A ferris wheel has height = 100 ft and completes 1 revolution in 3 minutes at a constant speed. Compute the speed of a rider in the ferris wheel.
Ferris wheel = circle modeled by x= cos t ; y = sin t
radius = 50 ft; freq = 3^-1 min circle equation--> $ x= 50 \cos{3t} $ ; $ y= 50 \sin{3t} $
speed (s) $ = \sqrt{ (\frac{dx}{dt})^{2}+ (\frac{dy}{dt})^{2}}$
$ x'(t) = -150 \sin{3t} $
$ y'(t) = 150 \cos{3t} $
thus
s = $\sqrt{ 150^{2}\sin^{2}{3t}+ 150^{2}\cos^{2}{3t} }$
s = $\sqrt{ 150^{2}(\sin^{2}{3t}+ \cos^{2}{3t}) }$
s = $ 150 \sqrt{\sin^{2}{3t}+ \cos^{2}{3t} }$
s = 150 feet/min
Please verify my interpretation (feet/min) as well. Thank you
$$v=r\omega=50ft\times\frac{2\pi}{3\times60}rad/s\approx0.531m/s\text{ or }104.52 ft/min$$
$$\omega=\frac{d\theta}{dt}=^*\frac{\Delta\theta}{\Delta t}\text{ *as it moves with constant velocity}$$