2 C only please, show work
Related questions
Question
2 C only please, show work
![2. (From an old exam) A toy train travels counterclockwise around a circular track of radius R. Taking the
center of the circle as the origin, the position vector of the train as a function of time is
F(t) = RCOS (wt)i + R sin (wt)ĵ
where is a constant.
a) Find an expression for the train's velocity vector as a function of time.
b) Find an expression for the train's speed at any given time. Simplify it as much as you can. (Recall that
cos² 0+ sin² 0 = 1, for any 0.)
c) Your result for the speed in part (b) should be a constant. What does this imply about the train's acceler-
ation?
d) Find an expression for the train's acceleration vector as a function of time. Is it consistent with your
answer to part (c)? (To check this, try sketching the two vectors at a couple of different times.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90ae1aab-f107-47ea-b140-00f7b3a4760d%2F83460be9-2f55-4324-9f8e-61c7f077dc10%2F7tiqsim_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2. (From an old exam) A toy train travels counterclockwise around a circular track of radius R. Taking the
center of the circle as the origin, the position vector of the train as a function of time is
F(t) = RCOS (wt)i + R sin (wt)ĵ
where is a constant.
a) Find an expression for the train's velocity vector as a function of time.
b) Find an expression for the train's speed at any given time. Simplify it as much as you can. (Recall that
cos² 0+ sin² 0 = 1, for any 0.)
c) Your result for the speed in part (b) should be a constant. What does this imply about the train's acceler-
ation?
d) Find an expression for the train's acceleration vector as a function of time. Is it consistent with your
answer to part (c)? (To check this, try sketching the two vectors at a couple of different times.)
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
Introduction:
The case here is case of uniform circular motion. We know that attributes of motion perpendicular to each other does not affect the motion in perpendicular directions. In uniform circular motion, the speed is constant. Only the direction of velocity changes keeping magnitude that is speed constant.
Step by step
Solved in 2 steps
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)