# The Canoe Race

Extracts from this document...

Introduction

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.

Middle

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

Conclusion

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.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month