a tire completes a revolution every 2 seconds. a nail gets stuck in the tire. the height y (in meter) the…

a tire completes a revolution every 2 seconds. a nail gets stuck in the tire. the height y (in meter) the nail is modeled by y = - 9/2cos(πt) + 4.5 where t is the time (in seconds). a) find the height of the nail after 1.3 seconds. b) find the height of the nail after 1/4 of a second. c) what is the time when the nail is 9 meters high? 11) a ferris wheel completes a revolution every 15 minutes. after you step onto the ferris wheel your height y (in feet) of can be modeled by y = 60sin(π/15(t - 10)) + 60, where t is the time (in minutes). a) how high are you off the ground after 25 minutes? b) how high are you off the ground after 300 seconds? c) how long have you been on the ride if you know you are at a height of 120 meters? 12: the depth of the water in a harbor, h, is modeled by h = 15sin(π/12(t + 5)) + 40 where measured in meters and t is the time (in hours) after midnight.
Answer
Explanation:
Step1: Solve part (a) of the nail - height problem
Substitute $t = 1.3$ into $y=-\frac{9}{2}\cos(\pi t)+4.5$. $y =-\frac{9}{2}\cos(1.3\pi)+4.5$. Since $\cos(1.3\pi)=\cos(\pi + 0.3\pi)=-\cos(0.3\pi)\approx - 0.9511$, then $y=-\frac{9}{2}\times(- 0.9511)+4.5=\frac{9\times0.9511}{2}+4.5 = 4.280 + 4.5=8.7805$ meters.
Step2: Solve part (b) of the nail - height problem
Substitute $t=\frac{1}{4}$ into $y =-\frac{9}{2}\cos(\pi t)+4.5$. $y=-\frac{9}{2}\cos(\frac{\pi}{4})+4.5$. Since $\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\approx0.7071$, then $y=-\frac{9}{2}\times0.7071 + 4.5=-3.182 + 4.5 = 1.318$ meters.
Step3: Solve part (c) of the nail - height problem
Set $y = 9$ in $y=-\frac{9}{2}\cos(\pi t)+4.5$. $9=-\frac{9}{2}\cos(\pi t)+4.5$. First, subtract 4.5 from both sides: $9 - 4.5=-\frac{9}{2}\cos(\pi t)$, so $4.5=-\frac{9}{2}\cos(\pi t)$. Then, $\cos(\pi t)=-1$. We know that $\cos(\pi t)=-1$ when $\pi t=(2n + 1)\pi$, $n\in\mathbb{Z}$. So $t = 2n+1$, $n\in\mathbb{Z}$. Since $t\geq0$, the first non - negative solution is when $n = 0$ and $t = 1$ second.
Step4: Solve part (a) of the Ferris - wheel problem
First, note that $t = 25$ minutes. Substitute $t = 25$ into $y = 60\sin(\frac{\pi}{15}(t - 10))+60$. $\frac{\pi}{15}(t - 10)=\frac{\pi}{15}(25 - 10)=\frac{\pi}{15}\times15=\pi$. Since $\sin(\pi)=0$, then $y=60\times0 + 60=60$ feet.
Step5: Solve part (b) of the Ferris - wheel problem
First, convert 300 seconds to minutes. Since 1 minute = 60 seconds, 300 seconds $=5$ minutes. Substitute $t = 5$ into $y = 60\sin(\frac{\pi}{15}(t - 10))+60$. $\frac{\pi}{15}(t - 10)=\frac{\pi}{15}(5 - 10)=\frac{\pi}{15}\times(-5)=-\frac{\pi}{3}$. Since $\sin(-\frac{\pi}{3})=-\frac{\sqrt{3}}{2}$, then $y=60\times(-\frac{\sqrt{3}}{2})+60=-30\sqrt{3}+60\approx60 - 51.96 = 8.04$ feet.
Step6: Solve part (c) of the Ferris - wheel problem
Set $y = 120$ in $y = 60\sin(\frac{\pi}{15}(t - 10))+60$. $120=60\sin(\frac{\pi}{15}(t - 10))+60$. Subtract 60 from both sides: $60=60\sin(\frac{\pi}{15}(t - 10))$. Then $\sin(\frac{\pi}{15}(t - 10)) = 1$. We know that $\sin\theta=1$ when $\theta=\frac{\pi}{2}+2k\pi$, $k\in\mathbb{Z}$. So $\frac{\pi}{15}(t - 10)=\frac{\pi}{2}+2k\pi$. Divide both sides by $\pi$: $\frac{1}{15}(t - 10)=\frac{1}{2}+2k$. Multiply both sides by 15: $t - 10=\frac{15}{2}+30k$. $t=\frac{15}{2}+10 + 30k=\frac{15 + 20}{2}+30k=\frac{35}{2}+30k$. For the first non - negative solution when $k = 0$, $t = 17.5$ minutes.
Answer:
a) (Nail) $8.7805$ meters; (Ferris - wheel) 60 feet b) (Nail) $1.318$ meters; (Ferris - wheel) $8.04$ feet c) (Nail) 1 second; (Ferris - wheel) $17.5$ minutes