A) 140 degrees
First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is
T = 32 s
So the angular velocity is
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s[/tex]
Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:
[tex]\theta= \omega t[/tex]
and substituting t = 75 seconds, we find
[tex]\theta= (0.20 rad/s)(75 s)=15 rad[/tex]
In degrees, it is
[tex]15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}[/tex]
So, the new position is 140 degrees from the initial position at the top.
B) 2.7 m/s
The tangential speed, v, of a point at the egde of the wheel is given by
[tex]v=\omega r[/tex]
where we have
[tex]\omega=0.20 rad/s[/tex]
r = d/2 = (27 m)/2=13.5 m is the radius of the wheel
Substituting into the equation, we find
[tex]v=(0.20 rad/s)(13.5 m)=2.7 m/s[/tex]
