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(D) Let the distance of the car from the point on the track closest to the camera be $x$. The perpendicular distance from the camera to the track is $

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$1$

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(D) Let the distance of the car from the point on the track closest to the camera be $x$. The perpendicular distance from the camera to the track is $d = 30\ m$.From the geometry,we have $x = d 	an 	heta = 30 	an 	heta$.Differentiating both sides with respect to time $t$,we get:$\frac{dx}{dt} = 30 \sec^2 	heta \cdot \frac{d	heta}{dt}$Here,$\frac{dx}{dt} = v_{	ext{car}} = 40\ m/s$ and $\frac{d	heta}{dt} = \omega$ (the angular speed of the camera).So,$40 = 30 \sec^2 	heta \cdot \omega$.At the given position,$	heta = 30^\circ$. Therefore,$\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$.Thus,$\sec^2 30^\circ = \left(\frac{2}{\sqrt{3}}ight)^2 = \frac{4}{3}$.Substituting these values into the equation:$40 = 30 \cdot \left(\frac{4}{3}ight) \cdot \omega$$40 = 40 \cdot \omega$$\omega = 1\ rad/s$.

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