a police car is parked 50 feet from an abandoned building (see figure). the light on top of the police car…

Explanation:

Step1: Convert angular velocity to radians per second

The rotation rate is given as 45 revolutions per minute. Convert this to radians per second. $\frac{d\theta}{dt} = 45 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{90\pi}{60} \text{ rad/s} = \frac{3\pi}{2} \text{ rad/s}$

Step2: Establish a relationship between x and $\theta$

Let $x$ be the distance along the wall from the point perpendicular to the car, and $d$ be the distance from the car to the wall (50 ft). From the diagram, we can form a right-angled triangle where $x$ is the opposite side to $\theta$ and $d$ is the adjacent side. $\tan(\theta) = \frac{x}{d}$ Given $d = 50$ ft, the relationship is: $x = 50 \tan(\theta)$

Step3: Differentiate the equation with respect to time

Differentiate both sides of the equation $x = 50 \tan(\theta)$ with respect to time $t$ to find $\frac{dx}{dt}$. $\frac{dx}{dt} = \frac{d}{dt}(50 \tan(\theta))$ $\frac{dx}{dt} = 50 \sec^2(\theta) \frac{d\theta}{dt}$

Step4: Substitute the given values and calculate

The angle $\theta$ is given as 60 degrees. First, find $\sec^2(\theta)$: $\sec(\theta) = \sec(60^\circ) = \frac{1}{\cos(60^\circ)} = \frac{1}{1/2} = 2$ $\sec^2(\theta) = (2)^2 = 4$ Now, substitute $\sec^2(\theta) = 4$ and $\frac{d\theta}{dt} = \frac{3\pi}{2}$ rad/s into the differentiated equation: $\frac{dx}{dt} = 50 \times 4 \times \frac{3\pi}{2}$ $\frac{dx}{dt} = 200 \times \frac{3\pi}{2}$ $\frac{dx}{dt} = 100 \times 3\pi$ $\frac{dx}{dt} = 300\pi \text{ ft/s}$

Answer:

The light beam is moving along the wall at a speed of $300\pi \text{ ft/s}$.