# The Canoe Race

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More ↓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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### Related Searches

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.