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# I was delegated to investigate the physics aspects and principles involved with a theme park ride.

Extracts from this document...

Introduction

Khaled HamidPage

Physics investigation

Introduction:

As part of a physics investigation, I was delegated to investigate the physics aspects and principles involved with a theme park ride. In the past it was usually roller coasters which were thoroughly examined, but I have chosen a ride that is a little far from the conventional; the ‘Tidal wave’, which is a giant log (wooden carriage) hurtling down from a high point down to a steep slope and in to water.

Upon investigating the roller coasters, the main aim was to identify two aspects of physics involved and to further discuss these principles/aspects in order to find the effect, its purpose and how it influences the ride itself and people on board it.

Identification of physics involved:

There are many aspects of physics involved in the ride but I am only set to discuss two. However, it would be relevant to mention how the other aspects of physics involved and how they affect each other.

The first aspect of physics which I noticed on the tidal was the acceleration and deceleration involved needed to bring the ride to a rest position. I also have chosen to discuss the force involved to ‘push’ the carriage down the steep slope.

Explanation of principle and its uses to the ‘Tidal wave’:

The acceleration is a measure of the rate at which the velocity of a particle is changing.

...read more.

Middle

S = ut + ½ at2

Account of one aspect (2): Where is the formula derived from and how is it related to ‘Tidal Wave’?

Think about the carriage moving along a straight line at the bottom of the hill with a constant acceleration (a). Suppose that its initial velocity, at a time is (u). After a further time (t), its velocity has increased to (v). From the definition of acceleration as (change in velocity)/(time taken) we have :

a = (v – u)/t, or re arranging to give v = u + at

From the definition of average velocity v = s/t we can find the distance by forming/rearranging to establish s = ut. The average velocity v is written in terms of the initial velocity u and final velocity v as:

V = u + v/2

And using the previous equations for v,

v = ( u + u + at )/2  =  u + at/2

Substituting this we have:

S = ut + ½ at2

The right hand side of the equation is the sum of two terms. The ut term is the distance the carriage has travelled in time t; if it had been travelling with a constant speed u, and the ½ at2 terms is the additional distance travelled as a result of the acceleration

Diagram and analysis:

v = 0ms-1     s = 17m   u = 14.13 ms-1   a = ?   t = 3.47 sec

mass of carriage = 1500kg

number of people on carriage = 11

15 m

27 deg

The figures are not 100% correct.

...read more.

Conclusion

However I cannot suggest anymore improvements because the ride is a ‘log’ ride and only needs slopes and hills to generate a greater kinetic energy and increase acceleration.

One problem that could arise is a safety problem. The train has no seatbelts or a safety function to make sure the rider will stay put in the train, therefore I suggest that they add waist bands to the carriage to prevent young children with smaller mass from actually falling out. Although this has not happened, and the engineers have carefully calculated out the physic involved in preventing this, I think this would be a safer thing to do. It will also generate some psychological safety, juts like rollercoaster’s with shoulder pads.

Bibliography:

For this investigation I used my background knowledge as the main source of reference and information but the following books were also of aid:

• Horner Salters
• AS/A2 physics – Moe Beijin
• Revise AS Physics - Letts

For the colour diagram, I got the jpeg from the following url;

- http://www.glenbrook.k12.il.us

...read more.

This student written piece of work is one of many that can be found in our GCSE Forces and Motion section.

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