ,
##
\begin{align*}
\vec v_{avg}
&\equiv
\frac{\Delta \vec x_{total}}{\Delta t_{total}}\\
&\stackrel{\scriptsize\rm const \ a }{=}\frac{\frac{1}{2}\vec a(\Delta t)^2 +\vec v_i\Delta t}{\Delta t}\\
&\stackrel{\scriptsize\rm const \ a }{=}\frac{1}{2}\vec a(\Delta t) +\vec v_i\\
&\stackrel{\scriptsize\rm const \ a }{=}\frac{1}{2}(\vec v_f -\vec v_i) +\vec v_i \\
&\stackrel{\scriptsize\rm const \ a }{=} \frac{1}{2}(\vec v_f+ \vec v_i)
\end{align*}
##
which looks like a straight-average of the velocities of two piecewise-constant-velocity-legs,
but it's really the straight-average of the starting and ending velocities.
(If this rendered poorly, try reading it here:
https://www.physicsforums.com/threa...nics-kinematics-comments.813456/#post-5106894 )