/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Over a time interval of 2.16 yea... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Over a time interval of 2.16 years, the velocity of a planet orbiting a distant star reverses direction, changing from \(+20.9 \mathrm{km} / \mathrm{s}\) to \(-18.5 \mathrm{km} / \mathrm{s}\) . Find (a) the total change in the planet's velocity (in \(\mathrm{m} / \mathrm{s}\) s and average acceleration (in \(\mathrm{m} / \mathrm{s}^{2} )\) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.

Short Answer

Expert verified
Change in velocity: -39,400 m/s; Average acceleration: -0.000578 m/s².

Step by step solution

01

Convert velocities to m/s

Convert the given velocities from km/s to m/s. To convert, multiply the velocity in km/s by 1000.Original velocity: \(+20.9\, \text{km/s} = 20.9 \times 1000 = +20900\, \text{m/s}\)Reversed velocity: \(-18.5\, \text{km/s} = -18.5 \times 1000 = -18500\, \text{m/s}\)
02

Calculate the total change in velocity (Δv)

The total change in velocity is given by the final velocity minus the initial velocity.\[\Delta v = v_{final} - v_{initial} = -18500 \, \text{m/s} - (+20900 \, \text{m/s})\]Calculate:\[\Delta v = -18500 - 20900 = -39400 \, \text{m/s}\]
03

Convert time from years to seconds

In order to find acceleration, the time interval must be in seconds since velocities are in meters per second. Convert 2.16 years to seconds using the conversion 1 year = 365.25 days, 1 day = 24 hours, 1 hour = 3600 seconds.\[\text{Time in seconds} = 2.16 \times 365.25 \times 24 \times 3600 = 68169600 \, \text{seconds}\]
04

Calculate average acceleration

Acceleration is the change in velocity divided by the time over which the change occurs. Using the formula \( a = \frac{\Delta v}{\Delta t} \):\[a = \frac{-39400 \, \text{m/s}}{68169600 \, \text{s}}\]Calculate:\[a \approx -0.000578 \, \text{m/s}^2\]

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 91Ó°ÊÓ!

Key Concepts

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

Planetary Motion
Planetary motion refers to the movement of planets around a star, like the Earth orbiting the Sun. This concept is crucial for understanding how celestial bodies like planets or satellites move in the cosmos. In our exercise, the motion involved is that of a planet reversing its velocity over a long period. Planets orbit in a largely elliptical path, and their velocities can vary significantly due to gravitational forces exerted by the central star.
  • The velocity can change in both magnitude and direction.
  • Planets do not orbit at a constant speed; their velocities can increase or decrease based on their position in the orbit.
Understanding planetary motion helps explain the dynamic forces such as gravity and inertia that keep the planets in their paths. This knowledge is essential for fields ranging from astronomy to space exploration.
Velocity Change
Velocity change is the difference between the final and initial velocity of an object. In the given problem, a planet's velocity changes from +20.9 km/s to -18.5 km/s, indicating a reversal.
Change in velocity (\(\Delta v\)) is computed as:\[\Delta v = v_{\text{final}} - v_{\text{initial}}\]
In practical terms, when velocity reverses, it implies that the object was stopped and then started moving in the opposite direction. This change plays a pivotal role in understanding motion dynamics, instructing that a decrease in velocity (or a negative change) denotes slowing down or reversing direction. The sign, either positive or negative, indicates the direction of motion relative to an initial reference point.
  • Positive velocity means motion is in the initial direction.
  • Negative velocity indicates a reverse direction.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the causes of this motion (like forces). The main objective is to describe the position, velocity, and acceleration of a body over time.
In this exercise, we focus on the average acceleration:\[a = \frac{\Delta v}{\Delta t}\]
  • It measures how quickly velocity changes over a given time interval.
  • A negative acceleration can indicate that an object is slowing down or changing direction.
  • A key part of kinematics is connecting graphs of position, velocity, and acceleration over time.
By determining average acceleration, we can better understand the changes in the motion of the planet. Kinematics provides a foundation for predicting future motion and analyzing the past movement of any object in straight-line or curved paths.
Unit Conversion
Unit conversion is fundamental in physics to ensure that all measurements are in a consistent system for accurate calculation and analysis. In the problem, both velocity and time require conversion:
  • Velocity: From km/s to m/s. Multiply by 1000 to change kilometers to meters.
  • Time: From years to seconds. Convert based on specific time units (365.25 days/year, 24 hours/day, 3600 seconds/hour).
Correct conversion is essential because using consistent units allows for correct application of physics formulas, which often depend on specific units. Inconsistencies in units can lead to errors in calculations and misunderstandings of physical phenomena. For students, mastering unit conversion is crucial for problem-solving across various topics in physics. The process enhances comprehension and accuracy, offering a reliable means to tackle more complex problems in science and engineering.

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

Two motorcycles are traveling due east with different velocities. However, four seconds later, they have the same velocity. During this four-second interval, cycle \(\mathrm{A}\) has an average acceleration of 2.0 \(\mathrm{m} / \mathrm{s}^{2}\) due east, while cycle \(\mathrm{B}\) has an average acceleration of 4.0 \(\mathrm{m} / \mathrm{s}^{2}\) due east. By how much did the speeds differ at the beginning of the four-second interval, and which motorcycle was moving faster?

A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.0 s later it is rising at a speed of 15 \(\mathrm{m} / \mathrm{s}\) . Assuming air resistance has no effect on the rock, calculate its speed \((\mathrm{a})\) at launch and (b) 5.0 \(\mathrm{s}\) after launch.

In 1954 the English runner Roger Bannister broke the four-minute barrier for the mile with a time of \(3 : 59.4 \mathrm{s}(3 \mathrm{min} \text { and } 59.4 \mathrm{s}) .\) In 1999 the Moroccan runner Hicham el-Guerrouj set a record of \(3 : 43.13\) s for the mile. If these two runners had run in the same race, each running the entire race at the average speed that earned him a place in the record books, el-Guerrouj would have won. By how many meters?

A Boeing 747 Jumbo Jet has a length of 59.7 \(\mathrm{m}\) . The runway on which the plane lands intersects another runway. The width of the inter- section is 25.0 \(\mathrm{m}\) . The plane decelerates through the intersection at a rate of 5.70 \(\mathrm{m} / \mathrm{s}^{2}\) and clears it with a final speed of 45.0 \(\mathrm{m} / \mathrm{s}\) . How much time is needed for the plane to clear the intersection?

Due to continental drift, the North American and European continents are drifting apart at an average speed of about 3 \(\mathrm{cm}\) per year. At this speed, how long (in years) will it take for them to drift apart by another 1500 \(\mathrm{m}\) (a little less than a mile)?

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.