Problem 66 - Precalculus Mathematics for Calculus

Question

Velocity A woman walks due west on the deck of an ocean liner at \(2 \mathrm{mi} / \mathrm{h}\). 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 approximately 25.08 mi/h, directed 4.57° west of north.

Step 1: Understand the Problem

We need to determine the woman's resultant velocity relative to the water, given her walking velocity towards the west and the ship's velocity towards the north.

Step 2: Define the Velocity Vectors

Express the woman's walking velocity as a vector. Since she walks west, this can be represented as \( \vec{v}_w = -2\hat{i} \) (miles per hour) because west is the negative x-direction. Next, express the ship's velocity moving north as \( \vec{v}_s = 25\hat{j} \)(miles per hour) since north is the positive y-direction.

Step 3: Calculate Resultant Velocity Vector

Add the two vectors to find the woman's velocity relative to the water:\[ \vec{v}_{r} = \vec{v}_w + \vec{v}_s = -2\hat{i} + 25\hat{j} \]

Step 4: Find the Magnitude of the Resultant Velocity

To find the speed of the woman relative to the water, calculate the magnitude of \( \vec{v}_{r} \):\[ |\vec{v}_{r}| = \sqrt{(-2)^2 + (25)^2} = \sqrt{4 + 625} = \sqrt{629} \approx 25.08 \, \text{mi/h} \]

Step 5: Determine the Direction of the Velocity

Find the angle \( \theta \) of the resultant velocity with respect to the north:\[\theta = \tan^{-1}\left(\frac{|-2|}{25}\right) = \tan^{-1}\left(\frac{2}{25}\right) \approx 4.57^\circ \]The direction is \(4.57^\circ\) west of north.

Key Concepts

Resultant Velocity

When two velocities combine, we refer to the overall result as a "resultant velocity." Imagine it as the final direction and speed you get when you put two forces together. In this exercise, we look at how the woman's velocity on the ship and the ship's velocity combine. She walks west, and the ship moves north.

• The woman's walking velocity is given as 2 miles per hour (west).
• The ship's velocity we know is 25 miles per hour (north).

When we talk about resultant velocity, we're trying to capture her total motion in relation to the water around them. This is sometimes imagined as the diagonal path she takes due to both the walking and the ship's movement. The resultant is a mixture of these two motions, which is determined by adding their vectors.

Velocity Vectors

Vectors describe both a direction and a magnitude, which is really just a fancy way to say how fast something is going in which direction. Using vectors can help us simplify complex motion problems like the one in this exercise. For the woman walking due west, we define her velocity vector as \( \vec{v}_w = -2\hat{i} \). This is because west is typically seen as the negative x-direction in vector terms. For the ship moving due north, the velocity vector is \( \vec{v}_s = 25\hat{j} \). Here, north is the positive y-direction. To find how the woman moves in relation to the water, we need to combine these vectors: \[ \vec{v}_r = \vec{v}_w + \vec{v}_s = -2\hat{i} + 25\hat{j} \] This resultant vector tells us that the woman's direction is a mix of her walking west and the ship moving north.

Magnitude and Direction

When we talk about magnitude in vector problems, this is basically the "length" of the velocity vector, showing us the speed. To find it, we use the Pythagorean theorem in two dimensions, treating it like finding the hypotenuse of a right-angled triangle. For this problem:\[|\vec{v}_r| = \sqrt{(-2)^2 + (25)^2} = \sqrt{4 + 625} = \sqrt{629} \approx 25.08 \, \text{mi/h} \]This gives us the speed of the woman's resultant velocity.After establishing the speed, we need to find the direction. The direction shows which way she is heading from north. We calculate the angle \( \theta \) using the inverse tangent function:\[\theta = \tan^{-1}\left(\frac{|{-2}|}{25}\right) = \tan^{-1}\left(\frac{2}{25}\right) \approx 4.57^\circ \]This angle indicates she is moving 4.57 degrees west of due north.