A rotating beacon is located 1 kilometer off a straight shoreline. If the beacon rotates at a rate of 3 revolutions per minute, how fast (in kilometers per hour) does the beam of light appear to be moving to a viewer who is 1/2 kilometer down the shoreline.

I need to show work, so formatting answers in this manner would be most appreciated. Thanks in advance! :) :)

Reiny is correct; I did not fully read the problem and assumed the shore was perpendicular to the beam

I beg to differ with the usually correct drwls.

This is a related rate problem dealing with angular velocity, so trig will be needed.

Let the point on the shore be x km down the shoreline.
so we have a right angled triangle, let the angle at the lighthouse be ß
we are given dß/dt = 6pi/min

The tanß = x/1
and sec^2 ß(dß/dt) = dx/dt

when x = 1/2 and using Pythagoras I found sec^2 ß = 1.25

so 1.25(6pi) = dx/dt

so dx/dt = 23.5629 km/min
= 1413.72 km/h

Well, let's break this down step by step:

1. First, we need to determine the speed of the beam of light at the viewer's location. Since the beacon completes 3 revolutions per minute, we know that the beam of light traces out a circle with a circumference of 1 kilometer every minute.

2. We can calculate the speed of the beam by dividing the circumference (1 kilometer) by the time taken to complete one revolution (1 minute):
Speed of beam = Circumference / Time
= 1 kilometer / 1 minute

3. Now we need to convert the unit of time from minutes to hours, since we want the speed in kilometers per hour. There are 60 minutes in an hour, so:
Time = 1 minute / 60
= 0.0167 hours (rounded to four decimal places)

4. Finally, we can calculate the speed of the beam by dividing the circumference (1 kilometer) by the time (0.0167 hours):
Speed of beam = 1 kilometer / 0.0167 hours
= 60 kilometers per hour (rounded to two decimal places)

So, the beam of light appears to be moving at a speed of approximately 60 kilometers per hour to the viewer who is 1/2 kilometer down the shoreline.

Hope that brings a ray of light to your question! 😄

To solve this problem, we need to relate the rotations per minute of the beacon to the speed at which the beam of light appears to be moving to the viewer.

Let's start by finding the circumference of the circle formed by the beam of light when it reaches the viewer 1/2 kilometer down the shoreline. The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle.

In this case, the radius of the circle is the distance between the beacon and the viewer, which is 1 kilometer. Therefore, the circumference C is:

C = 2π * 1 kilometer = 2π kilometers

Next, we need to find the distance covered by the beam of light in one revolution. Since the beacon rotates at a rate of 3 revolutions per minute, we can divide the circumference by the number of revolutions to find the distance covered in one revolution. Let's call this distance d.

d = C / 3 revolutions = (2π kilometers) / 3 = (2/3)π kilometers

Finally, we need to find the speed at which the beam of light appears to be moving to the viewer. The speed is given by the distance covered divided by the time taken. Since we know the distance covered in one revolution and the beam of light completes 3 revolutions per minute, we can find the speed in kilometers per minute.

Speed = d / 1 minute = (2/3)π kilometers / 1 minute = (2/3)π kilometers/minute

To find the speed in kilometers per hour, we can multiply the speed in kilometers per minute by 60, since there are 60 minutes in one hour.

Speed = (2/3)π kilometers/minute * 60 minutes = 40π kilometers/hour

Therefore, the beam of light appears to be moving at a speed of 40π kilometers per hour to the viewer who is 1/2 kilometer down the shoreline.

3 rev/min = 6 pi radians/min

Multiply that by the 1 km radial distance to get the speed the beam apparears to move at that distance:
6 pi x 1 = 18.85 km/min
Multiply that by 60 min/hr to get the spped in km per hour