Posted by Josh on Wednesday, September 13, 2006 at 10:56pm.
An athlete starts at point A and runs at a constant speed of 5.40 m/s around a round track 250 m in diameter.
Find the x component of this runner's average velocity between points A and B. (answer is 3.44 m/s)
Find the y component of this runner's average velocity between points A and B. (answer is 3.44 m/s)
Find the x component of this runner's average acceleration between points A and B. (answer is 0.149 m/s^2)
Find the y component of this runner's average acceleration between points A and B. (answer is 0.149 m/s^2)
Find the x component of this runner's average velocity between points A and C. (answer is 3.44 m/s)
Find the y component of this runner's average velocity between points A and C. (answer is 0 m/s)
Find the x component of this runner's average acceleration between poitns A and C. (answer is 0 m/s^2)
Find the y component of this runner's average acceleration between points A and C. THIS IS THE PART THAT IS A PROBLEM FOR ME!
POINTS START ON THE LEFT AND GO CLOCKWISE (so, the top point is B, the right is C, the bottom is D, and the left is A)
Ok speed = magnitude of velocity = sqrt((v_x)^2 + (v_y)^2) = 5.40m/s
Average velocity is (v_f+v_0)/2
Since A and C are opposite one another the velocities should cancel when you add them to find average acceleration. The same should be true for B and D.
Does that seem plausible?
This is a vector problem, you have to work it in hoizontal and vertical components. THe part you are looking for is vertical only.
Start with the definition of acceleration:
acceleration= (finalvelocityinitialvelocity)/time
In your question you only want the y component, so
ay= (vfyviy)/time If I have the picture correct, then (check it)
Viy is 5.40m/s (minus means down)
Vfy is 5.40m/s.
ay=(5.4 (5.4))/time...
so, do the math
For this function it probably would help to have a position, velocity and acceleration function.
If we place the track so that the center coincides with the origin,
then the position function looks like (x,y)=(r*cos(f(t)),r*sin(f(t))). Where r is the radius and f(t) is some expression so that when t=0 the runner is at A, or (r,0), B is (0,r), C =(r,0) and D is (0,r).. Since d=250m, r=125m
The velocity function looks like
(v_x,v_y)=(r*sin(f(t)),r*cos(f(t)))f'(t), and acceleration looks like
(a_x,a_y)=(r*cos(f(t)),r*sin(f(t)))f'(t) + (r*sin(f(t)),r*cos(f(t)))f"(t)
The f(t) function is most likely linear so the second deriv. should be 0 and the second term is 0.
I'm not sure if you're using calculus in your physics course or not, but you should definitely be using the trig functions. These expressions should look 'vaguely' familiar. I didn't recall them until I had logged off yesterday.

Physics  me, Sunday, March 6, 2016 at 9:34pm
Where'd you get 3.44m/s
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