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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

A dog in an open field is at rest under a tree at time \(t=0\) and then runs with acceleration \(\vec{a}(t)=\left(0.400 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\imath}-\left(0.180 \mathrm{~m} / \mathrm{s}^{3}\right) t \hat{\jmath}\) How far is the dog from the tree \(8.00 \mathrm{~s}\) after it starts to run?

An airplane pilot wishes to fly due west. A wind of \(80.0 \mathrm{~km} / \mathrm{h}\) (about \(50 \mathrm{mi} / \mathrm{h}\) ) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is \(320.0 \mathrm{~km} / \mathrm{h}\) (about \(200 \mathrm{mi} / \mathrm{h}),\) in which direction should the pilot head? (b) What is the speed of the plane over the ground? Draw a vector diagram.

A physics book slides off a horizontal tabletop with a speed of \(1.10 \mathrm{~m} / \mathrm{s} .\) It strikes the floor in \(0.480 \mathrm{~s}\). Ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor. (d) Draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

An airplane pilot sets a compass course due west and maintains an airspeed of \(220 \mathrm{~km} / \mathrm{h}\). After flying for \(0.500 \mathrm{~h},\) she finds herself over a town \(120 \mathrm{~km}\) west and \(20 \mathrm{~km}\) south of her starting point. (a) Find the wind velocity (magnitude and direction). (b) If the wind velocity is \(40 \mathrm{~km} / \mathrm{h}\) due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of \(220 \mathrm{~km} / \mathrm{h}\).

A baseball thrown at an angle of \(60.0^{\circ}\) above the horizontal strikes a building \(18.0 \mathrm{~m}\) away at a point \(8.00 \mathrm{~m}\) above the point from which it is thrown. Ignore air resistance. (a) Find the magnitude of the ball's initial velocity (the velocity with which the ball is thrown). (b) Find the magnitude and direction of the velocity of the ball just before it strikes the building.
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