huc369 said:
Summary: How is the intermediate displacement speed derived?
I don't know if I wrote it correctly
View attachment 315157
Since SUVAT involves constant-acceleration,
and constant-velocity (zero acceleration) is a special case,
consider your proposed formula in the zero-acceleration case, where ##v_0=v_B=v_T##
(where B represents the intermediate displacement: ##(x_B-x_0)=\frac{1}{2}(x_T-x_0)## in your time interval ##0\leq t\leq T##).
Assume ##v_B>0##.
Your proposed formula would read
$$v_B=\frac{\sqrt{(v_B)^2+(v_B)^2}}{2}=\frac{v_B}{\sqrt{2}}\qquad\mbox(false).$$
Instead, it should (based on the symbols available in your recollection) be
$$v_B=\sqrt{ \frac{(v_0)^2+(v_T)^2}{2}}$$
so that $$v_B=\sqrt{ \frac{(v_B)^2+(v_B)^2}{2}}=v_B.$$
(Technically, to be a speed, the left-hand side should be ##|v_B|##.)
Now, following
@PeroK 's suggestion to your question,
this could be derived using SUVAT (the constant acceleration equations).