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= (finalvelocity-initialvelocity)/time

In your question you only want the y component, so

ay= (vfy-viy)/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.

NOT FACTUAL

To find the y component of the runner's average acceleration between points A and C, we need to determine the change in vertical velocity and divide it by the time taken.

The initial vertical velocity, Viy, is -5.40 m/s (as we take downwards as the negative direction).
The final vertical velocity, Vfy, is 5.40 m/s (as the runner returns to the same height).

The average acceleration is given by:
ay = (Vfy – Viy) / time

Since the runner completes a full lap around the track to return to point C, the time taken would be the same as the time taken to complete one lap. We can calculate the time taken using the equation: time = distance / speed.

The distance covered in one lap is equal to the circumference of the track: C = πd = π * 250 m = 785.4 m.

Therefore, the time taken for one lap is:
time = distance / speed = 785.4 m / 5.40 m/s = 145.22 s (approx.)

Now we can calculate the average acceleration:
ay = (Vfy – Viy) / time = (5.40 - (-5.40)) / 145.22 = 10.80 / 145.22 ≈ 0.0744 m/s².

Therefore, the y component of this runner's average acceleration between points A and C is approximately 0.0744 m/s².

To find the y component of the runner's average acceleration between points A and C, we first need to find the y component of the runner's initial velocity and final velocity at those points.

Given that the runner is starting at point A and running at a constant speed around the track, the y component of the initial velocity (Viy) at point A can be determined as -5.40 m/s (since downward is considered negative in this case).

To find the y component of the final velocity (Vfy) at point C, we need to consider that the track is a round track with a diameter of 250m. Point C is on the right side of the track. Assuming the runner completes one full lap of the track from point A to point C, the runner would have reached the top point B with a y component of velocity of 5.40 m/s. Then, as the runner continues from point B to point C, the y component of velocity gradually decreases until it reaches 0 m/s at point C. Therefore, the y component of the final velocity (Vfy) at point C can be considered as 0 m/s.

Now we can calculate the average acceleration in the y direction (ay) between points A and C using the formula:

ay = (Vfy - Viy) / time

Substituting the values we found,

ay = (0 - (-5.40)) / time
ay = 5.40 / time

The value of time is not provided in the question, so we cannot calculate the exact value of the y component of the runner's average acceleration between points A and C without the time. However, we can say that the y component of the average acceleration will be positive and equal to 5.40 m/s^2 divided by the time.

Where'd you get 3.44m/s