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
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! :) :)
3 answers
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
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
Reiny is correct; I did not fully read the problem and assumed the shore was perpendicular to the beam
2 answers