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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
thats not right
ive been thinking about this one for a few days :)
is the 15 miles a perpendicular distance from the wall?
|dw:1332801016388:dw|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
all this has me so confused
im wondering if the rotation speed can be component based to measure the wall speed. |dw:1332801109026:dw| 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 :)
yeah thats not it either. sheesh
it would help if you could gives us reasons as to why not ... other than simply becasue the program your inputing it into says its wrong :/
Yeah me too. But webworks doesnt give a reason. It just tells me its wrong. Helpful, right??
yeah, there aint a "how to do a similar" problem option is there?
nope. The calc department couldnt have found a worse program to use.
They reference a problem in the book but thats about it.
I think it's reasonable to use the instantaneous speed r dtheta when r= 15 sqrt(2) dtheta= 6pi rad/1min or 360pi rad / hour so r dtheta = 15*360*sqrt(2)*pi in mph about 23991 mph
yeah that one isnt right either.
'course I could be wrong.
i think thats abt what i had last time tan(t) = w/15 15sec^2(t) t' = w' 15*2*t' = w' 30*t' = w'
180pi*60 = 10800pi mph was last time
maybe the program is wrong
is it spose to be exact? or estimated with 3.14?
How about starting with the wall modeled as a vertical line in polar coords r cos(A)= 15 and take the derivative
#27, same method I used the first time around
15*6pi*(sqrt(2))^2 = 180pi * 60 = 10800pi mph
ironically, thats twice the speed i thought of with the circle strategy :)
then the program is wrong
ok got it. forgot pi
..... i think i just had a synapse burst :/
So what is the final answer??