## anonymous 3 years ago describe the motion of a point on a wheel that has a centre 4m off the ground, has radius of 15 cm, makes a full rotation every 10 seconds and starts at its highest point. use y=asink(t-b)+c model or cosine function.

1. anonymous

c=4 and a=.15 ... is that part clear?

2. anonymous

a=15 not .15..

3. anonymous

false. :) continuing... $f=\frac{ 1 }{10}$ $\omega = 2 \pi f$ $w= \frac{ \pi }{ 5 }$

4. anonymous

agree so far?

5. anonymous

yes i got that part

6. anonymous

so the phase...

7. anonymous

we want it to be at the highest point when t=0... sin =1 when theta = pi/2 pi/5 *(t-b) = pi/2 pi/5(-b) = pi/2 b= 5/2

8. anonymous

$y = .15 \sin (\frac{ \pi }{ 5} (t+\frac{ 5 }{2})) +4$

9. anonymous

would it be appropriate to keep(t-b) it at pi/5 *(t) + pi/2 ?

10. anonymous

you can... that's not the form they ask you to use though...

11. anonymous

whoops, should say b=-5/2 up there... typo

12. anonymous

oh okay. & also for cosine model it would just be $y=.15\cos(\frac{ \pi }{ 5 }t) +4$ right??

13. anonymous

yes

14. anonymous

why would it be -5/2 and not 5/2 ? :S

15. anonymous

why is b= -5/2? is that what you're asking?

16. anonymous

OH WAIT.. nvm.. you were referring to your reply before the equation. nvm. lol

17. anonymous

thanks!

18. anonymous

sure:)