Circular Track (First Meeting and Second Meeting)

[Chaim]

Natsu and Gray both run around a circular track. They both started to run at the same time. Gray started at the westernmost point of the track and runs counterclockwise, and Gray takes about ninety-five seconds to run one lap of the track. Natsu runs clockwise and he passes Gray for the 1st time after eleven seconds. Natsu runs at 3 meters per sec (speed) and takes 80 seconds to run 1 lap of the track.

What are Natsu's coords when he passes Gray for the 1st time?
What I did:
I knew that it took Gray about 95 seconds to run 1 full lap (which is equal to 2pi radians). So 2pi/95 rad/sec is his angular speed.
While Natsu's angular speed is then 2pi/80 rad/sec
Therefore speed = angular speed * radius
I knew one of the speed was 3, so 3 = (2pi/80)*r, and solved for r to be 240/2pi = 120/pi
Then I knew θ = angular speed * time, so I found the θ to be 11 * 2pi/95

There I put everything into this: rcos(θ+pi), rsin(θ+pi)
Where I got the coords: (-28.52, -25.40) for when Natsu meets Gray for the first time. (So I got the answers correct here, the second question is where I need help).

What are Natsu's coords when he passes Gray for the second time?
This is the part I got lost on:
Ok, so I know Natsu's angular speed is 2pi/80
While Gray's angular speed is 2pi/95

What I tried doing was to find the time, through θ = angular speed * time
So then θ = 2pi/80 * t (for Natsu)
And 2pi - θ = 2pi/95 * t (for Gray)

Though I intended to solve for t (time) and plug in t+11 to the coordinate equation: rcos(θ+pi), rsin(θ+pi)
Though I got lost on how to solve for the angle this time... which then gave me troubles finding time.

I knew that the angle had to be something like 2pi - θ at least... which just basically means the angle they swept through to meet the second time.
Anyway, what should I do from here? Unless there's another easier way.

Answers for the second part: (34.256, 16.898)

Thanks

 

[Ishuda]

The position of Gray is
\(\displaystyle P_g = (r cos(a_g t + \pi), r sin(a_g t + \pi)\)
with s_g equal to Gray's angular speed. That is, at time zero, Gray is at (x,y)=(-r,0) or (r,[FONT=MathJax_Math]θ[/FONT][FONT=MathJax_Main]) [/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]π[/FONT][FONT=MathJax_Main])[/FONT]. I believe that is what you have/meant.

You can do the same kind of equation for Natsu
\(\displaystyle P_g = (r cos(a_n (11-t) + 11 a_g + \pi), r sin(a_n (11-t) + 11 a_g + \pi)\)

Now when they first met they were on the same cycle, that is both angles were between 0 and 360 degrees. However the next time they meet they may be on a different cycle. That is Gray may be in the 360 to 720 degree cycle and Natsu still in the 0 to 360. if that were the case their actual angles would differ by 2\(\displaystyle \pi\) but, because of the cyclic nature of the sine and cosine, their position would be the same. So, second meeting
\(\displaystyle a_g t + \pi - (a_n (11-t) + 11 a_g + \pi) = 2\pi\)
\(\displaystyle (a_g + a_n) t - 11 (a_g + a_n) = 2 \pi\)
or
\(\displaystyle t = 11 + \frac{2 \pi}{a_g + a_n}\)

 

[Chaim]

The position of Gray is
\(\displaystyle P_g = (r cos(a_g t + \pi), r sin(a_g t + \pi)\)
with s_g equal to Gray's angular speed. That is, at time zero, Gray is at (x,y)=(-r,0) or (r,[FONT=MathJax_Math]θ[/FONT][FONT=MathJax_Main]) [/FONT][FONT=MathJax_Main]= [/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]r[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math]π[/FONT][FONT=MathJax_Main])[/FONT]. I believe that is what you have/meant.

You can do the same kind of equation for Natsu
\(\displaystyle P_g = (r cos(a_n (11-t) + 11 a_g + \pi), r sin(a_n (11-t) + 11 a_g + \pi)\)

Now when they first met they were on the same cycle, that is both angles were between 0 and 360 degrees. However the next time they meet they may be on a different cycle. That is Gray may be in the 360 to 720 degree cycle and Natsu still in the 0 to 360. if that were the case their actual angles would differ by 2\(\displaystyle \pi\) but, because of the cyclic nature of the sine and cosine, their position would be the same. So, second meeting
\(\displaystyle a_g t + \pi - (a_n (11-t) + 11 a_g + \pi) = 2\pi\)
\(\displaystyle (a_g + a_n) t - 11 (a_g + a_n) = 2 \pi\)
or
\(\displaystyle t = 11 + \frac{2 \pi}{a_g + a_n}\)

Ah ok, thanks
Though I was wondering, what would their angle be then at the second meeting?

 

[Ishuda]

Ah ok, thanks
Though I was wondering, what would their angle be then at the second meeting?

You know the time, the angular velocity, the fact that it linear and the angle at a particular time. In fact, the Position vector is
(x, y) = P(r,\(\displaystyle \theta\)) = r [cos(\(\displaystyle \theta\)), sin(\(\displaystyle \theta\)) ]
as given previously.

 

[Chaim]

Oh oops.

Ah I got it now.

Thanks