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Mathematics 20 Online
OpenStudy (anonymous):

Help please: Point P moves with angular velocity "w" on a circle of radius "r". In each case, find the distance "s" traveled by the point in time "t" w= 10rad/sec, r = 6 ft, t = 2 min

OpenStudy (anonymous):

v= w*r

OpenStudy (mertsj):

\[2 \min \times\frac{10rad}{1 \sec}\times\frac{1revolution}{2pi rad}\times\frac{12pi feet}{1 revolution}\times\frac{60 \sec}{1 \min}=7200 feet\]

OpenStudy (anonymous):

ok hold on let me think ^^!

OpenStudy (anonymous):

is there an easier way to do this?

OpenStudy (anonymous):

in my book, the angular velocity formula is: w=theta/t (where theta measured in radians)

OpenStudy (anonymous):

can i use that formula some how ?

OpenStudy (anonymous):

@Mertsj hi ?

OpenStudy (mertsj):

Yes you can. I was trying to help you understand why it is what it is.

OpenStudy (mertsj):

They gave you the angular velocity when they said w=10 rad/sec

OpenStudy (anonymous):

i think i got the answer but im not sure

OpenStudy (anonymous):

just plug it in the formulas i guess, to find v then use v=rw which is 6x10=60

OpenStudy (anonymous):

then when u got v=60 just multiply that with t=2, 60x2=120

OpenStudy (anonymous):

i dont know, cuz there's a similar problem to that in here, but the t is in second

OpenStudy (mertsj):

The linear velocity can be found by the formula linear velocity = angular velocity times radius. In your case linear velocity = 10 rad/sec times 6 feet. That would be 60 feet per second. in two minutes, there are 120 seconds so multiply that by 60 feet per second. 60(120)=7200 feet

OpenStudy (anonymous):

oh 60x120

OpenStudy (anonymous):

so if t=1 minute then i just take 60(60) ?

OpenStudy (mertsj):

You have to pay attention to the units, yes. If you have feet in 1 second and you want feet in 1 minute you would multiply by 60

OpenStudy (anonymous):

oh i see.. cool :D thanks Mertsj

OpenStudy (mertsj):

yw

OpenStudy (anonymous):

one more thing..

OpenStudy (mertsj):

You must be watching "The Five" That's what they always say at the end of their show.

OpenStudy (anonymous):

lol ^^!

OpenStudy (anonymous):

Is there anyway that i can graph this using the calculator? y=cos1/3x ?

OpenStudy (anonymous):

in radian btw

OpenStudy (mertsj):

I can on mine. Doesn't yours have a setting for radians or degrees?

OpenStudy (anonymous):

I have TI-84 and yes it does have radian and degrees

OpenStudy (mertsj):

Then enter it like y =cos ((1/3)x)

OpenStudy (mertsj):

Otherwise the calculator might think it is 1 over 3x

OpenStudy (anonymous):

oh nice

OpenStudy (mertsj):

Did it work?

OpenStudy (anonymous):

but it look different from the answer

OpenStudy (anonymous):

how do i zoom in to fit 1cycle only ?

OpenStudy (mertsj):

Define the domain

OpenStudy (anonymous):

it look totally different >.<

OpenStudy (anonymous):

btw the period of the graph tells when the graph stop/end right ?

OpenStudy (mertsj):

The graph goes on and on forever but the period covers 1 complete cycle. In the case of y = cos(1/3)x, the period is 6 pi or 1080 degrees

OpenStudy (mertsj):

So you could set your domain from -3pi to 3 pi or from 0 to 6 pi. That would give you a complete cycle.

OpenStudy (anonymous):

for all the graphs ? or just this one ?

OpenStudy (mertsj):

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