Problem 16 - Applied Mathematics

Question

A speedboat which can travel at 20 knots in still water starts from the corner \(X\) of an equilateral triangle \(X Y Z\) of side 10 nautical miles and describes the complete course XYZX in the least possible time. A tide of 5 knots is running in the direction \(\overrightarrow{Z X}\). Find: (a) the speed of the boat along XY, (b) to the nearest minute the time taken by the speedboat to traverse the complete course XYZX. (1 knot is one nautical mile per hour).

Step-by-Step Solution

Answer

Speed of boat along XY: 20 knots. Time to traverse XYZX: Total time\)

Step 1: Identify directions and given data

Identify the given values and directions. The speed of the boat in still water is 20 knots, the tide speed is 5 knots in the direction \(\rightarrow ZX\), and the side length of the equilateral triangle is 10 nautical miles.

Step 2: Determine boat speed along XY

First consider the speed of the boat along XY. Since there is no tide effect in the direction XY (tide is perpendicular to XY), the speed of the boat along XY is the same as its speed in still water.Speed along XY = 20 knots.

Step 3: Calculate the time to travel from X to Y

Calculate the time taken to travel from point X to point Y. The distance from X to Y is 10 nautical miles.Time along XY = \(\frac{10 \text{ nautical miles}}{20 \text{ knots}}= 0.5 \text{ hours}= 30 \text{ minutes}\).

Step 4: Determine boat speed and effective speed along YZ

For the path Y to Z, the tide effect is perpendicular to the direction of travel. Therefore, use the Pythagorean theorem to determine the boat's effective speed.Effective speed along YZ = \(\text{sqrt}(20^2-5^2) = \text{sqrt}(400-25) = \text{sqrt}(375) = 5\text{sqrt}(15) \text{ knots}\).

Step 5: Calculate the time to travel from Y to Z

Calculate time along YZ. The distance YZ is also 10 nautical miles.Time along YZ = \(\frac{10 \text{ nautical miles}}{5\text{sqrt}(15) \text{ knots}} = \frac{10}{5\text{sqrt}(15)} \text{ hours} = \frac{2}{1\text{sqrt}(15)} \text{ hours} \).

Step 6: Determine the boat speed along ZX

Along ZX, the tide and boat speeds add up vectorially. The tide assists the boat, making the resultant speed:Effective speed along ZX = 20 knots + 5 knots = 25 knots.

Step 7: Calculate the time to travel from Z to X

Calculate the time taken to travel from Z to X. The distance from Z to X is still 10 nautical miles.Time along ZX = \(\frac{10 \text{ nautical miles}}{25 \text{ knots}} = 0.4 \text{ hours}=24 \text{ minutes}\).

Step 8: Sum times for total trip

Sum the time taken for each leg of the trip.Total time = Time along XY + Time along YZ + Time along ZX \(\text{Total time} = 30 \text{ minutes} + \text{some minutes} + 24 \text{ minutes}\)

Key Concepts

Kinematics in two dimensions

Kinematics in two dimensions involves understanding the motion of objects in a plane. This means analyzing how an object moves in both horizontal and vertical directions simultaneously. For example, in the problem with the speedboat, we are looking at how the boat moves through water and how different factors, like the tide, affect its motion. To solve such problems, we must consider the speed and direction in each dimension.

Key components include:

• Initial speed and direction
• The effect of external forces, such as tides or wind
• Calculating resultant speeds
• Using the Pythagorean theorem to resolve components

Understanding these components allows us to predict the path and duration of travel accurately.

Vector addition

Vector addition is crucial to solving problems in two-dimensional kinematics. Vectors have both magnitude and direction. In the given exercise, our vectors include the boat's speed and the tide's speed. To find the resultant speed or effective speed of the boat, we add these vectors together.

Here's an example from the problem:

• The boat's speed in still water is 20 knots, represented by a vector.
• The tide runs at 5 knots towards ZX, another vector.
• To find the effective speed along ZX, we sum these vectors' magnitudes (20 + 5 = 25 knots) because they are in the same direction.

We often use techniques like the Pythagorean theorem to help resolve these Vector components into a resultant vector.

Pythagorean theorem

The Pythagorean theorem is a fundamental principle used to solve problems involving right triangles. It's expressed mathematically as \( a^2 + b^2 = c^2 \), where c represents the hypotenuse and a and b the other sides. This theorem helps find the resultant speed when the boat's speed and the tide's speed are perpendicular.

For example, along the path YZ in the problem:

• The boat travels at 20 knots, and the tide perpendicularly affects it at 5 knots.
• Using the Pythagorean theorem, the effective speed is \( \text{sqrt}(20^2 - 5^2) = \text{sqrt}(375) = 5\text{sqrt}(15) knots \).

This calculation ensures that we have the correct effective speed needed to determine the travel time.

Triangular path navigation

Navigating along a triangular path implies moving along multiple straight-line segments to form a triangle. In the given exercise, the speedboat travels along three sides of an equilateral triangle: XY, YZ, and ZX. Each segment has unique attributes based on how the tide interacts with the boat's speed.

Considerations include:

• The direction of each leg of the journey (e.g., XY, YZ, and ZX)
• The impact of the tide on each segment
• Using vector addition and the Pythagorean theorem to determine effective speeds
• Summing the travel times for each segment to find the total time

By understanding these factors, one can accurately navigate the triangular path and calculate the time required for each part of the journey.