Particle going in a circle with speed U problem

AI Thread Summary
The discussion focuses on a physics problem involving a particle moving in a circle with speed U, requiring the drawing of graphs for position, velocity, and acceleration over time. Participants clarify that x(t) and y(t) represent the particle's position, while u(t) and v(t) denote the x and y components of velocity, respectively. The derivatives du/dt and dv/dt correspond to the acceleration components. The equations for circular motion are introduced, including the parametric equations x = Rcos(θ) and y = Rsin(θ), along with the relationship between angular speed and time. Understanding these concepts is crucial for solving the problem effectively.
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Homework Statement


Hello all,
I am having some trouble with answering the problem below, mostly because I do not know what the letters stand for and what kind of graph is meant to be drawn. Any help on this would be greatly appreciated. Thanks

For a particle going in a circle with speed U, draw a graph showing:
x(t)
y(t)
u(t)=dx/dt
v(t)=dy/dt
du/dt
dv/dt


Homework Equations





The Attempt at a Solution


 

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In your diagram you've indicated x and y axes. x(t) will be the x position with respect to time, y(t) the y position with respect to time. So to begin, draw two more graphs with t as the horizontal axis and place x on the vertical axis of one and y on the vertical axis of the other.

Clearly u(t) and v(t) are meant to be the x-velocity and y-velocity. du/dt is the derivative of y-velocity with respect to time (what do we usually call the derivative of a velocity?). You'll want to draw similar graphs for these items.
 
Are you serious? You don't know what the letters stand for? That should have been the first thing you learned! x and y are the coordinates of that particle in some coordinate system. u and v, as is said in the problem are the derivatives (do you know what a derivative is?) of x and y and so are the x and y components of the velocity vector. du/dt and dv/dt are the x and y components of acceleration.

It helps to know that a circle, with center at (0, 0) and radius R has equation x^2+ y^2= R^2. And, of course, sin^2(x)+ cos^2(x)= 1. That means that parametric equations for the circle are x= Rcos(\theta), y= R sin(\theta). Now, the circumference of the circle is 2\pi R and if the particle is moving at constant speed U, it will complete the circle (moving through an angle of 2\pi radians) in time \frac{2\pi R}{U} so has angular speed \frac{U}{R}. That means the angle, at time t, is \theta= \frac{U}{R}t.
 
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