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12MME-1

Mathematical Methods IA1: Problem Solving and Modelling Task Introduction The purpose of this problem solving, and modelling task is to evaluate the appropriateness of using mathematical functions to model the distance travelled, velocity and acceleration of an amusement park ride. Three functions with unknown parameters have been provided to model the path of a ride (in terms of height and time) that drops a vertical distance of 10m and then ascends 10m to its original starting position. The application of these models will be evaluated in terms of feasibility and the level of thrill. The models provided represent the distance travelled by the ride over time. They are; 1. (= (=( )(− − ) 2 2. (((t) = () – em (t− j )+ e−m(t − j) −k 3. 𝐻 (t) = 2

Observations and Assumptions -

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The models presented are functions that show the changes in the ride’s height over time. ‘H’ represents the rides height above the ground in meters and ‘t’ represents the time since the start of the ride. B, C, D, F, M, J and K are variable parameters. The ride starts from a platform positioned exactly 10m above the ground and descends 10 metres before ascending 10m to return to the starting position. When graphed on a cartesian plane, the height-time graph will pass through the known point (0,10). The time taken for the ride to complete has not been specified by the provided models. Thus, each model may take a different time to complete. This will affect the velocity and acceleration and thus the overall excitement of the ride. Assume that weather conditions do not affect the path or time taken for the ride to complete. The task context states that the ride may provide thrill to riders when acceleration is above 8m/ s2. The ride provides a maximum level of thrill when the rides acceleration is 16m/ s2. The acceleration must not exceed 16m/ s2 to keep to strict ride safety requirements.

Mathematical Concepts and Procedures The unknown parameters in each model will be determined using algebraic procedures, and when this is not possible, technology will be used. The online graphing software Desmos will be used to visually see how changing the model’s parameters visually transform the equation.

12MME-1 Calculus procedures will also be used to determine the first and second derivate of each model, which represent the velocity and acceleration functions respectively. The maximum and minimum acceleration and velocity values will be calculated by: Min/ max velocity Min/max acceleration

Solve for ‘t’ when acceleration = 0 Solve for ‘t’ when the derivate of acceleration = 0

Determining the Models H ( t )=b(t−c )

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Model 1;

The first and second derivate of the distance function were found using calculus procedures to show the velocity and acceleration functions respectively. Velocity: dh b (t−c)2=2∗B(t−c )2−1 dt =2� (�−�)

Acceleration; dh 2 b (�−�) dt

=� * � �� (�−�)�−� =2� (�−�)0 = 2�

The second derivate provides a constant value for acceleration as the equation does not include variable parameters. Therefore, a value for acceleration can be set and the equation can be rearranged to find the parameter value ‘b’. The acceleration equation was set to equal 16m/ s2 as this value is provides the highest level of thrill whilst fitting to safety requirements. The equation was solved to find b; 16 = 2 b 8 =b The known point (0,10) was substituted into the distance equation with the b-value to find the parameter values for c; ((𝐻) 10 108 -C C

=(− (− ) 2 =−8 (0− )2 = (−� )2 8 =± 10 = ± 1.11803398875

√

The ‘C’ value determines the horizontal translation applied to the graph. The negative value for ‘C’ was disregarded as a negative value translates the graph by ‘c units’ in the negative direction (to the left). A positive c value is required to position the curve on the positive side of the x-axis. Therefore, the model is;

H (t )=8( t −1.11803398875)2

The distance-time graph and the acceleration graph for the first model is shown below;

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Figure 1: Distance-Time Graph (red line) and Acceleration Graph (blue line)

Height (m)

Time (s) Time (s)

Whilst this model produces a distance-time graph that is fitting to the task requirements, the model’s acceleration equation is unrealistic. The acceleration equation will always produce a constant value for acceleration of 16m/ s2 as the equation does not include any variable parameters. In theory, this ride would be very thrilling as the acceleration is equal to the maximum possible value for the entire duration of the ride. However, it is physically impossible for the ride to immediately reach maximum acceleration directly after the start of the ride. The rides acceleration must build over time as there are currently no launch mechanisms capable of overcoming the gravitational forces and resistance on the ride instantly after it is started. Even the world’s fastest roller coaster takes 1.8 seconds to reach maximum acceleration (BCC, 2012). Thus, model 1 is not feasible and should not be considered in the evaluation moving forward.

12MME-1 Model 2;

H ( t )=d sin (ft )– g

The distance equation was derived twice using calculus procedures to find the equations for velocity and acceleration; Velocity: dh dsin ( ft ) −g=d cos ( ft ) × f ' (t) dt ¿ d cos (ft ) × f =�� ���((�� )

Acceleration: dh dfcos ( ft )=−df cos ( ft ) ×f ' ( t ) dt ¿−dfsin (ft ) × f = −��� ��� (�� )

The equations were entered into Desmos to search for the parameter values required produce a model that fits the required path of the ride and has an acceleration between 816m/ s2. The graphing software found the equation; H ( t )=−10 sin( 1.26 t ) +10 The distance-time graph (red line) and the equations for velocity (Black Line) and acceleration (Blue Line) are shown below in Figure 2;

Height/Velocity/Acceleration

Time (s)

Figure 2

12MME-1 The time taken for the ride to complete was found using Desmos. The curve intersects with the line y = 10 for the second time (this point marks the end of the ride- travelled 10m down and then back up) at 2.493 seconds (shown below in figure 3).

Height (m)

Time (s) Figure 3: Intersection of the displacement equation with the point (y= 10)

Upon initial visual observation, this model is feasible as the displacement equation passes through the known point (0,10) and both the velocity and acceleration are variable over time. Furthermore, the ride meets safety requirements as the acceleration does not exceed 16m/ s2. Model 3;

H ( t) =

e m (t − j) + e−m(t− j) −k 2

Using Desmos, the following equation was generated and graphed; −0.94 (t−3.3 ) 0.94 (t−3.3 ) e +e H ( t) = −1 2 This equation was derived twice using calculus procedures to find acceleration and velocity: Velocity (first derivative): −m ( t − j) ' ' (t − j ) m( t− j ) × f (g )] [ e m × f ( u ) ]+[e +e−m (t − j) dh e − k= 2 dt 2 −m t− j m t− j [ e ( ) × m]+[e ( ) ×(− m )] ¿ 2 −m t− j m ( t − j) ]+[−m e ( ) ] [ me ¿ 2

*u = m(t-j) and g = -m(t-j)

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¿

[ mem (t − j) ]−[m em (t − j )]

2 Acceleration (second derivative): me ( t − j) × f ' (g)] [ ¿¿ m (t− j )× f ' (u)]−[m e−m 2 m(t − j) −m (t − j) − me dh me =¿ dt 2 me [ ¿¿ m (t− j )× m]−[me−m (t − j) ×(−m )] 2 ¿¿ 2 m e 2 [ ¿¿m (t− j)]−[−m e−m ( t − j) ] 2 ¿¿ 2 m(t− j) 2 m (t − j ) m e +m e ¿ 2

*u = m(t-j) and g = -m(t-j)

Model 3 – Distance-time graph (red line), velocity-time graph (blue line) and accelerationtime graph (green line)

Height/Velocity/Acceleration

Time (s) Found using Desmos, the model intersects y = 10 at (6.586, 10) and thus the ride completes 6.586 seconds.

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Again, this model presents a viable option as the path of the ride fits the task requirements. The acceleration does not exceed 16m/ s2.

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Evaluation Model 2 and 3 present models that suit the task requirements and therefore will be considered in the evaluation. The thrill of the ride will be evaluated in terms of the acceleration gained throughout the ride. Model 2 and 3’s acceleration-time graph is shown below. The red curve represents model 2 and the purple curve represents model 3.

Acceleration (m/s2)

Time (s) Model 2 Finish Time (2.493sec)

Model 3 Finish Time (6.586sec)

Both models complete in different times, as shown by the green line on the graph above. Obviously, the time taken for the ride to complete affects the rides velocity (change in speed over time) and then affects the rides acceleration (change in velocity over time).

12MME-1 Percentage of Ride Time above 8m/ s2 acceleration: To evaluate the thrill of the ride, the time that each model produces an acceleration above 8m/ s2 will be evaluated. Neither model produces an acceleration above 16m/ s2 within the restricted domain of [0,10], so there are no issues with the safety of the ride. Model 2: The times that model 2’s acceleration equation produces a value above 8m/ s2 was found algebraically. The acceleration equation was set to equal 8m/ s2 to find the first time where the curve interests the minimum acceleration required for thrill. Acceleration=−df 2 sin ( ft ) −10 ×1.26 ¿ −¿ ¿¿ ¿ 15.876 sin (1.26 t ) Let acceleration = 8m/ s

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8=15.876 sin ( 1.26 t ) 8 sin−1 [ ]= ( 1.26 t ) 15.876 0.5281= ( 1.26 t ) 0.419=t The acceleration is 8m/ s2 after 0.419s. The time that the acceleration falls below 8m/ s2 can be found using the first intercept point (0.419) as a reference point. The sine graph originates from (0,0), and the intercept points with 8m/ s2 occur an equal distance from the maximum value (turning point). The graph falls below an acceleration of 8m/ s2 ‘0.419 units’ before the graph completes ½ of a period. The period of the sine graph of the form y = sin b(x); 2π B Therefore, the period of the acceleration graph for model 2: 2π ������ = f 2π ������ = 1.26 ������ = 4.986655 ������ =

Therefore, acceleration falls below 8m/ s2 0.419 units before ½ the cycle (period) has been completed: 4.986655 −0.419 2 = 2.074

=

12MME-1 Therefore, the rides acceleration is above 8m/ s2 between [0.419, 2.074]. Desmos was used to validate these calculations. This period is equal to 1.655 seconds out of the rides total time of 2.493 seconds (ride total ride time found using demos). Ride time above 8m/ s2 = second intercept – first intercept = 2.074 – 0.419 = 1.655 seconds As a percentage, model 2’s acceleration is above the minimum threshold for thrill for 66.39% of the total time of the ride. Percentage of ride where acceleration >8m/ s2 =

time above threshold ∗100 total ridetime 1.655 = ∗ 100 2.493 = 66.39%

Model 3: Desmos was used to determine that model 3 produces an acceleration above 8m/ s2 between [0, 0.222] and [6.378, 6.617]. This is equal to 0.430 seconds above the threshold out of a possible 6.586 seconds of total ride time. Percentage of ride where acceleration >8m/ s2 =

time above threshold ∗100 total ridetime 0.461 ∗ 100 = 6.5856 = 6.92%

Therefore, model 2 has a much greater level of thrill than model 3 as the ride surpasses the minimum threshold for riders to experience thrill for a higher percentage of the ride time. Ride 3 only provides thrill for a small percentage of the ride. Max/min. acceleration Each model’s global minima and maxima for acceleration were then evaluated. These values were found algebraically and by analysing the graphs on Desmos. Maximum and minimum acceleration both occur when the gradient of the acceleration curve (the derivative of acceleration) is equal to 0 (stationary point). To find the coordinates of model 2’s stationary points, the derivate of acceleration was found and set it equal to 0. 2

A=df sin (ft ) dA =−df 3 cos (ft) dt

12MME-1 −10∗1.26 −(¿ ¿ 3)cos (1.26 t ) ¿¿ ¿ 20.003 cos(1.26 t) t=0 Desmos was then used to find that when t = 0, Acceleration = 10m/ s2. This represents a maximum value. Desmos was also used to find the global minima for model 2 acceleration at (1.2465, 0). The minimum velocity of model 3 was read using Desmos and found to be (3.3, 0.94) and the maximum value was found to be (0.05, 10). This information is summarised below; Model 2 Time (s) Acceleration (m/s2) Minimum 0.0 0 Acceleration Maximum 1.249 15.876 Acceleration

Model 3 Time (s) 3.30

Acceleration (m/s2) 0.884

0.00

9.847

The second model produces a maximum acceleration that is extremely close to the safety limit of 16m/s2 This ride provides the most thrilling experience 1.249 seconds into the ride when the acceleration reaches a maximum of 15.876. Model 3, however, only has a maximum acceleration of 9.847m/ s2 at 0 seconds. This value is only just above the minimum level for thrill of 8m/ s2. The minimum acceleration value of model 3 is lower than that of model 2. This value, however, should not be used to determine the overall level of thrill of each ride. The minimum value for acceleration of each model is the turning point of the acceleration function. The acceleration only occurs at a minimum for a split second and would have negligible effect on the overall thrill of the ride.

Conclusion The mathematical models 2 and 3 present feasible model of the distance, velocity and acceleration of an amusement park ride. Model 1 was disregarded early in the investigation due to its inability to produce a variable acceleration value over time. The models were obtained using both graphing software and algebraic procedures, and then evaluated for practicality (in real-world application) and by the level of thrill they provide to riders (based on acceleration). Model 2 provides the best model of the 3 that were evaluated. The model produces an acceleration curve that is above the minimum threshold for riders to experience thrill (8m/ s2) for the longest time of all models and produces the greatest maximum acceleration value.

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References Sood, S. (2012). The human limits of roller coasters. [online] BBC. Available at: http://www.bbc.com/travel/story/20120627-travelwise-the-thrill-of-death-defying-coasters [Accessed 1 Mar. 2020]....