Average Velocity of a particle moving in a circle over a given interval


by IAmPat
Tags: circle, particle, vectors
IAmPat
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#1
Feb17-11, 01:32 PM
P: 29
1. The problem statement, all variables and given/known data




2. Relevant equations

d = 2.5
c = pi*d = 7.854
velocity/s ?= c * 2 = 15.7079


3. The attempt at a solution
Since PR is 1/4 of the circle and the particle moves around the circle 2 times per second, I thought the average velocity would be 1/8th of the velocity that it's traveling. I'm really confused on this one.

EDIT: I only need to solve for (a), that's why I only have data pertaining to that. Not worried about B and C, but any intuition on those is welcome.
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zhermes
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#2
Feb17-11, 01:40 PM
P: 1,262
What is the definition of average velocity?
IAmPat
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#3
Feb17-11, 01:44 PM
P: 29
Quote Quote by zhermes View Post
What is the definition of average velocity?
The change in x over time. The change in X is -1.25 right? and the time is 1/8 of a second... .125

dx/dt = -1.25/.125 = 10?

I'm not following

gneill
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#4
Feb17-11, 01:51 PM
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Average Velocity of a particle moving in a circle over a given interval


If someone were to show you two snapshots, the first showing the particle at P at time zero, and the second at point R at a time 1/8 second later, and you had no idea about the circular course, or what path the particle took to get from P to R, how would you go about finding the average velocity?
gneill
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#5
Feb17-11, 01:52 PM
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Quote Quote by IAmPat View Post
The change in x over time. The change in X is -1.25 right? and the time is 1/8 of a second... .125
Why just x?
IAmPat
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#6
Feb17-11, 02:11 PM
P: 29
Quote Quote by gneill View Post
Why just x?
What do you mean why just x?

The change in x is -1.25 and the change in time is 0.125 seconds, right?
gneill
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#7
Feb17-11, 02:13 PM
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Quote Quote by IAmPat View Post
What do you mean why just x?

The change in x is -1.25 and the change in time is 0.125 seconds, right?
So the particle didn't move along the circular path, but instead traveled a straight line along the x-axis and ended up at the origin? Is point R at the origin?
IAmPat
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#8
Feb17-11, 02:20 PM
P: 29
Quote Quote by gneill View Post
So the particle didn't move along the circular path, but instead traveled a straight line along the x-axis and ended up at the origin? Is point R at the origin?
It's at Y=1.25 and X=0. If I were to draw a straight line from P to R, the angles would be 45, 90 and 45 right? and the equation for the line would be:

1.25^2 + 1.25^2 = C^2
1.5625 + 1.5625 = C^2
3.125 = C^2
C = 1.767767

So .... The vector is (1.767767, 135 degrees)?
gneill
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#9
Feb17-11, 02:26 PM
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Okay! So you've got a distance vector. That gives you the displacement (magnitude of the vector) and the direction. Apparently the displacement happened in 1/8 second. So what's the average velocity vector?
IAmPat
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#10
Feb17-11, 02:38 PM
P: 29
Quote Quote by gneill View Post
Okay! So you've got a distance vector. That gives you the displacement (magnitude of the vector) and the direction. Apparently the displacement happened in 1/8 second. So what's the average velocity vector?
So if the displacement is 1.767767 and happened in 1/8th. I assume I multiply the displacement by 8 to get the velocity per second...

So 14.1421 m/s?
gneill
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#11
Feb17-11, 02:43 PM
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Yup!

Remember though, that velocity is a vector. So you want to retain the angle information in your answer.
IAmPat
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#12
Feb17-11, 02:45 PM
P: 29
Quote Quote by gneill View Post
Yup!

Remember though, that velocity is a vector. So you want to retain the angle information in your answer.
Awesome. Your socratic method of teaching pissed me off to no end but it worked, and I'm very thankful for your help
gneill
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#13
Feb17-11, 02:55 PM
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Quote Quote by IAmPat View Post
Awesome. Your socratic method of teaching pissed me off to no end but it worked, and I'm very thankful for your help
Heh. Bet you'll remember how to find the average velocity though!

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