Unit 23: Applied mathematics for engineering
LO2, 3, 4, 5
Roller Coaster
Instructions and answers for teachers
These instructions should accompany the OCR resource ‘Roller Coaster’ activity which supports Cambridge
Technicals in Engineering Level 3.
Version 1
The Activity:
In this lesson element, learners are provided with a description of the profile of a simple roller coaster track. In Activity
1, learners are required to produce a mathematical model and solve this using particular values given. In Activity 2,
learners are expected to use standard energy equations to estimate the speed of a roller coaster carriage as it
descends the track.
The mathematics required for this task is fairly basic; however, several LOs are involved.
The spreadsheet ‘Rollercoaster_Data’ will plot the profile of the roller coaster track based on calculated parameters.
This activity offers an
This activity offers an
opportunity for English
opportunity for maths
skills development.
skills development.
!
Suggested
timings:
!
Task 1: 2–3 hours
Task 2: 1–2 hours
Activity 1
This task demonstrates the process of creating a mathematical model of a problem from a written
narrative.
Solutions
(i)
It should be clear from the description that the first section of track is a straight line starting at the origin
and with a gradient of 1 i.e. h1 = x Since the second section has a positive gradient (=1) at x = 15 and a
negative gradient at x = 30 this must be part of an inverted parabola with a maximum point between
x = 15 and x = 30 The gradient of the third section of tack starting at x = 30 continually diminishes to zero
when x = 50
The sketch therefore should be as follows. At this stage the sketch need not be accurate; however,
salient features and values should be indicated.
Version 1
Height (m)
Profile of roller coaster track
20
18
16
14
12
10
8
6
4
2
0
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50
Horizontal distance (m)
(ii)
We are given the height of the second section of track at x = 15 and at x = 30 This provides two
equations:
a152  b15  c  15
a302  b30  c  12
We also know that the gradient of the track at x = 15 is equal to 1.
The gradient of the second section of track is found by differentiating as follows.
Since h2  ax2  bx  c , gradient =
dh2
 2ax  b
dx
This provides a third equation:
30a  b  1
These three equations can be solved to find a, b and c.
One way to do this is to eliminate c from the first two equations by subtracting one from the other:
a(152  30 2 )  b(15  30)  3
 675a  15b  3
Dividing through by 3 gives:
 225a  5b  1
Version 1
We now have two linear simulations equations in two unknowns, a and b.
30a  b  1
 225a  5b  1
Now eliminate b.
150a  5b  5
 225a  5b  1
Adding the two equations gives:
 75a  6
and so
a
6
 0.08
 75
Substituting a into
30a  b  1 gives b  1  30 (0.08)  1  2.4  3.4
Learners should understand that these equations could also be solved using matrix methods as follows.
1  a 1
 30
 225  5 b   1

   
1
1  1
a  30
b    225  5 1
  
 
a 1   5  1 1
b  75 225 30 1
 

 
a
1
(5  1)  0.08
75
b
1
(225  30)  3.4
75
We can now use the values of a and b in h2  ax2  bx  c to calculate c
Since h = 15 when x =15, we have:
15  0.08152  3.4 15  c
c  15 0.08152  3.4 15  18
So the equation for the second section of track is :
h2  0.08x 2  3.4x  18
Version 1
An alternative method for calculating a, b and c is to directly solve the three simultaneous equations:
a15 2  b15  c  15
a30 2  b30  c  12
a30  b  1
Using matrix notation this becomes:
225 15 1  a  15
900 30 1  b   12

   
 30 1 0  c   1 
1
a  225 15 1 15
b   900 30 1 12
  
  
 c   30 1 0  1 
a   0.001 0.001  0.015  15
b    0.030  0.030 0.675  12
  
 
 c   0
0.225  6.750   1 
a    0.08 
b    3.40 
  

 c   18.00
For the third section of track we are given:
h3  dx3  ex 2  fx  g
We are given h3  12 when x  30 and h3  0 when x  50 . This provides two equations:
d 30 3  e30 2  f 30  g  12
d 50 3  e50 2  f 50  g  0
The gradient of the third section of track is given by:
dh3
 3dx 2  2ex  f
dx
The gradient of the second section of track at x  30 is:
dh2
 2ax  b  2(0.08)30  3.40  1.4
dx
Version 1
Since the gradients are continuous at the joins we have:
3d 302  2e30  f  1.4
We also are given that the gradient at x = 50 is zero and so:
3d 502  2e50  f  0
We now have four linear equation in four unknowns, d and e, f and g.
d 30 3  e30 2  f 30  g  12
d 50 3  e50 2  f 50  g  0
3d 30 2  2e30  f  1.4
3d 50 2  2e50  f  0
By elimination methods the solutions are found to be:
d  0.0005
e  0.095
f  5.75
g  112.5
The equation for the third section is therefore:
h3  0.0005x 3  0.095x 2  5.75x  112.5
(It is suggested that solution by elimination and matrix methods is demonstrated.)
(iii)
The highest point on the track occurs in the second section. This is where the gradient is zero and so:
dh2
 2ax  b  0
dx
With a  0.08 and b  3.4 we have:
 0.16x  3.4  0 giving x  21.25
The highest point on the track is therefore:
Version 1
h  0.08  21 .25 2  3.4  21 .25  18  18 .125 m.
(iv) See spreadsheet Rollercoaster_Data.
Activity 2
Learners should be taught about kinetic and potential energy and the principle of energy conservation.
Solutions
For this part of the lesson the following formulae are required:
Potential energy PE  mgh J
Kinetic energy
KE 
1 2
mv J
2
Where:
m kg is mass
g ms–2 is acceleration due to gravity
h m is height above a fixed reference level
v ms–1 is speed
We shall assume that g = 9.81.
It is also assumed that the conservation of energy principle applies. Since no energy is lost due to friction
or aerodynamic drag PE+KE is constant at all points on the track.
(i)
At the highest point on the track PE  1000  9.8118.125  177806.25 J
(ii)
1
2
At the highest point on the track (when v = 4 ms–1) KE  1000  4 2  8000 J
(iii)
Total energy at the highest point on the track = PE  KE  177806 .25  8000  185806 .25 J
Version 1
1
2
Total energy at height 12 m: 1000  9.8112  1000v 2
1
2
Since PE + KE is constant 1000 9.81 12  1000v 2  185806.25
Solving this gives v = 11.6693 ms–1 correct to 4 decimal places.
(iv)
When the carriage reaches the end of the track PE = 0 and so:
1
1000v 2  185806 .25
2
Solving this gives:
v  19.2773 ms–1 (about 43 mph)
A possible extension to this task is to replace the assumption that the carriage is modelled by a point
mass with a model involving sphere, a disk or a ring. This will introduce further energy into the equations
due to rotational inertia. Learners will of course need to be provided with relevant formulae.
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