Question

A searchlight rotates at a rate of 3 revolutions per minute. The beam hits a wall located 15 miles away and produces a dot of light that moves horizontally along the wall. How fast (in miles per hour) is this dot moving when the angle \theta between the beam and the line through the searchlight perpendicular to the wall is pi/4? Note that dtheta/dt3(2pi)=6pi

Answer 1

The horizontal distance is 11 miles * tan(pi/6) = 6.35 miles [remember pi/6 is radians not degrees]

At 6pi rad/min, the angle pi/6 is swept out in pi/6 /(6pi) min = 1/36 min = 1.67 sec

So the dot speed is 6.35 miles per 1.67 sec = 3.8 miles / sec = 228 miles per minute = 13,690 mph

Answer 2

thats not right

Answer 3

ive been thinking about this one for a few days :)

Answer 4

is the 15 miles a perpendicular distance from the wall?

Answer 5

yes

Answer 6

ok, one thing I thought of was relating the linear speed of the circumference of the circle created by the beam

15sqrt(2) * 2 pi = 30sqrt(2)pi, at 3 times a minute = 90sqrt(2) pi

this isnt the speed along the wall, this is the speed along the circle that hits the wall at that point

Answer 7

all this has me so confused

Answer 8

im wondering if the rotation speed can be component based to measure the wall speed.



which would make the linear speed to be 90pi miles per minute.

if its to be in hours we would then times that by 60

5400pi mph

IF thats a workable method. which I aint all that sure about :)