Question

A cruise ship moves southward in still water at a speed of $20.0\mathrm{km}/\mathrm{h},$ while a passenger on the deck of the ship walks toward the east at a speed of $5.0\mathrm{km}/\mathrm{h}$. The passenger's velocity with respect to Earth is a) $20.6\mathrm{km}/\mathrm{h}$, at an angle of ${14.04}^{\circ }$ east of south. b) $20.6\mathrm{km}/\mathrm{h}$, at an angle of ${14.04}^{\circ }$ south of east. c) $25.0\mathrm{km}/\mathrm{h}$, south. d) $25.0\mathrm{km}/\mathrm{h}$, east. e) $20.6\mathrm{km}/\mathrm{h}$, south.

Short answer

Question: A passenger on a cruise ship is walking eastward at 5.0 km/h while the ship is going southward at 20.0 km/h. Determine the velocity of the passenger with respect to Earth. Answer: The velocity of the passenger with respect to Earth is 20.6 km/h at an angle of 14.04° east of south.

Step-by-step solution

Create the velocity vectors

We need to create two velocity vectors: the velocity of the cruise ship (${\overrightarrow{v}}_C$) and the velocity of the passenger with respect to the cruise ship (${\overrightarrow{v}}_P$). The magnitude of these two vectors are given as: $v_C=20.0\mathrm{km}/\mathrm{h}$ (southward) and $v_P=5.0\mathrm{km}/\mathrm{h}$ (eastward). Since the directions are orthogonal to each other, their corresponding unit vectors will be: ${\hat{v}}_C=-\hat{j}$ (south) and ${\hat{v}}_P=\hat{i}$ (east).

Construct the resulting velocity vector

Now we can construct the resulting velocity vector of the passenger with respect to Earth by adding ${\overrightarrow{v}}_C$ and ${\overrightarrow{v}}_P$: ${\overrightarrow{v}}_{PE}={\overrightarrow{v}}_C+{\overrightarrow{v}}_P$. In terms of magnitudes and unit vectors: ${\overrightarrow{v}}_{PE}=(20.0\mathrm{km}/\mathrm{h})(-\hat{j})+(5.0\mathrm{km}/\mathrm{h})(\hat{i})$.

Calculate the magnitude of the passenger's velocity with respect to Earth

To calculate the magnitude of the resulting velocity vector, we will use the Pythagorean theorem: $v_{PE}=\sqrt{(5.0\mathrm{km}/\mathrm{h}{)}^2+(20.0\mathrm{km}/\mathrm{h}{)}^2}=20.6\mathrm{km}/\mathrm{h}$.

Calculate the angle between resulting velocity vector and south direction

We will now calculate the angle $\theta$ between the resulting velocity vector ${\overrightarrow{v}}_{PE}$ and the southward (-$\hat{j}$) direction. We can use the tangent function: $\tan \theta =\frac{(5.0\mathrm{km}/\mathrm{h})}{(20.0\mathrm{km}/\mathrm{h})}$. Solving for $\theta$, we get $\theta =\arctan (\frac{5}{20})={14.04}^{\circ }$.

Choose the correct option

The magnitude of passenger's velocity with respect to Earth is $20.6\mathrm{km}/\mathrm{h}$ and the angle of the velocity vector is ${14.04}^{\circ }$ east of south. Hence, the correct answer is: a) $20.6\mathrm{km}/\mathrm{h}$, at an angle of ${14.04}^{\circ }$ east of south.

Key concepts

Velocity Vectors

Velocity vectors are fundamental in understanding motion in physics. They not only quantify how fast an object is moving but also specify the direction of the movement. In the context of our cruise ship example, we introduced two critical vectors: the ship's velocity, ${\overrightarrow{v}}_C$, directed southward, and the passenger's velocity, ${\overrightarrow{v}}_P$, directed eastward.

When expressing velocity vectors, we use magnitude and direction. The ship moves at 20.0 km/h south, which we can denote with a vector pointing down on a diagram ($-\hat{j}$ direction). Conversely, the passenger's eastward walk at 5.0 km/h is represented by a vector pointing right ($\hat{i}$ direction).

By combining these vectors using vector addition, we can determine the passenger's overall velocity relative to the Earth. This holistic approach accounts for both individual motions to provide a complete picture of the passenger's trajectory. Understanding how to solve problems involving velocity vectors is critical, as it can be applied to countless real-world situations beyond just passengers on a cruise ship.

Frame of Reference

The frame of reference is a crucial aspect when discussing velocity, as it dictates the 'point of view' from which the motion is observed and measured. This can drastically change how we perceive an object's movement. For instance, a passenger walking on a cruise ship might appear stationary to another observer on the ship but is actually moving in relation to the Earth or another external point.

In our example, the frame of reference for the velocity of the cruise ship, ${\overrightarrow{v}}_C$, is the Earth since the water is still. The frame of reference for the passenger's velocity, ${\overrightarrow{v}}_P$, is the ship itself. To understand the passenger's velocity with respect to the Earth, we must combine these two vectors, effectively shifting the passenger's frame of reference from the ship to the Earth. Frames of reference are a foundation for more advanced physics concepts, including Relativity, where they take on even greater significance.

Pythagorean Theorem

The Pythagorean theorem is a staple in geometry, providing a way to determine the length of the hypotenuse of a right-angled triangle based on the lengths of the other two sides. In the context of physics and specifically in our problem, it is used to calculate the magnitude of the resultant velocity vector. This theorem is expressed as $a^2+b^2=c^2$, where $c$ is the hypotenuse.

Applying the Pythagorean theorem to the southward and eastward components of the passenger's velocity, we calculated the overall speed as $\sqrt{{5.0}^2+{20.0}^2}$ km/h. The ability to apply this theorem is immensely valuable, as it is a simple yet powerful tool to solve various problems involving right triangles, which are rather common in physics.

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent are crucial for analyzing the relationships between the angles and sides of triangles. These functions extend beyond triangles, aiding in the analysis of periodic functions and waves in physics. In our cruise ship scenario, the tangent function was used to find the angle between the resultant velocity vector and the southward direction.

The tangent of the angle is the ratio of the opposite side to the adjacent side in a right triangle. For the passenger's velocity, the tangent of the angle ($\theta$) represents the ratio of the velocity eastward (5.0 km/h) to the velocity southward (20.0 km/h). Solving for this angle using the inverse tangent, we found $\theta =\arctan (\frac{5}{20})={14.04}^{\text{o}}$. Trigonometric functions are indispensable for interpreting and solving many problems in physics, where angles and distances are often intertwined.