Question

Please help! An athlete starts at point A and runs at a constant speed of 5.80m/s around a round track 250m in diameter, as shown in figure.

1. Find the x component of this runner's average velocity between points A and B.
2. Find the y component of this runner's average velocity between points A and B.
3. Find the x component of this runner's average acceleration between points A and B.
4. Find the y component of this runner's average acceleration between points A and B.

Answer 1

let say the center of the circle is O, then the x component from a to b is aO = 125m
and y component is Ob = 125 m, too.
since the runner runs at a constant speed , the acceleration =0 at both axises

Answer 2

I tried 0 for the answer but it says thats not it.

Answer 3

oops, sorry, friend , I forgot, the acceleration of a circle moving is a= v^2/2 = you calculate it, ok?

Answer 4

So, 5.8^2/2?

Answer 5

yes, I think so.

Answer 6

Hm, well its saying 16.82 is not the answer either. Thanks for helping me though!

Answer 7

hihihi... sorry for my helpless.

Answer 8

It's okay, I've tried answering the problem 5 times and haven't found a solution yet so any help is beneficial :)

Answer 9

(1) The x velocity in A is 0, and in B is 5.80, so the average velocity in this path is,
\[v_a=\frac{5.80+0}{2}=2.90\ m/s\]
Could you check it?

Answer 10

The answer to question 1 and 2 are both 3.69 m/s. Now I'm trying to solve question 3 and 4, please let me know if you have any input, my assignment is due by midnight. Thanks!

Answer 11

Not sure how it is 3.69, I just failed the first question and it gave me the answer.

Answer 12

If you know the velocities from the previous questions, then accelerations should be easier, using,
\[a=\frac{v_f+v_0}{2}\]if I'm not wrong.

Answer 13

So, 3.69+0/2? That equals 1.845 and it is saying that's incorrect.

Answer 14

Let me try something.

Answer 15

Oh, man. I just realized. For the first and second they want you calculate via this formula,
\[v_x=\frac{\Delta x}{\Delta t}\]
If you go from A to B, the distance in the x axis is diameter/2=R=125 m. But you need to calculate the time to travel in one quarter of the circle. So you need the period,
\[T=\frac{2\pi R}{v}\]
And then divide by four, t=33.85 s. So the average x velocity is,
\[v_x=125/33.85=3.69\ m/s\]

Answer 16

For the third, the average acceleration should use the true x velocities and the time needed to travel the quarter circle,
\[a_x=\frac{5.80-0}{33.85}=0.171\ m/s^2\]
For the fourth,
\[a_y=\frac{0-5.80}{33.85}=-0.171\ m/s^2\]

Answer 17

I hope it helps ;).

Answer 18

Thanks, that was correct!