Contains

883 words

883 words

Category

Science & Nature

Science & Nature

The modelling of the segment of roller coaster

Rational :

Over the world there are many amusement park accident cause by roller coaster ,the probability of serious injuries caused by motorised cruises is estimated to be1 over 24 million ,however according to a research report, there are 4,423 children were injured in roller coaster every year between 1990 and 2010. Most of them suffered from tissue injuries. As a big fans of roller coaster I very concern of my safety when I play roller coaster ,I would like to design a model of the safest roller coaster in the world, so I will investigate on 3 types of roller coaster which are Accelerator Coasters ( Desert Race), Impulse roller coaster (Screaming Condor), Shuttle roller coaster. In order to letting me know more about the mathematics factor of roller coaster safety.

Methodology:

I am going to research on mechanics formula for the speed ,impact force and safety on the roller coaster, at the same time I will be using factors, differentiation and integration, Quadratic function find out the energy, speed changes, in order to design the safest roller coaster .

Aim to exploration

The properties of roller coasters can be used in mathematical ways, especially the application and knowledge of differentiation. The roller coaster also can be using functions. This requires careful and particular arranging with safe, even a pleasant experience for riders. This investigation aims to design a rollercoaster comprising of different polynomials - the application and knowledge of calculus processes of derivation and function required in this paper. The first and the second derivative of an equation is used to describe the slop, changes, and the shape of the function. These derivative can be used to discover the point of the function. Algebra is required in this, and it can be displayed of the rollercoaster.

To guarantee that the track is smooth, the lines should change easily. The changes between the direct stretches and the parabola happen at focuses W and Z . For the track to be smooth, the straight stretches of the rollercoaster must be digressions to the parabola at the changing focuses. The incline of the last rising is intended to be 0.8 and the slant of the last drop is intended to be - 1.3. With the end goal to disentangle the conditions, P is put at the inception of the chart and P and Q is viewed as on a level plane 40 meters separated. It ought to be noticed that every unit is equal to 1m and that neither the x-axis or y-axis speak to limitations or ground levels?

The first linear equation can be find the the data above.

y=L1(x)=m1 X+K1 where m is the slope of y intercept and k

the slope of L1, m1 is 0.8

L1(x)= 0.8 X+K1

and using the coordinate of W

L1(x)= 0.8 X+K1

0=0.7(0)+K1

0=K1

Hence L1(X)=0.7x

L1 is the line segment which is the W is the point that the rollercoaster up., when the parabola is on going ,the parabola of the function should be X?0.8 so that the equation of the line segment is L1(X)=0.8(X)for X?0.

The second derivative of L1 is :

L1 (X) =0.8 (X)

L1'(X) = 0.8

L1"(X)= 0

So the second derivative is 0 which show that the line is constant slope also mean it is a straight line.

The application of derivatives can be used when the equation need to join together . In this segment of rollercoaster the equation is f(x)=ax^2+ b(x) +C f(x)=ax^2+ b(x) +C

f(0)=a0^2+ b(0) +C

0=C

the original equation should be :

f(x)=ax^2+ b(x) +C

so f(x)=ax^2+ b(x)

L1 , touching at the point W(0,0). the slope of this function and F(X) is the same ,so what ever the point is the , the first derivative of L1's function is: f(x)=ax^2+ b(x)

f'(x)=2ax+ b

At pint W the slope of the graph is 0.8:

f'(x)=2ax+b

f'(0)=2a(0)+b

f'(0)=0.8

0.8=2x(0)+b

0.8=b

?

The original equation should be :

f(X)= ax^2+0.8x

The value f a can be find by using the data of L2 segment. The point of Z is 40 metres horizontally to the right of W(0,0) and the y coordinates of L2 is 40. but we don't know the y value. The second line segment L2is a tangent to the parabola F(X)=ax^2+0.8X, meaning that the slope is the some and the slope of L2 is -1.3, so the slope of L2's parabola is

(x)=ax^2+ 0.7(x)

f'(x)=2ax+ b(x)

The slope of sub -1.3 into the equation is :

f'(X)= -1.3

-1.3=2ax+0.8

-2=2ax

-1= ax

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