For how many minutes is a passenger above 100 feet?

Question

The function h below gives the height above the ground of a passenger on a Ferris wheel t minutes after the ride begins. During one revolution of the Ferris wheel, for how many minutes is the passenger at least 100 feet above the ground?

h(t) = 64 - 46cos[(π/5)t], where 0 ≤ t ≤ 0

How would I go about figuring this out? In case you need it to check, the answer is 2.14.

? · Accepted Answer

When t = 0, we have cos(0) = 1

so, h(0) = 64 - 46 = 18

When cosx = -1.....i.e. it's minimum value we gain a maximum height of 64 + 46 = 110

Now, cos(π) = -1....so, we require when πt/5 = π.....i.e. t = 5 mins

This is the time taken to travel half of the complete revolution....so the total time taken is 10 minutes.

We need to find the time when the passenger is first at 100 feet and the time when they are at 100 feet for a second time.

=> 100 = 64 - 46cos(πt/5)

=> cos(πt/5) = -36/46

=> πt/5 = 2.4696

i.e. t = 3.93 minutes

So, up to this time the passenger is rising up to 100 feet and during the last 3.93 minutes they are falling from 100 feet

=> they are at least 100 feet from the ground for 10 - (2 x 3.93) = 2.14 minutes.

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For how many minutes is a passenger above 100 feet?

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