In summary, the conversation discusses the use of polar coordinates and their unit vectors in expressing position, velocity, and acceleration. The unit vectors er and eθ are used to denote the direction of the vector, with er being the unit vector in the direction of r and eθ being perpendicular to er. The position vector is denoted as r=1(er, θ), and to find velocity and acceleration, the derivatives of the position equation with respect to time must include the derivatives of the unit vectors as well.
#36
Panphobia
435
13
Yea after I wrote the wrong answer, I thought about it and changed it. But the velocity would just be this right? v = 5*(dθ/dt)*Eθ and the acceleration would be 0?
Yes, the velocity vector is [tex]\vec v = 10 \hat e _{\theta} [/tex] in this case.
The acceleration is the time derivative of the velocity. It is not zero, as the velocity changes direction. You need the derivative of Eθ now. Go back to #20 and figure out how is it related to Er.
ehild
#38
Panphobia
435
13
I only needed the velocity vector in polar coordinate form. Thank you so much for the help!