Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Your average velocity over a 10 -min period is \(2.2 \mathrm{~m} / \mathrm{s}\). Is it possible that you were at rest at some point during the \(10 \mathrm{~min}\) ?

Short Answer

Expert verified
Yes, it is possible to be at rest at some point.

Step by step solution

01

Understand the Average Velocity

Average velocity is calculated by dividing total displacement by total time. In this case, your average velocity over 10 minutes is given as \(2.2 \mathrm{~m/s}\), meaning over the entire 10-minute period, your displacement divided by 600 seconds (since 10 minutes = 600 seconds) is 2.2 meters per second.
02

Calculate Total Displacement

Since average velocity \( v_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} \), we can rearrange to find Total Displacement = \(v_{avg} \times \text{Total Time} = 2.2 \frac{m}{s} \times 600 \text{ seconds} = 1320 \mathrm{~m}\).
03

Consider Possibilities for Rest

To determine if it is possible to be at rest at some point, think about the nature of average velocity. Average velocity can still be \(2.2 \mathrm{~m/s}\) even if the speed varies. This could mean periods of faster travel and periods of being at rest, as long as the average equates to the given value.
04

Conclusion Based on Variability

Since it's possible to have moments of rest if periods of greater speed make up for it, the average velocity of \(2.2 \mathrm{~m/s}\) does not exclude the possibility of being at rest at some points in the 10-minute interval.

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.

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

Key Concepts

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

Displacement
Displacement is a fundamental concept in understanding motion and average velocity. It refers to the overall change in position of an object from its starting point to its final position. Unlike distance, which accounts for the total path traveled, displacement considers only the shortest path between the initial and final positions. This measurement is vectorial, which means it has both magnitude and direction. For instance, if you start at point A, move to point B, and return to point A, your displacement is zero because you ended up at your starting location.
To calculate displacement when given average velocity and time, use the formula for average velocity: \( v_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} \). Rearrange it to find total displacement:
  • \( \text{Total Displacement} = v_{avg} \times \text{Total Time} \)
In the given exercise, the total displacement over 10 minutes is determined by multiplying the average velocity of \( 2.2 \text{ m/s} \) by the total time of \( 600 \text{ seconds} \), resulting in 1320 meters.
Total Time
Total time is the complete duration over which a process, such as motion, occurs. It's a crucial component in calculating average velocity because it provides the time span during which displacement takes place. In this exercise, the specified total time is 10 minutes, which converts into 600 seconds. Knowing how to change minutes into seconds is vital for problems involving different units of time.
The conversion from minutes to seconds is straightforward:
  • 1 minute = 60 seconds
  • Thus, 10 minutes = 10 x 60 = 600 seconds
This total time is used in the calculation of average velocity to determine the total displacement. The consistency in units across all measurements, including time, ensures that calculations are accurate and meaningful.
Rest
Rest is the condition where an object's speed is zero, meaning it temporarily stops moving. Despite having an average velocity, it is possible for motion to include periods of rest. Think of average velocity as an overall summary of speed over time—it doesn’t capture each moment’s speed but rather provides a general output over the whole period.
For instance, if an object has a period of rest, it might also have times of greater speed to maintain the average velocity. This variability in speed, including rest periods, can still lead to an average velocity of \(2.2 \text{ m/s}\) as long as faster movements compensate for the times at rest.
  • Period of rest contributes 0 m/s to the velocity
  • To balance out, faster speeds at other times must average to maintain the overall velocity
Understanding rest and its impact on average velocity demonstrates how measurements in physics can be 'average'—taking into account variations to present a coherent overview of an object's motion over a set time frame.

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

Which has the greater displacement, object 1 , which moves from \(5.0 \mathrm{~m}\) to \(7.0 \mathrm{~m}\) in \(2.0 \mathrm{~s}\), or object 2 , which moves from \(15 \mathrm{~m}\) to \(16 \mathrm{~m}\) in \(25 \mathrm{~s}\) ? Explain.

Two dragonflies have the following equations of motion: $$ \begin{aligned} &x_{1}=2.2 \mathrm{~m}+(0.75 \mathrm{~m} / \mathrm{s}) t \\ &x_{2}=-3.1 \mathrm{~m}+(-1.1 \mathrm{~m} / \mathrm{s}) t \end{aligned} $$ (a) Which dragonfly is moving faster? (b) Which dragonfly starts out closer to \(x=0\) at \(t=0\) ?

Sketch a position-time graph for an object that starts at \(x=1.5 \mathrm{~m}\), moves with a velocity of \(2.2 \mathrm{~m} / \mathrm{s}\) from \(t=0\) to \(t=1 \mathrm{~s}\), has a velocity of \(0 \mathrm{~m} / \mathrm{s}\) from \(t=1 \mathrm{~s}\) to \(t=2 \mathrm{~s}\), and has a velocity of \(-3.7 \mathrm{~m} / \mathrm{s}\) from \(t=2 \mathrm{~s}\) to \(t=5 \mathrm{~s}\).

Think \& Calculate You drive in a straight line at \(20.0 \mathrm{~m} / \mathrm{s}\) for \(10.0 \mathrm{~min}\), then at \(30.0 \mathrm{~m} / \mathrm{s}\) for another \(10.0 \mathrm{~min}\). (a) Is your average speed \(25.0 \mathrm{~m} / \mathrm{s}\), more than \(25.0 \mathrm{~m} / \mathrm{s}\), or less than \(25.0 \mathrm{~m} / \mathrm{s}\) ? Explain. (b) Verify your answer to part (a) by calculating the average speed.

Can you take a bicycle ride in such a way that the distance you cover is less than the magnitude of your displacement? Give an example if your answer is yes; explain why not if your answer is no.

See all solutions

Recommended explanations on Physics Textbooks

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