Hello MATHDUDE2 Originally Posted by

**MATHDUDE2** Hi, I'm not sure how to approach this question.

Q: A Ferris wheel has a diameter of 56m, and one revolution took 2.5 min to complete. Riders could see Niagara Falls if they were higher than 50m above the ground. Sketch cycles of a graph that represents the height of a rider above the ground, as a function of time, if the rider gets on at a height of 0.5m at t=0 min. Then determine the time intervals when the rider could see Niagara Falls.

I have the graph sketched and the resulting function I got is y= 28sin(4pi/5(x+pi/2))+28.

If a rider gets on at 0.5m at t=0, the equation would be y=28sin(4pi/5(x+pi/2))+28.5?

Where do I go from here? Can someone please show me what I have to do?

The answers are: 0.98min<t<1.52min, 3.48min<t<4.02min, 5.98min<t<6.52min.

Thanks!!!

Your second equation is better than the first, assuming that the rider gets on at the lowest point of the cycle, for this is now $\displaystyle y = 0.5$, whereas it is $\displaystyle y = 0$ in your first equation. You have correctly used the period of 2.5 to give the factor $\displaystyle \frac{4\pi}{5}$, but the 'phase angle' that gives the initial value is not correct. So if we write:$\displaystyle y = 28 \sin \Big(\frac{4\pi}{5}t+c\Big)+28.5$

When $\displaystyle t = 0, y = 0.5$, and hence:$\displaystyle 0.5=28\sin c+28.5$

$\displaystyle \Rightarrow -1=\sin c$

$\displaystyle \Rightarrow c=-\frac{\pi}{2}$ (or any equivalent angle)

So the function is$\displaystyle y = 28 \sin \Big(\frac{4\pi}{5}t-\frac{\pi}{2}\Big)+28.5$

So you now need to solve$\displaystyle 50 = 28 \sin \Big(\frac{4\pi}{5}t-\frac{\pi}{2}\Big)+28.5$

$\displaystyle \Rightarrow 28 \sin \Big(\frac{4\pi}{5}t-\frac{\pi}{2}\Big)= 21.5$

$\displaystyle \Rightarrow \sin \Big(\frac{4\pi}{5}t-\frac{\pi}{2}\Big)=\frac{21.5}{28}$

$\displaystyle \Rightarrow \frac{4\pi}{5}t-\frac{\pi}{2}=0.8755, \pi - 0.8755, 2\pi +0.8755, ...$

Can you complete it now?

Grandad