/*! 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 \(\mathrm{m} / \mathrm{s}\) ) and (b) its 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
(a) Total change in velocity is \(-39400\,\mathrm{m/s}\); (b) Average acceleration is approx. \(-5.78\times10^{-4}\,\mathrm{m/s^2}\).

Step by step solution

01

Convert Velocities to Meters Per Second

The initial velocity of the planet is given as \(+20.9\,\mathrm{km/s}\). Convert this to meters per second by multiplying by 1000. Thus, the initial velocity \(v_i\) is \(20.9\times1000 = 20900\,\mathrm{m/s}\). Similarly, the final velocity is \(-18.5\,\mathrm{km/s}\) which is \(-18.5\times1000 = -18500\,\mathrm{m/s}\).
02

Calculate the Total Change in Velocity

The change in velocity \(\Delta v\) is the final velocity \(v_f\) minus the initial velocity \(v_i\). Use the formula: \(\Delta v = v_f - v_i = -18500 - 20900 = -39400\,\mathrm{m/s}\).
03

Convert Time Interval to Seconds

The time interval is provided in years. First, convert years into seconds. There are approximately \(365.25\times24\times60\times60 = 31,557,600\,\mathrm{seconds}\) in a year. So, \(2.16\,\mathrm{years}\) is equivalent to \(2.16\times31,557,600 = 68,162,976\,\mathrm{seconds}\).
04

Calculate the Average Acceleration

The average acceleration \(a_{avg}\) is the change in velocity \(\Delta v\) divided by the time interval \(\Delta t\). So, \(a_{avg} = \frac{-39400}{68,162,976} \approx -5.78\times10^{-4}\,\mathrm{m/s^2}\).

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.

Change in Velocity
Change in velocity is an important concept in physics that describes how the speed and direction of an object, such as a planet, shifts over time. In this exercise, we observe a planet's velocity transitioning from \( +20.9 \, \mathrm{km/s} \) to \( -18.5 \, \mathrm{km/s} \).
This involves not just a change of speed, but also a reversal in direction. The formula used to calculate the change in velocity \( \Delta v \) is:
  • \( \Delta v = v_f - v_i \)
where \( v_f \) is the final velocity and \( v_i \) is the initial velocity.
This calculation helps determine how quickly the planet is decelerating or accelerating in the opposite direction. Here, substituting the given velocities:
  • \( \Delta v = -18,500 \,\mathrm{m/s} - 20,900 \,\mathrm{m/s} = -39,400 \,\mathrm{m/s} \)
The negative sign indicates not only a slow down but also a change in direction to the opposite.
Velocity Conversion
Converting velocity values from one unit to another is crucial, as it ensures consistency in measurements, especially in scientific calculations. In this context, converting from kilometers per second (km/s) to meters per second (m/s) involves a straightforward multiplication:
  • 1 kilometer = 1000 meters
To convert the initial planet's velocity from \( +20.9 \, \mathrm{km/s} \) to meters per second, multiply by 1000:
  • \( 20.9 \,\mathrm{km/s} \times 1000 = 20,900 \,\mathrm{m/s} \)
Similarly, for the final velocity \( -18.5 \, \mathrm{km/s} \):
  • \( -18.5 \,\mathrm{km/s} \times 1000 = -18,500 \,\mathrm{m/s} \)
Ensuring your units are consistent allows you to accurately compare and calculate changes in physical quantities.
Time Conversion
Accurately converting time units is a fundamental step when working with extended periods. Here, the time interval given is in years, which must be converted to seconds to properly compute average acceleration.
One year is typically approximated to contain:
  • 365.25 days (accounting for leap years)
  • Each day has 24 hours
  • Each hour has 60 minutes
  • Each minute has 60 seconds
To convert 2.16 years into seconds:
  • \( 2.16 \,\text{years} \times 31,557,600 \, \text{seconds/year} = 68,162,976 \, \text{seconds} \)
This conversion is essential for accurately determining rates of change, such as acceleration, over time.
Planetary Motion
Planetary motion refers to the movement of planets, often around a star, and involves complex interactions of forces, primarily gravity.
In this particular exercise, examining the velocity change of a planet helps us understand the dynamics of its orbit.Average acceleration in this context shows how the planet's speed and direction systematically change. The formula for average acceleration \( a_{avg} \) is:
  • \( a_{avg} = \frac{\Delta v}{\Delta t} \)
where \( \Delta v = -39,400 \,\mathrm{m/s} \) and time interval \( \Delta t = 68,162,976 \, \text{seconds} \).
Calculating gives:
  • \( a_{avg} = \frac{-39,400}{68,162,976} \approx -5.78 \times 10^{-4} \,\mathrm{m/s^2} \)
This negative acceleration indicates the reversal direction of the planet's motion. Understanding these concepts aids in studying the nature of celestial bodies and their movements in the universe.

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 cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position \(x=0 \mathrm{m}\) with an initial velocity of \(+5.0 \mathrm{m} / \mathrm{s}\) and a constant acceleration due to the fan. The direction to the right is positive. The cart reaches a maximum position of \(x=+12.5 \mathrm{m},\) where it begins to travel in the negative direction. Find the acceleration of the cart.

Two arrows are shot vertically upward. The second arrow is shot after the first one, but while the first is still on its way up. The initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. Suppose that the initial speed of the first arrow is \(25.0 \mathrm{m} / \mathrm{s}\) and that the second arrow is fired \(1.20 \mathrm{s}\) after the first. Determine the initial speed of the second arrow.

You are on a train that is traveling at \(3.0 \mathrm{m} / \mathrm{s}\) along a level straight track. Very near and parallel to the track is a wall that slopes upward at a \(12^{\circ}\) angle with the horizontal. As you face the window \((0.90 \mathrm{m}\) high, \(2.0 \mathrm{m}\) wide) in your compartment, the train is moving to the left, as the drawing indicates. The top edge of the wall first appears at window corner A and eventually disappears at window corner B. How much time passes between appearance and disappearance of the upper edge of the wall?

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 (a) at launch and (b) \(5.0 \mathrm{s}\) after launch.

At the beginning of a basketball game, a referee tosses the ball straight up with a speed of \(4.6 \mathrm{m} / \mathrm{s} .\) A player cannot touch the ball until after it reaches its maximum height and begins to fall down. What is the minimum time that a player must wait before touching the ball?

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.