A racing car is moving with velocity v = 40 m | vedclass

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Question

(A) Let $d = 30\ m$ be the perpendicular distance from the camera to the track.Let $r$ be the distance from the camera to the car at the given instant

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(A) Let $d = 30\ m$ be the perpendicular distance from the camera to the track.Let $r$ be the distance from the camera to the car at the given instant.From the geometry of the triangle,$\cos 30^{\circ} = \frac{d}{r}$,so $r = \frac{d}{\cos 30^{\circ}} = \frac{30}{\sqrt{3}/2} = \frac{60}{\sqrt{3}} = 20\sqrt{3}\ m$.The component of the car's velocity perpendicular to the line of sight (the line connecting the camera and the car) is $v_{\perp} = v \cos 30^{\circ}$.$v_{\perp} = 40 	imes \frac{\sqrt{3}}{2} = 20\sqrt{3}\ m/s$.The angular velocity $\omega$ is given by $\omega = \frac{v_{\perp}}{r}$.$\omega = \frac{20\sqrt{3}}{20\sqrt{3}} = 1\ rad/s$.

$A$ racing car is moving with velocity $v = 40\ m/s$ on a straight track as shown. We are recording it with a camera placed at a distance of $30\ m$ from the road. Find the angular velocity (in $rad/s$) with which we should rotate the camera to record the race at the instant shown in the diagram.

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