1
$\begingroup$

For this question I am mainly concerned about points A and B on the image below and the image below hopefully helps illustrate my question.

If point B is fixed and A has to move in a strait line in any direction at a specific velocity, 1 meter per second, then what is the average rate of A approaching or leaving point B for the entire time it is in the top or upper circle (fixed travel distance equal to the distance between A and B, lets say 10 meters)? For this average, please consider all directions A can go in.

The image below has the small circle in the center because in the scenario where A starts by moving towards B, once it passes the edge of the small circle then it is going away from point B.

If you can solve the above is there a formula that could be constructed to give this average rate of travel to or from B for different speeds of A or for different fixed distances that A travels?

Criteria

-Distance from Point A to B is for the sake of the question 10 Meters.

-Points traveling from point A to the edge of the circle are traveling 1m/s in a strait line

Help with Question

To further describe my question, in the image below a point leaves point V along line VD, while traveling that point will be getting closer to point Z at a until it hits point E then it will start traveling away from point Z. What is the rate that this point approached and left point Z? More so, what is the average rate of points going all directions from point V relative to point Z?

enter image description here

$\endgroup$
3
  • $\begingroup$ How familiar are you with calculus? $\endgroup$ Oct 3, 2014 at 0:10
  • $\begingroup$ Not familiar, it has been a while $\endgroup$
    – Joe
    Oct 3, 2014 at 0:14
  • $\begingroup$ But i do understand formulas i just have trouble putting them together myself $\endgroup$
    – Joe
    Oct 3, 2014 at 0:16

1 Answer 1

0
$\begingroup$

Well, here's one approach you can take, which will require a bit of trigonometry and calculus. If I use any unfamiliar terminology or techniques, let me know.

Note that, regardless of the direction of point $A$'s travel, it will remain within the top circle for exactly $10$ seconds. In order to find the average speed of $A$ relative to $B,$ we must find the average of the change in distance from $A$ to $B$ during those $10$ seconds--taken over all possible directions--and then divide by $10$ seconds. The latter of these two is simple enough. The calculus comes in on the first part.

Obviously, at the start, point $A$ is $10$ meters from point $B,$ but how can we find the final distance, so that we can find the average change in distance? Well, one way to go about it is to choose an angle $\theta$ in radians in the interval $[0,2\pi),$ measured counterclockwise from the segment from $A$ to $B$ (in their initial positions). Choosing a direction uniquely determines such an angle, and choosing such an angle uniquely determines a direction. Note, then, that after $10$ seconds, point $A$ will be $10\sin\theta$ meters to the right and $10-10\cos\theta$ meters above point $B$. The distance from $A$ to $B,$ then, after $10$ seconds, is given by $$\sqrt{(10\sin\theta)^2+(10-10\cos\theta)^2},$$ which, with some manipulation of trig identities, can be shown to be equal to $10\sqrt{2-2\cos\theta}.$ The change in distance from $B$, then, assuming that $A$ starts along the angle $\theta,$ is $$10\sqrt{2-2\cos\theta}-10.\tag{$\star$}$$ All that remains, then, is to integrate the function in $(\star)$ over $0\le\theta<2\pi,$ then divide the answer by $10$ seconds.

See what you can do with that. Again, let me know if there's anything you're stuck on, and I'll reply when I'm able.

$\endgroup$
3
  • $\begingroup$ i got (8/pi)-2 = .54648 $\endgroup$
    – Joe
    Oct 3, 2014 at 1:07
  • $\begingroup$ is this saying that the average rate a point is leaving point B is .54648m/s? $\endgroup$
    – Joe
    Oct 3, 2014 at 1:08
  • $\begingroup$ Not quite. I'm not quite sure why, but you divided by $\pi$ unnecessarily. The answer should be $8-2\pi$ meters per second. $\endgroup$ Oct 3, 2014 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.