Answer
$29\ sec$
Work Step by Step
Given $r=1800\ ft=\frac{1800}{5280}=\frac{15}{44}\ mi, \theta=40^\circ=\frac{2\pi}{9}\ rad, v=30.0\ mph=\frac{30}{3600}\ mi/sec=\frac{1}{120}\ mi/sec$, we have the arc length as $s=r\theta=\frac{15}{44}\times\frac{2\pi}{9}=\frac{5\pi}{66}\ mi$, thus the time is $t=\frac{s}{v}=\frac{\frac{5\pi}{66}}{\frac{1}{120}}\approx29\ sec$
(you can also calculate the angular speed $\omega=\frac{v}{r}$ first, then $t=\frac{\theta}{\omega}=\frac{r\theta}{v}$)
