Question
Asked Oct 24, 2019
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a roller coaster track is in the shape of the parabola y= 80 -(x^2/20), where the lengths are in meters. A roller coaster begins at height 0 with an inital speed of forty meters per second. Thereafter, the speed of the roller coaster satisfies conservation of energy. ( We ignore the effects of friction) Where on the track is the magnitude of the normal acceleration greatest?

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Expert Answer

Step 1

The vertical position causes the potential energy of the coaster car. Therefore, from conservation of energy.

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KE, PE KE, PE 1 mgh =mv + mgh, f v2ghv +2gh (40 m/s)+2(9.81 m/s)(0 m) - v2 +2(9.81 m/s)y (40 m/s)-2(9.81 m/s2)y 1

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Step 2

Consider the expression of the track of the roller coaster and differentiate it with respect to x.

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dy d 80 20 dx dx 2х = 0 - 20 x 10

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Step 3

Again differentiate with re...

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d2y 1 10

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