$17-$ A crest vertical curve is designed as shown. The station $($ft$)$ of the PVT is most nearly:

A$.44+06\mathrm{.}00$

B$.41+92\mathrm{.}00$

$a=\frac{g_2-9}{2L_{\text{s}\text{t}\text{a}\text{t}}}=\frac{-3\mathrm{.}5-5}{2(1600)}=$

C$.41+23\mathrm{.}00$

PVI or V

D$.40+75\mathrm{.}00$

Station $42+65$

$y_{\text{o}\text{v}\text{c}}=y_{\text{p}\text{v}\text{i}}-g_1(\frac{4m}{2})$

Elevation $447$ ft

$V{C}_{sm}=V{I}_{\text{s}\text{t}\text{a}}-(\frac{4m}{2})$

$Y_{PVt}=y_{VVL}+g_2(2\frac{L}{2})$

EVC