Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 14

A particle undergoes three successive displacements in a plane, as follows: \(4.13 \mathrm{~m}\) southwest, \(5.26 \mathrm{~m}\) east, and \(5.94 \mathrm{~m}\) in a direction \(64.0^{\circ}\) north of east. Choose the \(x\) axis pointing east and the \(y\) axis pointing north and find \((a)\) the components of each displacement, \((b)\) the components of the resultant displacement, \((c)\) the magnitude and direction of the resultant displacement, and \((d)\) the displacement that would be required to bring the particle back to the starting point.

Short Answer

Expert verified
The components of each displacement are Displacement 1: (-4.13m, -4.13m), Displacement 2: (5.26m, 0m), Displacement 3: \(5.94m \cdot cos(64)\,m, 5.94m \cdot sin(64)\,m\). The components of the resultant displacement are given by the sum of the respective x and y components from each displacement. The magnitude of the resultant displacement is calculated using Pythagoras' theorem and the direction is found using the tangent of the ratio of the y and x components. The displacement required to return to the start is just the negative of these resultant components.

Step by step solution

01

Calculate components for each displacement

Firstly, calculate the components for each displacement, bearing in mind that in the southwest direction, both x and y components are negative.\r\nFor displacement 1 (4.13m southwest), both x and y components are same: \(-4.13 m\).\r\nFor displacement 2 (5.26m east), \(x = 5.26m\) and \(y = 0m\).\r\nFor displacement 3 (5.94m, 64 degrees north of east), \(x = 5.94m \cdot cos(64^\circ)\) and \(y = 5.94m \cdot sin(64^\circ)\).
02

Calculate components of the resultant displacement

The resultant displacement can be calculated by simply adding the respective x and y components from each displacement. The x components are \(-4.13m\), \(5.26m\), \(5.94m \cdot cos(64^\circ)\). The y components are \(-4.13m\), \(0m\), \(5.94m \cdot sin(64^\circ)\). So the resultant displacement in the x direction is \(-4.13m + 5.26m + 5.94m \cdot cos(64^\circ)\) and in the y direction is \(-4.13m + 0m + 5.94m \cdot sin(64^\circ)\).
03

Calculate magnitude and direction of the resultant displacement

The magnitude of the resultant displacement will be \(\sqrt{{(Resultant\_x)}^2 + {(Resultant\_y)}^2}\) and direction can be calculated as atan(\(Resultant\_y / Resultant\_x)\) if Resultant_x > 0 else atan(\(Resultant\_y / Resultant\_x)\) + 180. This is computed in degrees.
04

Calculate displacement required to return to the starting point

The displacement needed to return to the start, or 'the return vector', is just the negative of the resultant vector. So, calculate \(-Resultant_x\) and \(-Resultant_y\), which will be the x and y component of the displacement that brings the particle back to the starting point.

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 had a velocity of \(18 \mathrm{~m} / \mathrm{s}\) in the \(+x\) direction and 2.4 s later its velocity was \(30 \mathrm{~m} / \mathrm{s}\) in the opposite direction. What was the average acceleration of the particle during this 2.4-s interval?

New York Yankees pitcher Roger Clemens threw a fastball at a horizontal speed of \(160 \mathrm{~km} / \mathrm{h}\), as verified by a radar gun. How long did it take for the ball to reach home plate, which is \(18.4 \mathrm{~m}\) away?

A ball is dropped from a height of \(2.2 \mathrm{~m}\) and rebounds to a height of \(1.9 \mathrm{~m}\) above the floor. Assume the ball was in contact with the floor for \(96 \mathrm{~ms}\) and determine the average acceleration (magnitude and direction) of the ball during contact with the floor.

The legal speed limit on a highway is changed from \(55 \mathrm{mi} / \mathrm{h}\) \((=88.5 \mathrm{~km} / \mathrm{h})\) to \(65 \mathrm{mi} / \mathrm{h}(=104.6 \mathrm{~km} / \mathrm{h})\). How much time is thereby saved on a trip from the Buffalo entrance to the New York City exit of the New York State Thruway for someone traveling at the higher speed over this \(435-\mathrm{mi}(=700-\mathrm{km})\) stretch of highway?

The single cable supporting an unoccupied construction elevator breaks when the elevator is at rest at the top of a \(120-\mathrm{m}\) high building. ( \(a\) ) With what speed does the elevator strike the ground? \((b)\) For how long was it falling? \((c)\) What was its speed when it passed the halfway point on the way down? ( \(d\) ) For how long was it falling when it passed the halfway point?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Famous Physicists

Read Explanation

Turning Points in Physics

Read Explanation

Nuclear Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Energy 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