Problem 66 - Algebra and Trigonometry

Question

Velocity A woman walks due west on the deck of an ocean liner at 2 milh. The ocean liner is moving due north at a speed of 25 \(\mathrm{mi} / \mathrm{h}\) . Find the speed and direction of the woman relative to the surface of the water.

Step-by-Step Solution

Answer

The woman's speed relative to the water is about 25.08 mi/h, heading 84.5° west of north.

Step 1: Understand the Problem

We need to find the resultant speed and direction of the woman relative to the water surface. This involves combining her velocity with the ship's velocity to get the total velocity vector.

Step 2: Identify Components of Velocity

The woman's velocity is 2 mi/h west (which is a negative direction on the x-axis if east is positive), and the ship's velocity is 25 mi/h north (positive direction on the y-axis). So, the velocity components are: \( \mathbf{v}_x = -2 \) mi/h and \( \mathbf{v}_y = 25 \) mi/h.

Step 3: Calculate the Resultant Speed

Use the Pythagorean theorem to calculate the magnitude of the resultant velocity vector: \[ v = \sqrt{(-2)^2 + 25^2} = \sqrt{4 + 625} = \sqrt{629} \approx 25.08 \text{ mi/h} \]

Step 4: Determine Direction Using Trigonometry

The direction \( \theta \) can be found using the arctangent function: \[ \theta = \tan^{-1} \left( \frac{25}{-2} \right) \] This gives approximately \( \theta = 95.5^\circ \) where \(0^\circ\) is east and \(90^\circ\) is north. Since the direction is west of north, it is more accurate to say that the direction is \(84.5^\circ\) west of north.

Key Concepts

Vector Addition

To solve problems involving relative velocity, it's important to understand vector addition. Vectors have both magnitude and direction, which makes them perfect for describing velocities. In the exercise, the woman and the ocean liner are moving in perpendicular directions:

• The woman's direction: 2 mi/h west
• The ocean liner's direction: 25 mi/h north

When combining these velocities, we use a process called vector addition. This involves adding the corresponding components of each vector.
To visualize it better, imagine drawing the vectors on a graph. The woman's westward speed would be a line pointing to the left (negative x-axis), while the ocean liner's northward speed is a line pointing upwards (positive y-axis). By connecting these arrows head-to-tail, you form a right triangle where the hypotenuse represents the resultant velocity of the woman relative to the water surface. Thus, vector addition helps us determine the new speed and direction when two vectors interact.

Pythagorean Theorem

The Pythagorean theorem is essential when dealing with right triangles formed by vector components. It allows us to find the magnitude of the resultant vector. In this problem, it's used to determine the speed of the woman relative to the water surface.

• The woman's westward velocity: \(-2 \text{ mi/h}\).
• The ship's northward velocity: \(25 \text{ mi/h}\).

These perpendicular vector components form a right triangle.
The Pythagorean theorem states that for any right triangle, the square of the hypotenuse is the sum of the squares of the other two sides. Here’s the formula in action: \[ v = \sqrt{(-2)^2 + 25^2} = \sqrt{4 + 625} = \sqrt{629} \approx 25.08 \text{ mi/h} \].This calculation tells us the speed of the woman relative to the ocean surface. It’s a great example of how this classic mathematical theorem is applied in physics to find resultant magnitudes.

Trigonometry

Trigonometry comes into play when we need to find the direction of a vector resultant from two perpendicular components. In this exercise, after finding the magnitude of the resultant velocity using the Pythagorean theorem, we now need to find its direction.To calculate the angle, trigonometry gives us tools like the tangent function. If given a right triangle with an opposite side of 25 and an adjacent side of -2, the direction angle \( \theta \) can be found using:\[ \theta = \tan^{-1} \left( \frac{25}{-2} \right) \].The arctan function returns an angle, \( \theta \), in the range of -90 to 90 degrees, which we then interpret relative to a specific compass direction.
Our calculated angle is about \( 95.5^\circ \), north of east generally, but since we are considering the angle where the vector shifts from north to west, we adjust this to \( 84.5^\circ \) west of north. This is how trigonometry reveals the precise direction of motion in terms of compass bearings, a crucial skill in navigation and physics.