Ryan Burdett – LSG07 : Maja – SACE#: 300930G
Maths Studies Folio Task 1
When designing vehicles it is important to analyse the capabilities of the vehicle in order to optimise
its functional efficiency and to make any improvements to the vehicle. Calculus can be used to
calculate the velocity and acceleration of a vehicle, given its displacement over time. The use of
optimisation through calculus can be used to find the optimal speed for minimum running costs over
a freight vehicle. The aim of this investigation is to analyse the specifications of the travel of a rocket
car and the running costs of a freight truck.
Part 1 – Rocket Cars
To find the specifications and capabilities of a Rocket Car, information regarding the car’s
performance when being run on a test track was recorded through its displacement from a starting
point over time. The function of the Rocket Car’s displacement was found to be as follows (where s
is displacement in metres and t is time in seconds):
𝑠(𝑡) = −5𝑡 3 + 20𝑡 2 + 15𝑡
Using first principles as a form of differential calculus it was possible to find both the velocity and
acceleration of the rocket car for the duration of the test, where velocity is the first derivative of
displacement over time (as velocity is the rate of change of displacement) and acceleration is the
second derivative (as acceleration is the rate of change of velocity). They were both calculated as
follows, where s’(t) is velocity of the Rocket Car over time and s’’(t) is the acceleration of the Rocket
Car over time:
𝑓 ′ (𝑥) = lim
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
ℎ→0
𝑠 ′ (𝑡) = lim
(−5(𝑡+ℎ)3 +20(𝑡+ℎ)2 +15(𝑡+ℎ))−(−5𝑡 3+20𝑡 2 +15𝑡)
ℎ
ℎ→0
𝑠
′ (𝑡)
= lim
ℎ→0
𝑠 ′ (𝑡) = lim
ℎ→0
−5(𝑡 3 +3𝑡 2ℎ+3𝑡ℎ2 +ℎ3 )+20(𝑡 2 +2𝑡ℎ+ℎ2 )+15𝑡+15ℎ+5𝑡 3−20𝑡 2−15𝑡
ℎ
−5𝑡 3 −15𝑡 2 ℎ−15𝑡ℎ2 −5ℎ3 +20𝑡 2+40𝑡ℎ+20ℎ2 +15𝑡+15ℎ+5𝑡 3 −20𝑡 2 −15𝑡
ℎ
−15𝑡 2ℎ−15𝑡ℎ2 −5ℎ3 +40𝑡ℎ+20ℎ2 +15ℎ
𝑠
′ (𝑡)
= lim
𝑠
′ (𝑡)
= lim − 15𝑡 2 − 15𝑡ℎ − 5ℎ2 + 40𝑡 + 20ℎ + 15
ℎ
ℎ→0
ℎ→0
𝑠 ′ (𝑡) = −15𝑡 2 + 40𝑡 + 15
__________________________________________________________________________________
𝑓 ′′ (𝑥) = lim
𝑓 ′(𝑥+ℎ)−𝑓 ′(𝑥)
ℎ
ℎ→0
(−15(𝑡+ℎ)2 +40(𝑡+ℎ)+15)−(−15𝑡 2 +40𝑡+15)
𝑠 ′′ (𝑡) = lim
ℎ
ℎ→0
𝑠 ′′ (𝑡) = lim
ℎ→0
𝑠
′′ (𝑡)
𝑠
′′ (𝑡)
= lim
−15(𝑡 2+2𝑡ℎ+ℎ2 )+40𝑡+40ℎ+15+15𝑡 2 −40𝑡−15
ℎ
−15𝑡 2 −30𝑡ℎ−15ℎ2 +40𝑡+40ℎ+15+15𝑡 2−40𝑡−15
ℎ
ℎ→0
= lim
ℎ→0
−30𝑡ℎ−15ℎ2 +40ℎ
ℎ
𝑠 ′′ (𝑡) = lim − 30𝑡 − 15ℎ + 40
ℎ→0
𝑠 ′′ (𝑡) = −30𝑡 + 40
Ryan Burdett – LSG07 : Maja – SACE#: 300930G
When graphed, the functions for displacement, velocity and acceleration over time would look like
this:
Figure 1.1 A graph of the rocket car's displacement, velocity and acceleration over time
To help with understanding the changes of displacement, velocity and acceleration over time, sign
diagrams can be made which show when the values are increasing or decreasing and the local
maxima and minima. Finding the maxima and minima are done by finding when the derivative of the
function (or slope) is equal to zero. Finding the intervals of increase and decrease in values can be
found by substituting values either side of the maxima/minima into the derivative (or slope) function
to find the slope.
𝑠 ′ (𝑡) = −15𝑡 2 + 40𝑡 + 15 = 0
𝑠 ′ (𝑡) = −3𝑡 2 + 8𝑡 + 3 = 0
𝑡=
𝑡=
8±√64+36
6
8±10
18
−2
= 6 𝑜𝑟 6
6
1
= 3 𝑜𝑟 − 3
Values less than zero are unimportant as the test would not have started, therefore they can be
omitted from the sign diagram. Time values above and below 3 (in this case 1 and 4) can be used
here to find the slope of the graph to find if displacement is increasing or decreasing in these time
ranges.
𝑠 ′ (1) = −15(1)2 + 40(1) + 15 = 40
∴ 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
𝑠 ′ (4) = −15(4)2 + 40(4) + 15
𝑠 ′ (4) = −15 × 16 + 160 + 15
𝑠 ′ (4) = −240 + 175 = −65
∴ 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
Figure 1.2 the sign diagram for change in displacement
over time.
The same could be done again to create a sign diagram for change in velocity over time.
𝑠 ′′ (𝑡) = −30𝑡 + 40 = 0
Ryan Burdett – LSG07 : Maja – SACE#: 300930G
30𝑡 = 40
𝑡=
40 4
=
30 3
Time values of 1 and 2 will be used to find where the slope is increasing or decreasing
𝑠 ′′ (1) = −30(1) + 40 = 10
∴ 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
𝑠 ′′ (2) = −30(2) + 40 = −20
∴ 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑠 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
Figure 1.3 The sign diagram for change in velocity over time
This sign diagram shows that interval at which the rocket car is increasing is between 0 seconds and
1.33 seconds.
Since the acceleration formula (second derivative) is linear with a negative slope (-30), acceleration
is constantly decreasing.
With this information the maximum velocity of the rocket car can be found. Having already found
the time where velocity is at its maximum, we can put that value into the formula for velocity to find
maximum velocity.
𝑠 ′ (𝑡) = −15𝑡 2 + 40𝑡 + 15
4
4 2
4
3
3
3
𝑠 ′ ( ) = −15 ( ) + 40 ( ) + 15 = 41.667 𝑚𝑠 −1
This gives a maximum velocity of 41.667 ms-1 or 150 km/h.
From figure 1.1 it can be seen that as time progresses the car travelled away from its starting
position, then at a certain time turned around and comes back towards the starting position. The
specific time for this can be found when the car has a velocity of 0, as this is when it stopped going
forwards and turned around. This happened at 3 seconds after starting. At 4 seconds after starting
the rocket car was heading back towards the starting position.
Figure 1.4 This is a motion diagram showing the speed, location and time of the rocket car during the tests
Ryan Burdett – LSG07 : Maja – SACE#: 300930G
Part 2 – Cost efficiency of a Freight Truck
Given a fuel efficiency and running costs for a freight truck it is possible to calculate the optimal
speed at which the truck should be run to minimise cost and to calculate the cost at any given
average speed. From testing it has been found that a freight truck travelling at a constant speed of
110km/h can get a fuel efficiency of 7/km per litre, losing 0.1km/l fuel efficiency for every km/h
increase in speed. Thus the efficiency can be calculated in terms of speed as follows:
7−0.1(𝑣−110)𝑘𝑚
𝐿
where v is speed in km/h
Given a cost of $1.49 per litre of diesel the cost of fuel for the trip can be calculated as
1.49
1.49
1.49
𝐶𝑓𝑢𝑒𝑙 = 7−0.1(𝑣−110) = 7−0.1𝑣+11 = 18−0.1𝑣
Adding to this is the cost of truck driver wages ($35/hour) and truck maintenance ($9.50/hour).
Since we need this in terms of speed, we can add these costs together and divide by speed v.
𝐶𝑜𝑡ℎ𝑒𝑟 =
35+9.50
𝑣
=
44.5
𝑣
Adding these costs together will result in a total cost for the freight delivery per kilometre. Since the
freight truck is usually used for transport between Adelaide and Sydney, the distance would be
1375km.
𝐶𝑡𝑜𝑡𝑎𝑙 = 1375(𝐶𝑓𝑢𝑒𝑙 + 𝐶𝑜𝑡ℎ𝑒𝑟 )
1.49
𝐶𝑡𝑜𝑡𝑎𝑙 = 1375 ((18−0.1𝑣) +
𝐶𝑡𝑜𝑡𝑎𝑙 =
𝐶𝑡𝑜𝑡𝑎𝑙 =
𝐶𝑡𝑜𝑡𝑎𝑙 =
44.5
)
𝑣
1375(1.49𝑣+44.5(18−0.1𝑣))
𝑣(18−0.1𝑣)
1375(1.49𝑣+801−4.45𝑣)
𝑣(18−01.𝑣)
1375(801−2.96𝑣)
𝑣(18−0.1𝑣)
Assuming a constant speed of 110km/h a total cost of the Adelaide to Sydney trip can be found.
𝐶=
𝐶=
𝐶=
1375(801−2.96(110))
110(18−0.1(110))
1375(801−325.6)
110(18−11)
1375(475.4)
653675
=
770
770
= 848.93
This gives a total cost for the trip as $848.93. Using differentiation it is possible to calculate a speed
which would result in a minimum cost for the trip. This is done by calculating the result when the
derivative is equal to zero.
𝑑𝐶
𝑑𝑣
=
−4070𝑣(18−0.1𝑣)−(24750−275𝑣)(801−2.96𝑣)
(18𝑣−0.1𝑣 2 )2
−4070𝑣(18−0.1𝑣)−(24750−275𝑣)(801−2.96𝑣)
(18𝑣−0.1𝑣 2 )2
=0
−4070𝑣(18 − 0.1𝑣) − (24750 − 275𝑣)(801 − 2.96𝑣) = 0
Use the quotient rule to differentiate
Find where it equals to zero
Multiply both sides by the denominator
Ryan Burdett – LSG07 : Maja – SACE#: 300930G
−73260𝑣 + 407𝑣 2 − 19824750 + 73260𝑣 + 220275𝑣 − 814𝑣 2 = 0
220275𝑣 − 19824750 − 407𝑣 2 = 0
𝑣=
220275±√2202752 −4×407×19824750
814
= 114 𝑜𝑟 427.2
Expand
Simplify
Solve the quadratic
The optimal speed for minimising cost would then be 114km/h as a truck would be incapable of
travelling at 427.2 km/h. As seen in figure 2.1, after a certain speed (around 170km/h) the model
breaks down and does not show logical costs. It shows that if the truck travels at a speed around
$250, the cost would be negative.
Figure 2.1 A graph showing the cost of freight depending on speed
Since it would likely be impossible to maintain a constant speed throughout the trip, the model
would better represent an average speed for the trip. Factors that would affect the reasonableness
of the model would include how accurately the change in speed would affect the change in fuel
efficiency, as it may not be a linear relationship. Another factor would be if the truck was travelling
up or down hills, or how heavy the load of the truck is.
For every additional freight container the truck carries, the fuel efficiency reduces by 0.05km/L,
which would reduce the optimum speed of travel required for minimum cost. This can be shown
mathematically by incorporating this into the cost for fuel formula as follows, where a is the number
of freight containers:
𝐶𝑓𝑢𝑒𝑙=
1.49
18 − 0.1𝑣 − 0.05(𝑎 − 1)
The other costs can then be added to this for a total cost function, then differentiated to give a
model for finding the values of either v to minimise costs.
𝐶𝑡𝑜𝑡𝑎𝑙=
1.49
44.5
+
18 − 0.1𝑣 − 0.05(𝑎 − 1)
𝑣
Ryan Burdett – LSG07 : Maja – SACE#: 300930G
This works best when given a number of freight containers beforehand. As an example we can find
the optimum travelling speed with 5 freight containers.
𝐶𝑡𝑜𝑡𝑎𝑙= 1375 (
1.49
44.5
+
)
18 − 0.1𝑣 − 0.05(5 − 1)
𝑣
𝐶𝑡𝑜𝑡𝑎𝑙= 1375 (
1.49
44.5
+
)
17.8 − 0.1𝑣
𝑣
𝐶𝑡𝑜𝑡𝑎𝑙= 1375 (
𝐶𝑡𝑜𝑡𝑎𝑙=
1.49𝑣 + 44.5(17.8 − 0.1𝑣)
)
𝑣(17.8 − 0.1𝑣)
1375(792.1 − 2.96𝑣)
17.8𝑣 − 0.1𝑣 2
𝑑𝐶 −4070𝑣(17.8 − 0.1𝑣) − (24475 − 275𝑣)(792.1 − 2.96𝑣)
=
=0
𝑑𝑣
(17.8𝑣 − 0.1𝑣 2 )2
−4070𝑣(17.8 − 0.1𝑣) − (24475 − 275𝑣)(792.1 − 2.96𝑣) = 0
−19386647.5 + 217827.5𝑣 − 407𝑣 2 = 0
𝑣=
217827.5 ± √217827.52 − 4 × 407 × 19386647.5
= 112.75
814
This shows a decrease in optimum speed for an increase in freight, meaning that in general an
increase in freight will require a decrease in optimum speed to minimise costs.
Ryan Burdett – LSG07 : Maja – SACE#: 300930G
optimal for each car type and to find a
distance suitable for a fair race between the
vehicles.
Part 3 - What is fast, and how do
different cars perform in different
distance races?
To find the optimum distance for each type of
car, the time taken to reach the maximum
speed would be found by using differentiation
on the supplied velocity models, or when the
value for the function is equal to 1 (this is
because the function gives percentage of top
speed as a decimal, where 1 is 100% speed).
The average acceleration over the time taken
to reach top speed can be used in a formula
for displacement to find the optimum race
distance (see appendix B for further detail).
The top speeds and velocity functions are
listed in table 3.1.
_________________________________________________________________________________
Fast can mean many things, but when it
comes to racing what matters more,
acceleration or top speed? To find a solution
to this question three automotive companies
have provided information describing the
velocities of their cars (Top Fuel Dragsters, Jet
cars and high-performance Streetcars). In this
investigation the acceleration and speed of
the cars will be analysed and compared to
determine which types of race would be
Type of Vehicle
Dragster
Jet Car
Top speed (ms-1)
147
144
Streetcar
120
Percent velocity function
𝑣(𝑡) = 0.733𝑡𝑒 −0.27𝑡
1.05
𝑣(𝑡) =
1 + 100𝑒 −.285𝑡
𝑣(𝑡) = 𝑒 0.025𝑡 − 1
Table 3.1 This table shows the top speed and function for velocity over time of the vehicles
Following this method, the optimum race distance for the vehicles would be calculated as follows
(see appendix B for calculations for other cars):
𝑣(𝑡) = 0.733𝑡𝑒 −0.27𝑡
Take the original equation
𝑣 ′ (𝑡) = (0.733 − 0.19791𝑡)𝑒 −0.27𝑡 = 0
Find the derivative and set to zero
0.733 − 0.19791𝑡 = 0
0.733
𝑡 = 0.19791 = 3.7𝑠
𝑠=
𝑣𝑡
2
=
147×3.7
2
Solve for time to reach maximum velocity
Use top speed and time taken to reach it to find distance travelled
𝑠 = 271.95𝑚
Type of Vehicle
Dragster
Jet Car
Streetcar
Time to reach top speed (s)
3.7
26.67
27.73
Optimum distance (distance to top speed in m)
271.95
1920.2
1663.8
A visual representation of the change in velocity of each of the cars can be seen in the graph in
appendix A.
Ryan Burdett – LSG07 : Maja – SACE#: 300930G
A fair race for the vehicles would be one where the vehicles cover approximately the same distance
in the same amount of time. One method of doing this could simply be by finding the mean of the
optimum race distances for the cars. In this case a fair race distance for the cars could be 1285
metres.
Results
From these calculations the optimum race distances for each have been calculated in terms of
distance taken for the car to reach their maximum speeds. For the Top Fuel dragster, the optimum
racing distance would be approximately 272 metres, where it can reach its top speed of 147 metres
per second, but any distance after this it would only reduce in speed. For the Jet Car, the optimum
race distance would be approximately 1920 metres, where it would reach its top speed of 144
metres per second. Finally, for the high-performance street car, the optimum racing distance would
be approximately 1664 metres, where it would reach a top speed of 120 metres per second.
With these distances it was possible to find a simple, reasonably fair distance for a race between the
vehicles merely by finding the mean of the optimum distances. This resulted in a distance of 1285
metres.
Assumptions
The assumptions made in this investigation were the top speeds of each of the vehicles, which were
found with the online resources in appendix C. Dragster: 147ms-1. Jet Car: 144ms-1. Street Car:
120ms-1.
Also assumed was that there would be no outside influence on the cars and that they would not act
differently to the supplied models.
.
Ryan Burdett – LSG07 : Maja – SACE#: 300930G
Appendix A
This graph shows the percentage of maximum velocities of the cars over time. It can be seen that the dragster reaches top speed much earlier than either
of the other cars, but the top speed is not sustained.
Ryan Burdett – LSG07 : Maja – SACE#: 300930G
Appendix B
The time taken for the cars to reach top speed can be found by finding t when the equations for
velocity are equal to 1. Alternatively, if the function cannot equal 1, the time taken to reach
maximum speed can be calculated by finding the derivative of the function and setting it to 0.
Acceleration is equal to change in velocity over time or:
𝑎=
(𝑣−𝑣0 )
𝑡
The displacement of the cars from their starting point can then be found by using the following
formula:
1
𝑠 = 𝑣0 𝑡 + 𝑎𝑡 2
2
Since the initial velocity of all of the cars is 0, that section can be removed from the equations. When
combined, the equations give the following equation:
1 𝑣
𝑣𝑡
𝑠 = ( ) 𝑡2 =
2 𝑡
2
Calculations for Jet Car and Streetcar:
Jet Car:
𝑣(𝑡) =
1.05
1+100𝑒 −.285𝑡
−0.285𝑡
1 + 100𝑒
𝑒
−0.285𝑡
=
=1
= 1.05
0.05
100
0.05
−0.285𝑡 = ln ( 100 )
𝑡=
ln(
0.05
)
100
−0.285
𝑠=
𝑣𝑡 144 × 26.67
=
2
2
𝑠 = 1920.2𝑚
= 26.67
Streetcar:
𝑣(𝑡) = 𝑒 0.025𝑡 − 1 = 1
𝑒 0.025𝑡 = 2
0.025𝑡 = ln(2)
ln(2)
𝑡=
= 27.73
0.025
𝑣𝑡 120 × 27.73
=
2
2
𝑠 = 1663.8𝑚
𝑠=
Ryan Burdett – LSG07 : Maja – SACE#: 300930G
Appendix C – Resources
Furey, E., n.d., Velocity as a function of acceleration and time, accessed 15 March 2015,
http://www.calculatorsoup.com/calculators/physics/velocity_a_t.php
Stamatis, M. 2008, Martin Stamatis 4.577 329.10 mph (529.63 kmh) drag race, video, accessed 15 March 2015,
https://www.youtube.com/watch?v=CIRSR10F-ws
StreetFire, 2008, 323 MPH TurboJet car blows away Texas Speed Record, video, accessed 15 March 2015,
https://www.youtube.com/watch?v=a6qvmbTiZXk
Wikipedia, list of fastest production cars, accessed 15 March 2015,
https://en.wikipedia.org/wiki/List_of_fastest_production_cars