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the height h meters of a seat on a Ferris wheel after t minutes is given by

 

h(t)= -23.5cos(0.4t) +25

For t ≥ 0

 

a) find the initial height of the seat

 

To find the initial height, t=0, so;

 

h(0)= -23.5cos(0.4(0)) +25

 

h(0)=1.5m

 

once a passenger's seat is more than 30m above the ground, there are no trees in view and they can take unobstructed photographs of the nearby city.

 

b) Given that passengers only complete one rotation on the Ferris wheel, calculate how long they will have to take unobstructed photographs of the nearby city

 

I'm unsure of this question. I first tried making h(t)=30m which ended up with me having two graphs

 

y=-23.5cos(0.4t) +25,

 

and y=30

 

so I decided to find the intersections between these for the first wave.

 

t=4.46, and t=11.3

 

but I end up hitting a wall, I don't know what to do now. Any ideas?

 Feb 4, 2020
 #1
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+2

h(t)= -23.5cos(0.4t) +25

 

Your answer to (a) is correct  !!!

 

 

(b)  we have this inequality

 

h(t)= -23.5cos(0.4t) +25  >  30

 

Look at the graph here :    https://www.desmos.com/calculator/utxpmoarvo

 

Notice that the graph is above 30m  between t ≈ 4.463 sec  and  t ≈ 11.245 sec

 

Subtracting.....

 

11.245 - 4.463   =

 

6.782  sec   =  the time the ride is above 30m

 

Your approach was correct  !!!

 

cool cool cool

 Feb 4, 2020

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