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CourseworkHelp
:The Canoe Race
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
 |
A B
 |
PREVAILLING CURRENT AT A SPEED OF 2 MS-1
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 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.
 |
|||
 |
|||
Adj
Cos-1q =
Hyp
 |
75
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
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
Time = seconds
6
STRETCH BC
 |
C
Figure 4:
4 ms-1
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.
C 2 ms-1
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)
 |
Figure 6: Diagram of model 2
 |
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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