# Problem with motion in a circular path

## Homework Statement

Calculate the average vector velocity between 0 and pi/4 sec.

## Homework Equations

x=rcos2t
y=rsin2t
Vx=2rsin(2t)
Vy=2rcos2t
Ax=4rcos(2t)
Ay=-4cos(2t)
Circular path is x^2+y^2+r^2

## The Attempt at a Solution

I'm not sure if I am missing something simple or not, What I need to know is what is r and so I initially set the square of the x and the square of the y equations equal to the square r. Pythagorean style. when I solved for r I felt just a little dumb because I got 1=1 which well duh I shoul've saw that coming. Now I'm trying to guess whether I even need r to do an average, or if there is another way to solve for r that I am missing. I also know that x=r when t=0 but when I sub r for x in the position equation I just get 1.

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Chestermiller
Mentor
What are the x and y coordinates of the particle at time t = 0?
What are the x and y coordinates of the particle at time t = π/4?
What is the vector displacement of the particle (in terms of the unit vectors in the x and y directions) between time t = 0 and time t = π/4?
What is the definition of average (vector) velocity?

Chet

At t=0 y=0 and x=r
At t=π/4 I an uncertain because I haven't been able to figure at what r is.
I guess my real problem isn't finding the average but figuring out what r is.

Chestermiller
Mentor
At t=0 y=0 and x=r
At t=π/4 I an uncertain because I haven't been able to figure at what r is.
I guess my real problem isn't finding the average but figuring out what r is.
r is the radius of the circle, which is treated as an algebraic parameter.

chet

ahh duh, ok so then it should look something like this, taking the root of the sum of the x and y coordinates for t=π/4 and the root of the sum of the x and y coordinates at t=0 subtract the final position from the initial, and divide by two. Giving (2-root2)/2.

Chestermiller
Mentor
That's not correct. But, it's hard to tell what you did. Please, just answer, in order, the 4 questions I asked in post #2.

Chet

Is 0.7458 correct?

#1 at t=0, x=r and y=0
#2 at t=pi/4, x=rcos(2pi/4) and y=rsin(2pi/4)
#3 Displacement at x=rcos(2pi/4)-x(or r) and y=rsin (2pi/4) - 0
#4 The average vector velocity is the (final position - initial position) dived by the (final time - initial time).

Chestermiller
Mentor
[
Is 0.7458 correct?

#1 at t=0, x=r and y=0
#2 at t=pi/4, x=rcos(2pi/4) and y=rsin(2pi/4)
What is cos(2π/4) equal to?
What is sin(2π/4) equal to?
#3 Displacement at x=rcos(2pi/4)-x(or r) and y=rsin (2pi/4) - 0
Please express this in vector form using unit vectors or with an ordered pair of components.
#4 The average vector velocity is the (final position - initial position) dived by the (final time - initial time).
please express this in vector form using unit vectors or with an ordered pair of components.
What is the magnitude of this average velocity vector?

Chet

#3 Position= r(0)i+r(1)j for final, and Position=r(1)i+r(0)j for initial

#4 ||v||=√((-1r/(π/4))2+(1r/(π/4))2) --> (4 √(2) √(r2))/pi

Chestermiller
Mentor
#3 Position= r(0)i+r(1)j for final, and Position=r(1)i+r(0)j for initial
This should be written as
Initial position = r i
Final position = r j

Displacement vector = - r i + r j
#4 ||v||=√((-1r/(π/4))2+(1r/(π/4))2) --> (4 √(2) √(r2))/pi
What is $\sqrt{r^2}$ equal to?

They did not ask for the magnitude of the average velocity vector. They asked for the average velocity vector itself. This is equal to the displacement vector divided by the time interval pi/4. What do you get for that?

Chet

The average velocity vector would be v=-(4r/π)i+(4r/π)j

Chestermiller
Mentor
The average velocity vector would be v=-(4r/π)i+(4r/π)j
OK. We're done here. Nice job.

Chet

yes! thank you so much for helping me through that problem, and for being so patient.