Question

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

 

विकल्प

• ${\displaystyle \frac{4}{3}}$ rad/s

• ${\displaystyle \frac{3}{4}}$ rad/s

• ${\displaystyle \frac{8}{3}\sqrt{3}}$ rad/s

• 1 rad/s

Answer 1

A racing car is travelling along a track at a constant speed of 40 m/s. A T.V. cameraman is recording the event from a distance of 30 m directly away from the track as shown in the figure. In order to keep the car under view in the position shown, the angular speed with which the camera should be rotated is 1 rad/s.

Explanation:

We have

tan θ = ${\displaystyle \frac{x}{30};\frac{\text{d}}{\text{dt}}\left[\tan \theta \right]=\frac{\text{d}}{\text{dt}}\left[\frac{x}{30}\right]}$

${\displaystyle \frac{\text{d}}{\text{d}\theta }\left(\tan \theta \right).\frac{\text{d}\theta }{\text{dt}}=\frac{1}{30}\frac{\text{dx}}{\text{dt}}}$

sec²θ ${\displaystyle \frac{\text{d}\theta }{\text{dt}}=\frac{1}{30}\times 40}$

⇒ ${\displaystyle \frac{\text{d}\theta }{\text{dt}}=\frac{4}{3{\sec }^2\theta }}$

At the given instant, θ = 30°

⇒ ${\displaystyle \omega =\frac{\text{d}\theta }{\text{dt}}=\frac{4}{3}}$ cos²30°

= ${\displaystyle \frac{4}{3}\times {\left(\frac{\sqrt{3}}{2}\right)}^2}$

= ${\displaystyle \frac{4}{3}\times \frac{3}{4}}$

= 1 rad s^-1

or the angular speed with which the camera should be rotated = 1 rad/s.