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**Numbersarehard****Novice**- Registered: 2024-05-05
- Posts: 4

This question is from the Pre-Calculus textbook by Nelson (previously McGraw-Hill Ryerson) on pager 279, question 19 (you can find the textbook online)

19. A Ferris wheel with a radius of 10 m rotates once every 60 s. Passengers get on board at a point 2 m above the ground at the bottom of the Ferris wheel. A sketch for the first 150 s is shown (in the textbook).

a) Write an equation to model the oath of a passenger on the Ferris wheel, where the height is a function of time.

b) If Emily is at the bottom of the Ferris wheel when it begins to move, determine her height above the ground, to the nearest tenth of a meter, when the wheel has been in motion for 2.3 min

Math is hard, enjoy it while it's easy

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**Numbersarehard****Novice**- Registered: 2024-05-05
- Posts: 4

sorry was in the wrong section

Math is hard, enjoy it while it's easy

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 10,588

I'm trying this without having looked up the textbook.

Normally I'd give you a diagram but without my sketching program at the moment so you'll have to make your own.

Draw a circle centre C and a line straight down to point A, at the bottom of the circle. A cannot be on the ground because the wheel would scrape and anyway the passenger is 2m up when they get on so A must be 2m up

Continue CA down to point B on the ground.

Now mark a point P to show where the passenger is after a small rotation . Let angle ACP be alpha degrees

If the passenger gets on at 2 metres (presumably the lowest point) then the centre is 12m above the ground

Draw a horizontal line from P to line CA meeting it at D.

CD = 10cos(alpha) so

DB = height = 12 - 10cos(alpha)

Wheel does 360 in 60s = 6 degrees per sec.

So angle rotated after t seconds is 6t = alpha

So you can put that together to get h in terms of t.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**mathxyz****Member**- From: Brooklyn, NY
- Registered: 2024-02-24
- Posts: 1,053

Bob wrote:

I'm trying this without having looked up the textbook.

Normally I'd give you a diagram but without my sketching program at the moment so you'll have to make your own.

Draw a circle centre C and a line straight down to point A, at the bottom of the circle. A cannot be on the ground because the wheel would scrape and anyway the passenger is 2m up when they get on so A must be 2m up

Continue CA down to point B on the ground.Now mark a point P to show where the passenger is after a small rotation . Let angle ACP be alpha degrees

If the passenger gets on at 2 metres (presumably the lowest point) then the centre is 12m above the ground

Draw a horizontal line from P to line CA meeting it at D.

CD = 10cos(alpha) so

DB = height = 12 - 10cos(alpha)

Wheel does 360 in 60s = 6 degrees per sec.

So angle rotated after t seconds is 6t = alpha

So you can put that together to get h in terms of t.

Bob

Nicely-done. I can use your notes here to solve similar trigonometric applications.

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