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Posted by on Sunday, November 11, 2007 at 2:32pm.

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! :) :)

  • Calc - , Sunday, November 11, 2007 at 3:09pm

    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

  • Calc - , Sunday, November 11, 2007 at 4:57pm

    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

  • Calc - , Sunday, November 11, 2007 at 6:30pm

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

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