# Ferris wheel!

• Jul 12th 2008, 11:40 PM
Mercy99
Ferris wheel!
A spider attached itself to the point P on the side of car C when the point P was at its lowest point at 1.00pm.

The height, h meters, of the point P above ground level, at the time t hours after 1.00pm is given by;

h(t) = 62 + 60sin( (5t-1)pi / 2)

Two questions to the information;
i. Find the time, after 1.00pm, when P first reaches a height of 92 meters above ground level.

ii. Find the number of minutes during one rotation that P is at least 92 meters above ground level.

All help is much appreciated.
• Jul 13th 2008, 12:04 AM
Chris L T521
Quote:

Originally Posted by Mercy99
A spider attached itself to the point P on the side of car C when the point P was at its lowest point at 1.00pm.

The height, h meters, of the point P above ground level, at the time t hours after 1.00pm is given by;

h(t) = 62 + 60sin( (5t-1)pi / 2)

Two questions to the information;
i. Find the time, after 1.00pm, when P first reaches a height of 92 meters above ground level.

ii. Find the number of minutes during one rotation that P is at least 92 meters above ground level.

All help is much appreciated.

For part (i), set $h(t)=92$:

$92=62+60\sin\bigg(\frac{(5t-1)\pi}{2}\bigg)$

Now find t:

$30=60\sin\bigg(\frac{(5t-1)\pi}{2}\bigg)$

$\implies \frac{1}{2}=\sin\bigg(\frac{(5t-1)\pi}{2}\bigg)$

$\implies \sin^{-1}\bigg(\frac{1}{2}\bigg)=\frac{(5t-1)}{2}$ **

$\implies \frac{\pi}{6}=\frac{(5t-1)\pi}{2}$

$\implies \frac{\pi}{3}=(5t-1)\pi$

$\implies \frac{1}{3}=5t-1$

$\implies \color{red}\boxed{t=\frac{4}{15} \ \text{hours} = 16 \ \text{minutes}}$

For part (ii), find the other time value when $h(t)=92$

Let's go back to **

$\implies \sin^{-1}\bigg(\frac{1}{2}\bigg)=\frac{(5t-1)}{2}$

This time, we will let $\sin^{-1}\bigg(\frac{1}{2}\bigg)=\frac{5\pi}{6}$

$\implies \frac{5\pi}{6}=\frac{(5t-1)\pi}{2}$

$\implies \frac{5\pi}{3}=(5t-1)\pi$

$\implies \frac{5}{3}=5t-1$

$\implies \color{red}\boxed{t=\frac{8}{15} \ \text{hours} = 32 \ \text{minutes}}$

Thus the total time that $h(t)$ is at least 92 meters above the ground is $\frac{8}{15}-\frac{4}{15}=\color{red}\boxed{\frac{4}{15} \ \text{hours}=16 \ \text{minutes}}$.

I hope you can follow this.

If you need me to clarify things, feel free to ask.

--Chris

EDIT : Used wrong arcsine value...
• Jul 13th 2008, 12:17 AM
Mercy99
Very well done. I appreciate your help Chris, again :)