## anonymous one year ago Two students start at the northernmost point of the pool and walk slowly around it in opposite directions.(a)If the angular speed of the student walking in the clockwise direction (as viewed from above) is 0.050 rad/s and the angular speed of the other student is 0.015 rad/s, how long does it take before they meet? By using (0.015)t+(0.05)t=2pi I got t to equal 96.7 seconds (b) At what angle, measured clockwise from due north, do the students meet? (c) If the difference in linear speed between the students is 0.23 m/s, what is the radius of the fountain?

1. anonymous

For c I tried using the equation v=2pir/T using 96.7 seconds but realized thats not the full period to get around the fountain. I need to know b but I have no idea how to get the angle in the problem. Any help is appreciated

2. fretje

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3. fretje

from use of following relation $\alpha = \omega * t$ angle alpha = angular velocity * time

4. fretje

for student 1, S1 i called him/her: 0.05rad/s * 96.7s = 4.835 rad that is in degrees 4.835 rad * 360°/(2pi) = 277° for answer c, i can use the relation for tangential speed and angular velocity $v _{\tan} = \omega * radius$ we know omega 1 and omega 2 from student S1 and S2

5. fretje

so with both equations and for both the same radius i get when i divide the one by the other: $\frac{ v _{1} }{ v _{2} } = \frac{ \omega _{1} }{ \omega _{2} }$ i also have $v_{1} - v _{2} = 0.23 m/s$ i can solve both equations for v2 and i get $v _{2} = 0.09857 m/s$ so v1 $v _{1} = 0.23 m/s +v _{2} = 0.3285 m/s$

6. fretje

in previous i didnt mention this: $\omega _{1}/\omega _{2} = 3.3333333$ but we knew that. Then from $r = \frac{ v _{2} }{ \omega _{2} } = 0.09857 / 0.015 = 6.57 m$

7. anonymous

Thank you so much!!!!!

8. fretje

i hope it is correct.

9. anonymous

it was! Thaank you!