Problem 19 - Algebra and Trigonometry with Analytic Geometry

Question

Distance between ships A ship leaves port at 1:00 P.M. and travels \(\mathrm{S} 35^{\circ} \mathrm{E}\) at the rate of \(24 \mathrm{mi} / \mathrm{hr}\). Another ship leaves the same port at 1:30 P.M. and travels \(\mathrm{S} 20^{\circ} \mathrm{W}\) at \(18 \mathrm{mi} / \mathrm{hr}\). Approximately how far apart are the ships at 3:00 P.M.?

Step-by-Step Solution

Answer

Approximately 39.32 miles.

Step 1: Determine Travel Time

First, we need to calculate the time each ship has been traveling by 3:00 P.M. The first ship leaves at 1:00 P.M. and travels until 3:00 P.M., so it travels for 2 hours. The second ship leaves at 1:30 P.M. and travels until 3:00 P.M., so it travels for 1.5 hours.

Step 2: Calculate Distance Traveled by each Ship

For the first ship traveling at 24 mi/hr for 2 hours, the distance is given by speed multiplied by time: \(24 \text{ mi/hr} \times 2 \text{ hr} = 48 \text{ miles}\). Similarly, for the second ship traveling at 18 mi/hr for 1.5 hours, distance is: \(18 \text{ mi/hr} \times 1.5 \text{ hr} = 27 \text{ miles}\).

Step 3: Decompose Distances into Components

Both ships travel in directions that form angles with the cardinal south direction. Decompose their distances into south (y) and east/west (x) components. For Ship 1 (S 35° E):- Southward component is \(48 \cdot \cos(35°)\)- Eastward component is \(48 \cdot \sin(35°)\) For Ship 2 (S 20° W):- Southward component is \(27 \cdot \cos(20°)\)- Westward component is \(27 \cdot \sin(20°)\)

Step 4: Calculate Components for Each Ship

Calculate the actual values:Ship 1:- Southward: \(48 \cdot \cos(35°) \approx 39.31 \) miles- Eastward: \(48 \cdot \sin(35°) \approx 27.55 \) milesShip 2:- Southward: \(27 \cdot \cos(20°) \approx 25.37 \) miles- Westward: \(27 \cdot \sin(20°) \approx 9.23 \) miles

Step 5: Compute the Relative Distance Apart

To find the distance between the ships, we need to find the difference in their south and east/west components:Total southward difference: \(39.31 - 25.37 = 13.94\) milesTotal east-west difference: \(27.55 + 9.23 = 36.78\) milesWe add the westward component as positive since Ship 2 goes westward instead of east.

Step 6: Calculate Actual Distance Between the Ships Using the Pythagorean Theorem

Use the Pythagorean theorem:Distance = \(\sqrt{(13.94)^2 + (36.78)^2}\)= \(\sqrt{194.19 + 1352.68} \approx \sqrt{1546.87} \approx 39.32\) miles.

Key Concepts

Trigonometry

Trigonometry is a branch of mathematics that helps us understand relationships within right triangles. These relationships involve the angles and lengths of the sides. When working with navigation and distance problems, trigonometric functions such as sine, cosine, and tangent allow us to decompose vectors into components. This involves using angles out of a base direction (like south).
For example, if a ship sails \(S 35^{\circ} E\), we can use the cosine function to determine how much of the journey is southward, and the sine function to find the eastward distance. Always associate cosine with the side adjacent to the angle and sine with the opposite side.

Vector Decomposition

Vector decomposition is a technique utilized to break a vector down into its component parts, usually expressed in horizontal (x-axis) and vertical (y-axis) directions. It's particularly useful in navigation problems where directions aren't aligned with primary compass points.
In our ship navigation problem, both ships travel at angles deviating from due south. By decomposing their paths into a south component and an east-west component, we can measure their relative movement more easily. This process involves multiplying their travel distance by the sine and cosine of their given angle relative to north or south. By calculating these components, we can work out how far they have moved in each direction.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle used to find the length of a side in a right triangle. It outlines how the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides:
\[ c^2 = a^2 + b^2 \]
This theorem comes into play in our ship navigation problem when calculating the straight-line distance between two points (two ships) that have traveled at different directions. By using the total southward and east-west distances as sides of a right triangle, we can employ the Pythagorean theorem to find the direct distance between the ships.

Ship Navigation

Ship navigation involves planning and managing the course a vessel takes to reach its destination. It's crucial for ships to consider not just the speed and distance, but also the precise direction, as misjudging angles can lead a ship off course.
In real-world scenarios, as well as in our textbook example, navigation requires knowledge of trigonometry and vector decomposition to predict a vessel's location over time. Calculating intersecting paths or distances between vessels is a common task - ensuring that they avoid collisions and follow the most efficient route possible. The exercise illustrates how these directions relative to south use angles to maintain precision, integrating both speed and direction for robust planning.