A racing car is travelling along a track at a constant speed of $40m/s$
A T.V. camera men is recording the event from a distance of $30m$ directly away from the track as shown in figure. In order to keep the car under view in the position shown, the angular speed with which the camera should be rotated, is

The correct option is D $1\text{rad}/sec$

Step 1:Draw a rough diagram of given problem

x is the distance travelled by car at any time t

Step 2: Find angular speed of car.
Formula Used: $\frac{d\theta }{dt}=\omega$
From figure
$\tan \theta =\frac{x}{30}$
$LHS=\frac{d}{dt}\left(tan\theta \right)$
$=\frac{d}{d\theta }\left(tan\theta \right).\frac{d\theta }{dt}=se{c}^2\theta .\omega$

As$\frac{d\theta }{dt}=\omega$
Step 3: Find angular speed of car.
$RHS=\frac{d}{dt}\left(\frac{x}{30}\right)=\frac{40}{30}$
As$\frac{dx}{dt}=v=30m/s$
$se{c}^2\theta .\omega =\frac{40}{30}$
$\theta ={30}^{\circ }$and $se{c}^2{30}^{\circ }=\frac{2}{\sqrt{3}}$
we get$\frac{4}{3}.\omega =\frac{40}{30}$
$\omega =1\text{rad}/s$
Final Answer: (d)