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Chapter 3: Problem 73

In a World Cup soccer match, Juan is running due north toward the goal with a speed of \(8.00 \mathrm{~m} / \mathrm{s}\) relative to the ground. A teammate passes the ball to him. The ball has a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) and is moving in a direction \(37.0^{\circ}\) east of north, relative to the ground. What are the magnitude and direction of the ball's velocity relative to Juan?

Short Answer

Expert verified
The velocity of the ball relative to Juan is 7.4 m/s, 77.3 degrees East of North.

Step by step solution

01

Subtraction of Vectors

Start by defining a direction of motion for Juan and the football relative to the ground. First, we subtract the velocity of Juan from the velocity of the ball. This gives us the velocity of the ball relative to Juan; represented by \(V_{B/J} = V_{B/G} - V_{J/G}\). When subtracting, we subtract the components of each vector. The velocity of Juan is only on the north, so it will be \(V_{J/G} = 8.00 \, m/s \, N\). For the velocity of the ball, it is given as \(12.0 \, m/s\) \(37.0^{\circ}\) east of north, we will have to determine the east and north components.
02

Split the Ball's Velocity into Components

The velocity of the ball is given in a direction that is east of north direction. We have to find the northern component (\(V_{BN}\)) and eastern component (\(V_{BE}\)). The northern component can be given by \(V_{BN} = V_{B} \cos(37.0^{\circ}) = 12.0 \, m/s \cos(37.0^{\circ}) = 9.6 \, m/s \, N\). The eastern component can be given by \(V_{BE} = V_{B} \sin(37.0^{\circ}) = 12.0 \, m/s \sin(37.0^{\circ}) = 7.2 \, m/s \, E\). So, \(V_{B/G} = 9.6 \, m/s \, N + 7.2 \, m/s \, E\).
03

Determine the velocity of the Ball relative to Juan

Perform the subtraction of \(V_{B/G}\) and \(V_{J/G}\) by subtracting the components, given \(V_{B/J} = V_{B/G} - V_{J/G} = (9.6-8.0) \, m/s \, N + 7.2 \, m/s \, E = 1.6 \, m/s \, N + 7.2 \, m/s \, E\). The negative in front of the north component describes the direction.
04

Calculate for Magnitude and Direction

The magnitude (\(v_{B/J}\)) of the velocity of the ball observed from the position of Juan can be calculated using Pythagoras theorem. \(v_{B/J} = \sqrt{(1.6)^{2} + (7.2)^{2}} \, m/s = 7.4 \, m/s\). The direction (\(胃_{B/J}\)) can be calculated using the tangent. \(胃_{B/J} = \arctan(\frac{7.2}{1.6}) = 77.3^{\circ}\). The angle is north of the east, which is opposite of our normal convention of measuring counterclockwise from the x-axis (east).

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Key Concepts

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

Vector Components
When dealing with vectors, it is often necessary to break them down into smaller pieces called components. This makes it easier to analyze them. In our soccer scenario, the ball's velocity vector is not aligned with the basic north or east directions, making component analysis crucial.

To find these components, we use trigonometric functions:
  • The **north component** of the velocity uses the cosine function: \[ V_{BN} = V_{B} \cos(37.0^{\circ}) = 12.0 \, \text{m/s} \times \cos(37.0^{\circ}) = 9.6 \, \text{m/s} \, \text{N} \]
  • The **east component** uses the sine function:\[ V_{BE} = V_{B} \sin(37.0^{\circ}) = 12.0 \, \text{m/s} \times \sin(37.0^{\circ}) = 7.2 \, \text{m/s} \, \text{E} \]
Using these components simplifies further calculations, such as subtraction or addition of vectors.
Vector Subtraction
Vector subtraction is a crucial operation when determining relative motion, such as the velocity of the ball relative to Juan. The formula used is:

\[ V_{B/J} = V_{B/G} - V_{J/G} \]
Here, it's important to subtract each component separately:
  • The **north component**: \[ 9.6 \, \text{m/s} - 8.0 \, \text{m/s} = 1.6 \, \text{m/s} \]
  • The **east component** remains the same: \[ 7.2 \, \text{m/s} \]
So, the velocity of the ball relative to Juan is broken down into:
- **North**: 1.6 m/s
- **East**: 7.2 m/s
This method allows for easy determination of more complex vector relationships.
Pythagorean Theorem
Once vector components are found, the next task is to combine these into a single magnitude of the vector. This is where the Pythagorean Theorem comes into play. We apply it to find the overall speed of the ball relative to Juan.

Use the following equation:
\[ v_{B/J} = \sqrt{(1.6)^{2} + (7.2)^{2}} \, \text{m/s} \]
Solving this, we get:
  • \[ v_{B/J} = \sqrt{2.56 + 51.84} \, \text{m/s} = \sqrt{54.4} \, \text{m/s} \]
  • \[ v_{B/J} \approx 7.4 \, \text{m/s} \]
The Pythagorean Theorem proves essential in consolidating information from vector components to get a clearer understanding of actual motion.
Angle Calculation
Determining the direction of the ball's movement relative to Juan requires calculating the angle from the north to the east components. We use the tangent function for this purpose:

\[ \theta_{B/J} = \arctan\left(\frac{7.2}{1.6}\right) \]
This gives an angle:
  • \( \theta_{B/J} \approx 77.3^{\circ} \) north of east
This angle helps us visualize the ball's path from Juan's perspective. Remember, the angle is measured from the north to the east, illustrating how the direction overlays on standard coordinate systems.

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Most popular questions from this chapter

At its Ames Research Center, NASA uses its large "20 G" centrifuge to test the effects of very large accelerations ("hypergravity") on test pilots and astronauts. In this device, an arm \(8.84 \mathrm{~m}\) long rotates about one end in a horizontal plane, and an astronaut is strapped in at the other end. Suppose that he is aligned along the centrifuge's arm with his head at the outermost end. The maximum sustained acceleration to which humans are subjected in this device is typically \(12.5 g .\) (a) How fast must the astronaut's head be moving to experience this maximum acceleration? (b) What is the difference between the acceleration of his head and feet if the astronaut is \(2.00 \mathrm{~m}\) tall? (c) How fast in rpm (rev/min) is the arm turning to produce the maximum sustained acceleration?

Henrietta is jogging on the sidewalk at \(3.05 \mathrm{~m} / \mathrm{s}\) on the way to her physics class. Bruce realizes that she forgot her bag of bagels, so he runs to the window, which is \(38.0 \mathrm{~m}\) above the street level and directly above the sidewalk, to throw the bag to her. He throws it horizontally \(9.00 \mathrm{~s}\) after she has passed below the window, and she catches it on the run. Ignore air resistance. (a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground? (b) Where is Henrietta when she catches the bagels?

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A dog running in an open field has components of velocity \(v_{x}=2.6 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=-1.8 \mathrm{~m} / \mathrm{s}\) at \(t_{1}=10.0 \mathrm{~s} .\) For the time interval from \(t_{1}=10.0 \mathrm{~s}\) to \(t_{2}=20.0 \mathrm{~s},\) the average acceleration of the dog has magnitude \(0.45 \mathrm{~m} / \mathrm{s}^{2}\) and direction \(31.0^{\circ}\) measured from the \(+x\) -axis toward the \(+y\) -axis. At \(t_{2}=20.0 \mathrm{~s},\) (a) what are the \(x\) - and \(y\) -components of the dog's velocity? (b) What are the magnitude and direction of the dog's velocity? (c) Sketch the velocity vectors at \(t_{1}\) and \(t_{2}\). How do these two vectors differ?

A railroad flatcar is traveling to the right at a speed of \(13.0 \mathrm{~m} / \mathrm{s}\) relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (Fig. E3.36). What is the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter's velocity relative to the observer on the ground is (a) \(18.0 \mathrm{~m} / \mathrm{s}\) to the right? (b) \(3.0 \mathrm{~m} / \mathrm{s}\) to the left? (c) zero?
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