- #1

Easy_as_Pi

- 31

- 0

## Homework Statement

Each of the following equations in parts (a)-(e) describes the motion of a particle having the same path, namely the unit circle x

^{2}+ y

^{2}= 1. Although the path of each particle in part (a)-(e) is the same, the behavior, or "dynamics," of each particle is different. For each particle, answer the following questions.

i. Does the particle have constant speed? If so, what is its constant speed?

ii. Is the particle's acceleration vector always orthogonal to it velocity vector?

iii. Does the particle move clockwise or counterclockwise around the circle?

iv. Does the particle begin at the point (1,0)?

a. r(t) = cos(t)i + sin(t)j, t[itex]\geq[/itex]0

## Homework Equations

v = dr/dt

a = d

^{2}r/dt

s=|v|

## The Attempt at a Solution

Well, I could scan the page of my textbook that this problem is on, but I'll just say it instead: there is not a single example, be it in the text or the practice problems, that deals with anything like this. So, the only attempt I could make was to find the first derivative of r(t), which represents velocity, and then take the second derivative of r(t), which represents acceleration. r'(t) = -sin(t)i + cos(t)j, r"(t) = -cos(t)i + -sin(t)j. I know that, for speed to be constant, velocity must be constant, which means acceleration must equal zero. I have no idea how to show or determine if acceleration equals zero for every value t[itex]\geq[/itex] 0. My book says the answer is yes, it does have a constant speed equal to 1. However, acceleration does not remain constantly equal to 0. If t = pi, accel is -1, if t = pi/4, accel is -sqrt2. I know I must be something wrong here, but I'm very lost.