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Chapter 3: Problem 54

A passenger walks from one side of a ferry to the other as it approaches a dock. If the passenger's velocity is \(1.50 \mathrm{m} / \mathrm{s}\) due north relative to the ferry, and \(4.50 \mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) west of north relative to the water, what are the direction and magnitude of the ferry's velocity relative to the water?

Short Answer

Expert verified
The ferry's velocity relative to the water is approximately 2.60 m/s, 20.2掳 west of north.

Step by step solution

01

Understand the Problem

We have two velocities to consider: the velocity of the passenger relative to the ferry, and the velocity of the passenger relative to the water. We need to find the boat's velocity relative to the water.
02

Set Up the Velocity Vector Equation

Let the velocity of the ferry relative to the water be \( \vec{v}_{fw} \). The velocity \( \vec{v}_{pw} \) of the passenger relative to the water is given as 4.50 m/s at 30.0掳 west of north. The passenger's velocity relative to the ferry, \( \vec{v}_{pf} \), is 1.50 m/s directly north. Using the vector addition formula \( \vec{v}_{pw} = \vec{v}_{pf} + \vec{v}_{fw} \), we can solve for \( \vec{v}_{fw} = \vec{v}_{pw} - \vec{v}_{pf} \).
03

Break Down the Velocity of the Passenger Relative to the Water

The passenger's velocity relative to the water can be broken into components: \( v_{py} = 4.50 \cos(30.0^{\circ}) \) for the northward component, and \( v_{px} = 4.50 \sin(30.0^{\circ}) \) for the westward component.
04

Express the Passenger's Velocity Relative to the Ferry

Since the passenger is only moving north on the ferry, \( \vec{v}_{pf} \) is represented as \( (0, 1.50) \) m/s in vector components, where 0 is the component in the west-east axis and 1.50 m/s is directed north.
05

Calculate Velocity Components for the Ferry Relative to the Water

Use the equation for vector subtraction, \( \vec{v}_{fw} = \vec{v}_{pw} - \vec{v}_{pf} \).1. North component: \( v_{fy} = 4.50 \cos(30.0^{\circ}) - 1.50 \)2. West component: \( v_{fx} = 4.50 \sin(30.0^{\circ}) \)
06

Compute Magnitude of the Ferry's Velocity

The magnitude of the velocity vector \( \vec{v}_{fw} = (v_{fx}, v_{fy}) \) can be calculated using Pythagorean theorem: \[|\vec{v}_{fw}| = \sqrt{(v_{fx})^2 + (v_{fy})^2}\] Substituting the calculated values gives the magnitude.
07

Determine the Direction of the Ferry's Velocity

The direction can be found by calculating the angle \( \theta \) with the north direction using the arctangent function: \[\theta = \tan^{-1}\left(\frac{v_{fx}}{v_{fy}}\right)\]. The angle will be west of north.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Velocity Components
Understanding velocity components is key in breaking down a complex motion into simpler and manageable parts. Motion in two dimensions can be described by separating it into horizontal and vertical components. This is done using trigonometric functions:
  • The velocity component in the northward direction is called the north or y-component.
  • The westward velocity component is in the x-direction.
In the exercise, the passenger's velocity of 4.50 m/s relative to the water is decomposed into these components. The angle provided, 30掳 west of north, allows us to use trigonometry to find these components:
  • The northward component is calculated as: \[v_{py} = 4.50 \cos(30.0^{\circ})\]
  • The westward component is: \[v_{px} = 4.50 \sin(30.0^{\circ})\]
These components are instrumental in determining the ferry's velocity by further calculations.
Vector Subtraction
Vector subtraction is a fundamental concept important for determining the relative velocity between objects in physics. Here's how it works in simple terms:
  • When we subtract vectors, we essentially find the difference between two motion states.
  • This can help determine the velocity of one object relative to another.
In our problem, the equation \(\vec{v}_{fw} = \vec{v}_{pw} - \vec{v}_{pf}\) is used. This helps find the ferry's velocity relative to the water by subtracting the passenger's velocity relative to the ferry. Each component of the velocity vector is subtracted individually:
  • North Component: \(v_{fy} = v_{py} - v_{pfy}\), where \(v_{pfy}\) is 1.50 m/s north.
  • West Component: \(v_{fx} = v_{px}\) as the ferry has no east-west movement from the passenger's perspective.
Vector subtraction allows us to see how these combined motions result in the ferry's movement.
Trigonometry in Physics
Trigonometry helps bridge the gap between angles and distances in physics, especially when dealing with directions and magnitudes. Here's how it comes into play:
  • It allows conversion between a vector's magnitude into its components that lie on perpendicular axes.
  • Using trigonometric functions like cosine and sine, any vector can be split based on the given angle, offering clarity in direction and magnitude.
In the exercise, finding the magnitude and direction of the ferry's velocity involves the Pythagorean Theorem:\[|\vec{v}_{fw}| = \sqrt{(v_{fx})^2 + (v_{fy})^2}\]And the direction is determined using the inverse tangent function:\[\theta = \tan^{-1}\left(\frac{v_{fx}}{v_{fy}}\right)\]These calculations simplify the problem into basic geometry, making it approachable and solvable using fundamental high-school trigonometry principles.

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Most popular questions from this chapter

The Longitude Problem In \(1755,\) John Harrison \((1693-1776)\) completed his fourth precision chronometer, the \(\mathrm{H} 4\) which eventually won the celebrated Longitude Prize. (For the human drama behind the Longitude Prize, see Longitude, by Dava Sobel.) When the minute hand of the H4 indicated 10 minutes past the hour, it extended \(3.0 \mathrm{cm}\) in the horizontal direction. (a) How long was the \(\mathrm{H} 4\) 's minute hand? (b) At 10 minutes past the hour, was the extension of the minute hand in the vertical direction more than, less than, or equal to \(3.0 \mathrm{cm}\) ? Explain. (c) Calculate the vertical extension of the minute hand at 10 minutes past the hour.

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As an airplane taxies on the runway with a speed of \(16.5 \mathrm{m} / \mathrm{s}\), a flight attendant walks toward the tail of the plane with a speed of \(1.22 \mathrm{m} / \mathrm{s}\). What is the flight attendant's speed relative to the ground?

Find the \(x\) and \(y\) components of a position vector \(\overrightarrow{\mathrm{r}}\) of magnitude \(r=75 \mathrm{m},\) if its angle relative to the \(x\) axis is (a) \(35.0^{\circ}\) and (b) \(65.0^{\circ}\).

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