The Canoe Race

The Canoe Race

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The Canoe Race


A group of canoeists on holiday at the seaside decide to have a race
offshore. They set up a triangular course using a buoy and two other
floats, with the start and finish at the buoy. They have been told
that the prevailing current flows parallel to the shore at a speed of
about 2 ms-1. If the total course is to be 300 metres long investigate
where they might place the other two floats.

Problem: How does the layout of the floats effect the time taken to
complete the race?

I would like to investigate two different models one being a
right-angled triangle and the other being isosceles triangle. When
investigating the isosceles triangle, an equilateral triangle would be
investigated because as the length of the isosceles triangle will all
equal, it becomes an equilateral triangle. I would first of all
investigate the right-angled triangle.

Model 1: Right-Angled Triangle

C

[IMAGE]

A B

[IMAGE]

PREVAILLING CURRENT AT A SPEED OF 2 MS-1

[IMAGE]

COASTLINE

Figure 1: The shape of the course (model 1).

The shape of the course is shown in figure 1. To simplify the problem
I am assuming that the angle CAB is a right angle. Even though the
lengths of B and C will change, angle CAB will always be a right
angle. ‘A’ in figure 1is where the buoy is positioned, and thus it is
the start and finish of the race. The race starts at point A, then it
continues on to point B, then to point C and finishes at the buoy,
which is point A. Point B and C are two floats, first and second float
respectively. I will refer to the lengths AB, BC and AC, throughout
the investigation.

It is more applicable to make assumptions; this would make the problem
simpler. I will use the same assumptions for both the models. It is
vital that we assume that the canoe is a particle and that it’s mass

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is concentrated at a single point. I will assume that there are no
obstacles that may overcome the canoeists to alter their speed and
direction. In a like manner they will not also bump into any object or
each other. If these factors were taken into account, it would
complicate the matter because these factors will slow down the speed
or velocity of the canoeists. This will complicate the velocity
diagrams. In reality the prevailing current will fluctuate in both
magnitude and direction. To simplify the scale drawings, which I would
use to solve the problems, there would not be any environmental
factors, e.g. weather, which would affect the speed or velocity of the
canoeists. Apart from the prevailing current, which flows parallel to
the shore at a speed of 2ms-1, there are no other currents, no driving
rain, no huge waves and no winds. The current has to have a direction,
therefore, the prevailing current flows parallel to the shore at a
speed of about 2 ms-1from left to right.

The total course will be 300metres. Hence AB + BC + CA = 300. When
canoeing the canoe travels extra distance when climbing and descending
waves, the extra distance will not be taken into account. The shape of
the triangle or the course will have to remain the same or otherwise
the canoeists will not travel the same distance. Therefore the shape
and the positioning of the course will have to remain the same. If the
course changed it will effect the time taken to complete the course.
Thus, it will also presumed that the floats and the buoy are stable,
therefore, they do not move with the current.

A sensible value for the speed of the canoeists would be 4ms-1 in
motionless water. Furthermore, all the canoeists maintain both the
same speed and direction; they will not be exhausted at any point of
the race. Therefore, it will be taken into account that the canoeists
will turn around a float at 4 ms-1 instantaneously, so that no extra
time is taken in slowing down and turning the canoe at the floats.
Therefore, the speed of the canoeists is constant. Furthermore, they
travel in a straight line along the lengths AB, BC and CA in figure 1,
so that no extra distance is travelled.

I will vary the length of AB in figure 1 in steps of 10 metres, i.e.
10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130 and 140. The
length of AB in figure 1 cannot be or more than 150m. When each side
is 75m and the base of triangle is 150m, the angle between each side
is 0 degrees. Thus, this is not possible for triangle of these
measurements to exist. This fact can be proved by trigonometry.

[IMAGE]

[IMAGE]

Adj

[IMAGE]
Cos-1q =

Hyp

[IMAGE]

75

[IMAGE]
Cos-1q =

75

q = 0

Figure 2:

Consequently, it is proved that this triangle of this particular
measurement cannot exist.

AB in figure 1 is constant; therefore I will use the letter K in
calculations to represent that it is a known constant. Therefore, the
equation AB + BC + CA = 300 can be written as K + BC + CA = 300. We
know from Phythagoras Theorem that for a right-angled triangle, like
the triangle ABC in figure 1 K2 + AC2 = BC2 . To find the values of BC
and CA, we can solve these two equations simultaneously:

AC + BC + AB = 300

ÞAC + BC + K = 300

AC2 = BC2 – K2

Þ AC = 300 - BC – K

Þ (AC)2 = (300 – BC –K)2

(300 – BC –K)2 = BC2 – K2

[(300 – K) - BC]2 = BC2 – K2

(300 – K)2 - 2 (300 – K) BC + BC2 = BC2 – K2

3002 – 600K + K2 – 2BC (300 – K) = - K2

2K2 – 600K + 3002 = 2BC (300 – K)

Therefore, BC = 2K2 – 600K + 3002 / 600 – 2K

Thus AC can be found, using the formula AC= 300 – BC – K.

From the above derived formulas we can find the distances for each
stretch of the course. This information can be used to find the time
taken to complete that particular stretch of the course. The time
taken for each of the three stretches will be added to give the total
time it will take to complete the course. We have to find the velocity
at which the canoes travels on each stretch of the course, the missing
part the velocity can be found, we know distance, thus we can
substitute into the equation Time = Distance / Velocity and find
velocity. This can be achieved by using scale velocity diagrams. The
following are the procedures that need to be taken in order to do the
scale drawings.

STRETCH AB

The model has to be convenient for the canoeists. The current of the
sea is an important factor and we have to take it into account, i.e.
the effect of the prevailing current. The prevailing current is a
constant opposing force. The canoe will be pushed away from its path,
regardless of the direction it is moving in.

Thus, it is meaningful to travel the path AB in figure 1 from left to
right where the velocity of the canoe (4ms-1) and the prevailing
current (2ms-1) would be added together. Thus the resultant velocity
on the stretch AB in figure 1 is 6ms-1 from left to right. This is
shown in the following diagram:

4 ms-1 2 ms-1

[IMAGE]
[IMAGE]
[IMAGE]
[IMAGE]
[IMAGE]
A B

Figure 3: Showing an illustration of the resultant velocity, when a
canoeists travels from A to B

Thus the time taken can be calculated from A to B:

K is the distance of AB

Resultant velocity is 6 ms-1

Therefore,

K

[IMAGE]
Time = seconds

6

STRETCH BC

[IMAGE]

[IMAGE]
[IMAGE]
[IMAGE]
[IMAGE]
[IMAGE]
[IMAGE]
[IMAGE]
C

[IMAGE]
Figure 4:

4 ms-1

[IMAGE]
2 ms-1

Angle

B

With the aid of the data we have on distances, we can find the time
taken at the stretch BC. Foremost we need to know the direction that C
is from B. Using trigonometry; angle ABC in figure 4 can be
calculated. Using the formula Cos q = AB / BC we can find the value of
q. The line BC on the angle as in figure 4 is drawn. As there is a
current flowing from left to right at 2ms-1, we have to draw this on
the diagram as 4cm. We have to acknowledge that the speed of the
canoeist is 4ms-1, which is always constant. Thus we set a compass to
a gap of 8cm. From the end of the 2ms-1 line we use the compass to
cross the line BC. The resultant velocity of the canoeist travelling
from B to C can now be found by measuring the length of BC. Therefore,
the time for this stretch BC, is found by distance of BC / velocity of
BC.

STRETCH CA

With aid of figure 5 I will calculate the time taken and the velocity
of this stretch CA.

[IMAGE]
C 2 ms-1

[IMAGE]
[IMAGE]
[IMAGE]
[IMAGE]
[IMAGE]
[IMAGE]
[IMAGE]

4 ms-1

Figure 5:

A

As for all the stretches we had to consider the current, we have to
approve that there is a current of 2ms-1 moving from left to right,
therefore, it has to be drawn in the diagram, as shown in figure 5.
Line CA is drawn in the diagram, as shown in figure 5, the direction
of CA is vertically down, as it is perpendicular to the stretch AB.
The compass is set to 8 cm to portray the velocity of the canoe, which
is 4ms-1 in still water. The compass is used from the end of the 2m/s
line to cut the line AC. Thus the velocity is known by measuring the
distance AC. Hence the time taken to get from C to A can be found by
dividing the distance of CA/ velocity of CA.

By adding up the three times that it takes to do the three stretches
the total time to complete the course is found.

As I mentioned before I am constructing seven different courses which
have the lengths for the stretch AB of 10, 20, 30, 40, 50, 60, 70, 80,
90, 100, 110, 120, 130 and 140 metres. Thus the following calculations
will show how the values of AC and BC will change with different
values of AB, note that all the calculations are to the nearest
integer, this is due to the assumptions. They limit the preciseness to
the extent that taking calculations to decimal places would not make
the calculations any more precise. I have also calculated the angle
ABC, which is required in the scale diagrams:

Calculation 1: When AB = 10 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 102 – 600*10 + 3002 / 600 - 20

Þ BC = 84200 / 580

Þ BC = 145.17

Thus, BC = 145 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 10 – 145

Þ AC = 145

Thus AC = 145 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1q = 10 / 145

Þ Cos-1q = 86.04o

Thus, q = 86 o

Calculation 2: When AB = 20 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 202 – 600*20 + 3002 / 600 - 40

Þ BC = 78800 / 560

Þ BC = 140.71

Thus, BC = 141 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 20 – 141

Þ AC = 139

Thus AC = 139 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos Ø = 20 / 139

Þ Cos-1q = 81.8o

Thus, q = 82 o

Calculation 3: When AB = 30 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 302 – 600*30 + 3002 / 600 - 60

Þ BC = 73800 / 540

Þ BC = 136.6

Thus, BC = 137 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 30 – 145

Þ AC = 133

Thus AC = 133 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 30 / 137

Þ Cos-1q = 77.3o

Thus, q = 77 o

Calculation 4: When AB = 40 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 402 – 600*40 + 3002 / 600 - 80

Þ BC = 69200 / 520

Þ BC = 133.08

Thus, BC = 133 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 40 – 133

Þ AC = 127

Thus AC = 127 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 40 / 133

Þ Cos-1q = 72.5

Thus, q = 73 o

Calculation 5: When AB = 50 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 502 – 600*50 + 3002 / 600 - 100

Þ BC = 65000 / 500

Þ BC = 130

Thus, BC = 130 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 50 – 145

Þ AC = 120

Thus AC = 120 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 50 / 130

Þ Cos-1q = 67.4o

Thus, q = 67 o

Calculation 6: When AB = 60 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 602 – 600*60 + 3002 / 600 - 120

Þ BC = 61200 / 480

Þ BC = 127.5

Thus, BC = 128 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 60 – 128

Þ AC = 112

Thus AC = 112 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 60 / 128

Þ Cos-1q = 62.04o

Thus, q = 62 o

Calculation 7: When AB = 70 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 702 – 600*70 + 3002 / 600 - 140

Þ BC = 57800 / 460

Þ BC = 125.7

Thus, BC = 126 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 70 – 126

Þ AC = 104

Thus AC = 104 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 70 / 126

Þ Cos-1q = 56.3o

Thus, q = 56 o

Calculation 8: When AB = 80 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 802 – 600*80 + 3002 / 600 – 2*80.

Þ BC = 54800 / 440

Þ BC = 124.6

Thus, BC = 125 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 80 – 125

Þ AC = 95

Thus AC = 95 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 80 / 125

Þ Cos-1q = 50.21o

Thus, q = 50 o

Calculation 9: When AB = 90 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 902 – 600*90 + 3002 / 600 – 2*90

Þ BC = 52200 / 420

Þ BC = 124.3

Thus, BC = 124 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 90 – 124

Þ AC = 86

Thus AC = 104 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 90 / 124

Þ Cos-1q = 43.5o

Thus, q = 44 o

Calculation 10: When AB = 100 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 1002 – 600*100 + 3002 / 600 - 200

Þ BC = 50000 / 400

Þ BC = 125

Thus, BC = 125 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 100 – 125

Þ AC = 75

Thus AC = 75 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 100 / 126

Þ Cos-1q = 36.9o

Thus, q = 37 o

Calculation 11: When AB = 110 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 1102 – 600*110 + 3002 / 600 – 2*110

Þ BC = 48200 / 380

Þ BC = 126.8

Thus, BC = 127 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 110 – 127

Þ AC = 63

Thus AC = 63 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 110 / 127

Þ Cos-1q = 29.9o

Thus, q = 30 o

Calculation 12: When AB = 120 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 1202 – 600*120 + 3002 / 600 – 2*120

Þ BC = 46800 / 360

Þ BC = 130

Thus, BC = 130 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 120 – 130

Þ AC = 50

Thus AC = 50 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 120 / 130

Þ Cos-1q = 22.6o

Thus, q = 23 o

Calculation 13: When AB = 130 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 1302 – 600*130 + 3002 / 600 – 130*2

Þ BC = 45800 / 340

Þ BC = 134.7

Thus, BC = 135 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 130 – 135

Þ AC = 35

Thus AC = 35 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 130 / 135

Þ Cos-1q = 15.64o

Thus, q = 16 o

Calculation 14: When AB = 140 metres then from the formula

BC = 2K2 – 600K + 3002 / 600 – 2K

Þ BC = 2 * 1402 – 600*140 + 3002 / 600 – 140*2

Þ BC = 45200 / 320

Þ BC = 141.3

Thus, BC = 141 metres

Therefore, AC = 300 – AB – BC

Þ AC = 300 – 140 – 141

Þ AC = 19

Thus AC = 19 metres.

The angle required for the triangle, is calculated as follows:

Using trigonometry: Cos-1q = AB / BC

Þ Cos-1 q = 140 / 141

Þ Cos-1q = 6.83o

Thus, q = 7 o

All the results have been put into the following table:

Length AB

Length BC

Length AC

Angle ABC

In metres

In metres

In metres

In degrees

10

145

145

86

20

141

139

82

30

137

133

77

40

133

127

73

50

130

120

67

60

128

112

62

70

80

126

125

104

95

56

50

90

124

86

44

100

125

75

37

110

127

63

30

120

130

50

23

130

135

35

16

140

141

19

7

Table 1: The values of AC and BC with the different values of AB.

The scale diagrams follow this page. All the scale drawings have been
drawn for every length of stretch AB. The results of the scale
drawings are summarised in table 2.

Distance AB

Total Time taken

In metres

In seconds

10

87

20

88

30

89

40

89

50

91

60

90

70

92

80

93

90

95

100

94

110

96

120

96

130

99

140

99

Table 2: Summary of the total time taken in the different models.

From the scale drawings it can be seen that out of the three stretches
it is evident that stretch AB takes the least time to complete. This
is due to the fact that the prevailing current is giving the canoeists
an advantage, both the velocities add up to give a higher yield of
velocity. At the other stretches the canoeists have to work harder as
they are fighting against the resistance of the prevailing current.
Thus, it is obvious it takes more time to complete those stretches.
But, if you examine table 2, it can be clearly seen that even the time
taken to complete stretch AB is the quickest, as the distance of AB
increases the time taken to complete the course is increases. On the
other hand it is seen that as the distance of AB decreases the time
taken to complete the race decreases.

Model 2: Isosceles Triangle

As I have mentioned before, that I would use the same assumptions as
model 1. The following diagram, is the shape of model 2, which is an
isosceles triangle. Concerning the diagram A is the buoy. At this
point the canoeists start and finish the race. The race direction will
be clockwise and consequently the canoes paddle along the line AB. On
reaching B, which is the first float, the canoes then travel along the
line BC. On reaching C, which is the second float, the canoes then
travel along the line CA and finish the course as they reach A again.
I will be referring to the lengths AB, BC and CB as stretches in
future. Note that stretch AB = AC.

A (Buoy)

[IMAGE]

Figure 6: Diagram of model 2

[IMAGE]

C (2nd Float) B (1st Float)

As the total length of the race is 300 metres, therefore AB+BC+CA=300.
I will vary the length of BC, the base, in steps of 20 metres with
values of 60, 80, 100, 120 and 140 metres. As I have mentioned before
that the length of the base cannot be more than 150 metres as the
course is limited to a total length of 300 metres.

We know the distances for each stretch of the course. Thus the time
taken to complete the particular stretches of the course can be found.
The total time can be found by adding the time taken for each of the
three stretches, this will give the time taken to complete the whole
course. Now we have to find the velocity at which the canoes travels
on each stretch of the course, as we know the distance, we can
substitute into the formula, to find time:

Time = Distance / Velocity

Subtracting the base length from 300, and then dividing the answer by
2 gives you the lengths of the stretches of AC and AB, because it is
an isosceles triangle two of the sides will have the same length. An
example of the calculation is shown below:

300 – BC / 2 = AB and AC

Thus, when BC = 80m

Then AB and AC = 300 – 80 / 2 = 110m

Therefore AB and AC = 110m, when BC = 80m.

The Length Of Stretch BC In Metres

The lengths of stretches AC and AB In Metres

60

120

80

110

100

100

120

90

140

80

Table 3: The lengths of the stretches of AC and AB are summarised in
the above table:

The scale drawings follow this page. The scale drawings were used to
calculate firstly the velocity and then finally the time. All the
times were added up together to give the total time taken for a
particular course. The results from the calculation are all summarised
in table 4.

The Length Of Stretch BC In Metres

Total Time Taken In Seconds

60

90

80

92

100

93

120

95

140

99

Table 4: Summaries of the results for model 2.

It can be clearly seen in the scale drawings that out of the three
stretches it is evident that stretch BC takes the most time to
complete, not like model 1, that the base stretch (AB) is the
quickest. It is also evident that stretches AB and AC were equal, both
the stretches took the same time to complete. If you examine table 4,
it can be clearly seen that as the distance of BC increases the time
taken to complete the course is increases and the shorter the distance
of stretch BC the time taken to complete is the quickest.

Referring to model 1; one may think that the longer the distance of
AB, the time required to finish the race will decrease, due to the aid
of the current. Actually, this is not the case; the time taken to
complete the race actually increases. This is probably due to the fact
that, as the distance of AB increases, the distance of stretches BC
and AC joint together is larger. And at stretches BC and AC, the
canoeists have to travel against the current. So they have to travel a
greater distance with the resistance of the current. This slows the
canoeists down drastically. These two factors increase the time taken
to complete the race.

It is evident from the results in table 2 that as the distance of AB
increases the time taken also increases. There are a few exceptional
cases where the time taken decreases, from the previous time, e.g. the
courses where AB is equal to 50 and 60 metres. It is probably due to
human error or because I have rounded the results, and therefore, this
could have effected the accuracy. But most of the time the time taken
increases as AB increases.

It can be concluded in model 1 that as the distance of AB increases
the time taken to complete the race increases, and as the distance of
AB decreases the time taken to complete the race decreases. Thus, if
the canoeists want to maximise the time taken to complete the race
then they should maximise the distance of AB, which is 140m, but, if
they want to minimise the time taken to complete the race, they should
minimise the distance of AB.

But if you look at model 2, it is a different story to a certain
extent. It is seen that stretch BC takes the most time to complete. In
model two the other two stretches were infact took less time to
complete. This is due to the fact that stretch BC is fighting against
the current and therefore reduces the velocity from 4ms-1 to 2ms-1. At
stretches AB and AC the canoeists were travelling with current at an
angle. Therefore, stretches AB and AC had the aid of the current.
Thus, as the distance of stretch BC increased the more the canoeists
had to travel against the current, therefore, this increased the time
taken. So, I can conclude that as the distance of BC increases the
time taken to complete the race increases, and as the distance of BC
decreases the time taken to complete the race decreases. Consequently,
if the canoeists want to maximise the time taken to complete the race
then they should maximise the distance of BC, which is 140m, but, if
they want to minimise the time taken to complete the race, they should
minimise the distance of BC.

I have found out by investigating the two models that the positions of
the two floats and the buoy can drastically effect the time taken to
complete the race. It is also important in what direction you are
travelling in and from what place you start your race, because as in
model 1, it can be seen that as you travel along stretch AB with the
aid of the current, then time taken to complete the stretch is less.
But, on the other hand in model 2, when you are travelling against the
current in stretch BC it took more time to complete the stretch. I
believe that if I were to place the buoy at C in model 2, the whole
scenario would be different. Because, at the base you are travelling
with current, and when you come to the second stretch you would have
to travel against the current, and thus take more time tom complete
the stretch. When you come up to the last stretch you would have the
aid of the current at an angle, so this would aid the canoeists.

The assumptions that I made at the beginning in order to simplify the
problem will drastically affect the accuracy of the calculations. In
reality the canoeists would never be able to maintain a constant
speed. They would obviously be tired due to fatigue. It would be
impossible for them to paddle in a straight line and also turn around
a corner instantaneously. Thus the canoeists are actually travelling
further than 300 metres, as they do not paddle on a straight
horizontal path but they climb up and down the waves.

I also assumed that the canoeists were acting like a particle. In
reality the mass of the object would not be concentrated in one single
point. They would have to do turn around a corner in an arc; thus they
will travel an even further distance. Therefore, they would slow down
dramatically as they turn around the floats. As each canoeist would
turn around a float differently, therefore, each would take different
amount of time and would travel different distances. So in actual fact
all these assumptions have made the times calculated slightly faulty
and should be greater than what they are.

Also in reality there would be many obstacles. Such as the weather
conditions or there could also be objects in the water that slow down
the canoeists, such as litter. There is a possibility that the
canoeists may bump into each other. I assumed that the current was
parallel to the shore, which is very unlikely, currents come in at
different angles. There could have been huge waves. But I ignored all
these points, these points will invalidate my investigation.

It would be difficult to measure out the course accurately because the
buoy and floats are moving about. This would affect the time it takes
to complete the course. The only way to solve this problem would to be
have the buoy and floats tied by a rope, which is stuck to the ground
or have poles driven into the ground.

As far as I am concerned the analysis has been accurate, and they have
been done to the best of my ability. But it is not fully accurate as
possible. For example in Table 1 both the distances and angles were
rounded up to the nearest integer. I also rounded the values obtained
for the velocity from the scale drawing. Due to human error the scale
drawings were probably not always measured accurately. But these
errors have not changed the total time significantly. The reason I did
not make it accurate as possible because all the assumptions made
reduced the true accuracy of the calculations. Without the assumptions
that I made at the beginning of the investigation, it would be
extremely difficult or even impossible to solve the problem just by
theory work. It would need to be done practically. Doing it
practically would be also hard, e.g. getting the right weather
conditions, the equipment, the current to be parallel to the shore
etc.

I trust that the formulas I derived to calculate the distances of the
stretches are accurate. Nevertheless I could have made errors, which
have resulted the calculations to be inaccurate, if I had the second
chance to do this coursework I would check my answers several times.

If I had the opportunity to solve this problem again I would
investigate several other factors and extend it. I would use a wider
variety of different lengths for AB. I could use different types of
canoes, such as a two-man or a three-man canoe. This would give
different mass, and thus it will change the speed and the time. I
could also investigate the effect of currents coming in different
directions and different speeds. On the other hand I could change my
model into a scalene triangle.
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