= Solution
Here $b>0$ because $a=0$ and the two cost coefficients are not both zero. The interior <Euler-Lagrange equation> makes $V$ quadratic. Imposing each displayed pair of endpoint data and the distance constraint gives the following formal profiles and their costs:
$$
\begin{array}{c|c|c}
\text{choice}&V(t)&b\int_0^T\dot V^2\,dt\\ \hline
1&3Lt^2/T^3&12bL^2/T^3\\
2&6Lt(T-t)/T^3&12bL^2/T^3\\
3&3Lt(2T-t)/(2T^3)&3bL^2/T^3
\end{array}
$$
Thus \b[choice (3) is the best strategy], and its true minimum cost is
$$
\boxed{E_{\min}=\frac{3bL^2}{T^3}.}
$$
The qualification from part (b) applies to row (1): its formal quadratic is not a minimizer with free terminal speed, since $\dot V(T)\ne0$. Row (2) is a genuine minimizer for its fixed zero terminal speed, and row (3) is the genuine free-terminal-speed minimizer, satisfying its natural condition $\dot V(T)=0$.
A direct lower bound makes the comparison rigorous. For every admissible speed with $V(0)=0$,
$$
L=\int_0^T(T-t)\dot V(t)\,dt,
$$
by <integration by parts>. The <Cauchy-Schwarz inequality> gives $L^2\leq(T^3/3)\int_0^T\dot V^2\,dt$. Equality requires $\dot V$ to be proportional to $T-t$, giving exactly row (3). Row (1)'s zero initial acceleration excludes equality; however short initial layers, as in part (b), approach that same infimum. Hence choice (1) has infimum $3bL^2/T^3$ but does not attain it, while choice (3) does. This distinction resolves the boundary-condition issue in the printed minimization request.
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