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Chapter 3: Problem 54

Two friends, Barbara and Neil, are out rollerblading. With respect to the ground, Barbara is skating due south at a speed of \(4.0 \mathrm{m} / \mathrm{s}\). Neil is in front of her. With respect to the ground, Neil is skating due west at a speed of \(3.2 \mathrm{m} / \mathrm{s}\). Find Neil's velocity (magnitude and direction relative to due west), as seen by Barbara.

Short Answer

Expert verified
Neil's relative velocity is approximately 5.12 m/s at 51.34掳 south of west.

Step by step solution

01

Understand Relative Velocity

Relative velocity is used to determine how fast one object is moving with respect to another. In this scenario, we want to find Neil's velocity as seen by Barbara. We will denote this as \( \vec{v}_{N/B} \), which stands for Neil's velocity relative to Barbara.
02

Break Down the Problem

Barbara's velocity \( \vec{v}_B \) is due south at \( 4.0 \ \mathrm{m/s} \), which we can represent as a vector \( (0, -4.0) \) since it moves along the negative y-axis. Neil's velocity \( \vec{v}_N \) is due west at \( 3.2 \ \mathrm{m/s} \), represented as \( (-3.2, 0) \) because it moves along the negative x-axis.
03

Apply the Relative Velocity Formula

To find the relative velocity of Neil with respect to Barbara (\( \vec{v}_{N/B} \)), subtract Barbara's velocity vector \( \vec{v}_B \) from Neil's velocity vector \( \vec{v}_N \). The formula is \( \vec{v}_{N/B} = \vec{v}_N - \vec{v}_B \). Substitute the values: \[\vec{v}_{N/B} = (-3.2, 0) - (0, -4.0) = (-3.2, 4.0)\].
04

Calculate the Magnitude of the Relative Velocity

The magnitude \( |\vec{v}_{N/B}| \) of the relative velocity vector can be found using the Pythagorean theorem: \[ |\vec{v}_{N/B}| = \sqrt{(-3.2)^2 + (4.0)^2} \]. Calculating, we find: \[ |\vec{v}_{N/B}| = \sqrt{10.24 + 16} = \sqrt{26.24} \approx 5.12 \ \mathrm{m/s} \].
05

Determine the Direction of the Relative Velocity

The direction \( \theta \) of the relative velocity is given by \[ \theta = \tan^{-1}\left(\frac{-4.0}{-3.2}\right) \]. The negative signs cancel each other out, giving \[ \theta = \tan^{-1}\left(\frac{4.0}{3.2}\right) \approx \tan^{-1}(1.25) \approx 51.34\degree \]. Since we calculated this with respect to the west, Neil's relative direction as seen by Barbara is \( 51.34\degree \) south of west.

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Key Concepts

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

Vector Subtraction
Vector subtraction is crucial to understanding relative velocity because it allows you to see the motion of one object from the perspective of another. In this rollerblading scenario, we want to find the velocity of Neil as seen by Barbara. To do this, you need to subtract Barbara's velocity vector from Neil's. This is done by considering their individual velocities as vectors:
  • Barbara's velocity is due south: represented as \((0, -4.0)\) since it doesn't have a component in the x-direction.
  • Neil's velocity is due west: represented as \((-3.2, 0)\) because it doesn't have a y-component.
To find the result, compute: \[\vec{v}_{N/B} = \vec{v}_N - \vec{v}_B = (-3.2, 0) - (0, -4.0) = (-3.2, 4.0)\]This gives us the vector \((-3.2, 4.0)\), describing Neil鈥檚 velocity from Barbara's viewpoint.
Magnitude Calculation
The next step is to find out how fast Neil is traveling from Barbara's perspective. This involves calculating the magnitude of the vector, which tells us the speed. The magnitude of the vector \((-3.2, 4.0)\) can be found using the Pythagorean theorem. This theorem helps calculate the 'length' of the velocity vector, as the vector essentially forms a right triangle with the x and y components.First, square each component, sum them, and take the square root:\[|\vec{v}_{N/B}| = \sqrt{(-3.2)^2 + (4.0)^2} = \sqrt{10.24 + 16} = \sqrt{26.24} \approx 5.12 \, \text{m/s}\]Neil鈥檚 relative speed, as seen by Barbara, is approximately 5.12 meters per second.
Direction Determination
Understanding the direction is as important as knowing the speed in relative motion. It tells us where one object is going from another's perspective. For this purpose, we will calculate the angle \(\theta\), the direction of Neil's relative velocity.Use the tangent function which compares the sides of the velocity vector triangle:\[ \theta = \tan^{-1}\left(\frac{4.0}{-3.2}\right)\]Upon simplifying, the negative signs cancel. Calculate the angle:\[\theta = \tan^{-1}(1.25) \approx 51.34\degree\]This angle is measured relative to west, placing Neil's direction at approximately 51.34 degrees south of west, as visible to Barbara. Neil appears to be moving not only away from due west but also veering south from Barbara's position.

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Most popular questions from this chapter

An airplane with a speed of 97.5 m/s is climbing upward at an angle of \(50.0^{\circ}\) with respect to the horizontal. When the plane's altitude is \(732 \mathrm{m},\) the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

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A spacecraft is traveling with a velocity of \(v_{0 x}=5480 \mathrm{m} / \mathrm{s}\) along the \(+x\) direction. Two engines are turned on for a time of 842 s. One engine gives the spacecraft an acceleration in the \(+x\) direction of \(a_{x}=1.20 \mathrm{m} / \mathrm{s}^{2}\), while the other gives it an acceleration in the \(+y\) direction of \(a_{y}=8.40 \mathrm{m} / \mathrm{s}^{2}\) At the end of the firing, find (a) \(v_{x}\) and (b) \(v_{y^{*}}\)

A dolphin leaps out of the water at an angle of \(35^{\circ}\) above the horizontal. The horizontal component of the dolphin's velocity is \(7.7 \mathrm{m} / \mathrm{s}\). Find the magnitude of the vertical component of the velocity.
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