Answer
$-2~rad/s$ (decrease)
$1~rad/sec$ (increase)
Work Step by Step
Step 1. Given $y=132ft$, $v=\frac{dx}{dt}=-264ft/sec$
Step 2. $tan\theta=\frac{x}{y}=\frac{x}{132}$ and $sec^2\theta\frac{d\theta}{dt}=\frac{1}{132}\frac{dx}{dt}=\frac{-264}{132}=-2$
Step 3. When the car is right in front, we have $\theta=0, cos^20=1$, thus $\frac{d\theta}{dt}=-2cos^20=-2rad/s$ (decrease)
Step 4. After $0.5sec$ pass through the origin, we have $x=264(0.5)=132ft, \frac{dx}{dt}=264ft/sec$ . In this case $cos\theta=\frac{132}{\sqrt {132^2+132^2}}=\frac{\sqrt 2}{2}$. Thus $\frac{d\theta}{dt}=cos^2\theta\frac{1}{132}\frac{dx}{dt}=\frac{264}{132(2)}=1rad/sec$ (increase)
