Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 5

Two displacement vectors each have magnitude \(20 \mathrm{km}\) One is directed \(60^{\circ}\) above the \(+x\) -axis; the other is directed \(60^{\circ}\) below the \(+x\) -axis. What is the vector sum of these two displacements? Use graph paper to find your answer.

Short Answer

Expert verified
Answer: To find the vector sum, first find the x and y components of each vector. Then, add the x-components and y-components separately. Finally, convert the sum of the components back into polar form (magnitude and angle) using the Pythagorean theorem and arctangent function. The magnitude of the vector sum and the angle of the vector sum will give the final answer.

Step by step solution

01

Find the components of the first vector

To find the x and y components of the first vector, we will use trigonometry. Recall that the x-component of a vector is given by its magnitude multiplied by the cosine of the angle it makes with the x-axis, while the y-component is given by its magnitude multiplied by the sine of that angle. For the first vector, magnitude = 20km and angle = 60 degrees above +x-axis. So, the x-component is 20*cos(60°), and the y-component is 20*sin(60°).
02

Find the components of the second vector

Similarly, for the second vector, magnitude = 20km and angle = 60 degrees below +x-axis. For this vector, we will take the angle as -60 degrees to account for the negative direction of the angle. The x-component is 20*cos(-60°) and the y-component is 20*sin(-60°).
03

Add the x-components and y-components of the two vectors

To find the vector sum, we will add the x-components and y-components separately. Sum of x-components = 20*cos(60°) + 20*cos(-60°) Sum of y-components = 20*sin(60°) + 20*sin(-60°)
04

Convert back to polar form

To convert back to polar form(pa and coord), we will use the Pythagorean theorem and the arctangent function. Magnitude of the vector sum = sqrt[(sum of x-components)^2 + (sum of y-components)^2] Angle of the vector sum = arctan(sum of y-components)/(sum of x-components) The magnitude of the vector sum and the angle of the vector sum will give the final answer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle's constant acceleration is south at $2.50 \mathrm{m} / \mathrm{s}^{2}\( At \)t=0,\( its velocity is \)40.0 \mathrm{m} / \mathrm{s}$ east. What is its velocity at \(t=8.00 \mathrm{s} ?\)
Orville walks 320 m due east. He then continues walking along a straight line, but in a different direction, and stops \(200 \mathrm{m}\) northeast of his starting point. How far did he walk during the second portion of the trip and in what direction?
A particle's constant acceleration is north at $100 \mathrm{m} / \mathrm{s}^{2}\( At \)t=0,\( its velocity vector is \)60 \mathrm{m} / \mathrm{s}$ east. At what time will the magnitude of the velocity be $100 \mathrm{m} / \mathrm{s} ?$
A boat that can travel at \(4.0 \mathrm{km} / \mathrm{h}\) in still water crosses a river with a current of \(1.8 \mathrm{km} / \mathrm{h}\). At what angle must the boat be pointed upstream to travel straight across the river? In other words, in what direction is the velocity of the boat relative to the water?
You are working as a consultant on a video game designing a bomb site for a World War I airplane. In this game, the plane you are flying is traveling horizontally at \(40.0 \mathrm{m} / \mathrm{s}\) at an altitude of $125 \mathrm{m}$ when it drops a bomb. (a) Determine how far horizontally from the target you should release the bomb. (b) What direction is the bomb moving just before it hits the target?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Space Physics

Read Explanation

Quantum Physics

Read Explanation

Classical Mechanics

Read Explanation

Linear Momentum

Read Explanation

Turning Points in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free