Maths question help

Mario95

Today I did my Pre-Leaving Cert Maths (P 2) exam and I came across a very strange question:

Q7 The London Eye is the most popular paid tourist attraction in Britain with over 3.5 million visitors annually.

The design principle behind the London Eye is very similar to that of a Ferris wheel but its actually an observation wheel. The wheels 32 passenger capsules are attached to the external circumference of the wheel and are rotated by electric motors. Each excursion lasts the length of time it takes to complete one revolution.

The observation wheel has diameter 120m and the centre of each passenger capsule is 4m from the external circumference of the wheel. The centre of a passenger capsule is 6m above ground level at the point that passengers enter the capsule.

The wheel does not usually stop to allow passengers on, as the rate of movement is slow enough to allow visitors to walk on and off moving capsules. The wheel can complete 2.5 revolutions per hour if it does not stop.

(a) What is the maximum height that the centre of a passenger capsule reaches above ground level and how long does it take a capsule to reach this height if its wheel hasn't stopped?
(b) Find the angle of rotation of a passenger capsule from the point of entry to the capsule, to after 5 minutes
(c)(i) Write down the terms of t, the angle of rotation of a passenger capsule at any given time, where t is time in minutes after a person enters the capsule, if it hasn't stopped
(ii) Hence, by drawing a sketch or otherwise, find an expression for height h in metres of the centre of the capsule above ground level, in terms of t
(iii) The wheel does not normally stop to allow passengers to embark and disembark and, therefore, the speed of the wheel is constant. Suggest a reason why this is desirable

The weird question is c) part ii).
It looked super complicated so I just skipped it to save time.

I went home and after about an hour I came up with a solution which required some knowledge of vectors and integration (In P2 ).
That way is obviously too complex, so I am probably missing something.
I got

h(t) = -64cos((pi * t) / 12) + 70

.

Can anyone come up with a simpler solution (not involving integration/vectors)?

Tesco TripleChicken

I got 11.666t=h
i figured out it moves up 11.666m every minute. Not sure if it's what they were expecting but im pretty sure it's a right way of doing it.

RacingSilver

This looks fairly easy unless I'm completely missing the point!

Wheel does a full revolution in 24 minutes.
Therefore in 1 minute, it makes an angle of 360/24 degrees i.e. 15 degrees.
So at any time t, it makes an angle of 15t degrees.
This is the answer to c (i).

c(ii),
At any time t, angle = 15t degrees, so draw a diagram illustrating this fact.
The angle at the centre is 15t.
The hypotenuse is 64.
The adjacent side is (70 -h),

Therefore Cos(15t) = (70-h)/64
=> h = 70 -64Cos(15t)


Ps - the previous answer h = 11.666t is not right, this gives a height of 140m when the person is at the top of the wheel.
Also at t = 0, this answer gives h = 0, which is not correct.
The height is clearly not increasing at the same rate per minute.

Mario95

RacingSilver wrote: »

This looks fairly easy unless I'm completely missing the point!

Wheel does a full revolution in 24 minutes.
Therefore in 1 minute, it makes an angle of 360/24 degrees i.e. 15 degrees.
So at any time t, it makes an angle of 15t degrees.
This is the answer to c (i).

c(ii),
At any time t, angle = 15t degrees, so draw a diagram illustrating this fact.
The angle at the centre is 15t.
The hypotenuse is 64.
The adjacent side is (70 -h),

Therefore Cos(15t) = (70-h)/64
=> h = 70 -64Cos(15t)


Ps - the previous answer h = 11.666t is not right, this gives a height of 140m when the person is at the top of the wheel.
Also at t = 0, this answer gives h = 0, which is not correct.
The height is clearly not increasing at the same rate per minute.

I knew I must have missed something simple. Thanks! :pac: