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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, \(a_{avg} = \frac{-39400}{68,162,976} \approx -5.78\times10^{-4}\,\mathrm{m/s^2}\).

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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.

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Most popular questions from this chapter

The Kentucky Derby is held at the Churchill Downs track in Louisville, Kentucky. The track is one and one-quarter miles in length. One of the most famous horses to win this event was Secretariat. In 1973 he set a Derby record that would be hard to beat. His average acceleration during the last four quarter-miles of the race was \(+0.0105 \mathrm{m} / \mathrm{s}^{2}\). His velocity at the start of the final mile \((x=+1609 \mathrm{m})\) was about \(+16.58 \mathrm{m} / \mathrm{s} .\) The acceleration, although small, was very important to his victory. To assess its effect, determine the difference between the time he would have taken to run the final mile at a constant velocity of \(+16.58 \mathrm{m} / \mathrm{s}\) and the time he actually took. Although the track is oval in shape, assume it is straight for the purpose of this problem.

The Lost Drone. You and your team are exploring the edge of an Antarctic mountain range and you send a drone ahead to help navigate. After takeoff you lose sight of the drone and, a few seconds later, the controls malfunction and the drone stops sending visual images and navigational information except for speed and directional data. Changing speeds erratically, the drone heads west until it makes a drastic turn at the 5 -minute mark to \(35.0^{\circ}\) east of south. After nearly ten minutes, the speed drops to zero and the drone stops sending data. It has crashed. Using the speed directional data, the team draws up the graph shown in the drawing.. How far is the drone from you, and in what direction must you go to retrieve it? Express your result as a geographical direction (i.e., in the form \(30^{\circ}\) north of east, etc.).

A diver springs upward with an initial speed of \(1.8 \mathrm{m} / \mathrm{s}\) from a \(3.0-\mathrm{m}\) board. (a) Find the velocity with which he strikes the water. / Hint: When the diver reaches the water, his displacement is \(y=-3.0 \mathrm{m}\) (measured from the board ), assuming that the downward direction is chosen as the negative direction. \(J\) (b) What is the highest point he reaches above the water?

Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest and maintains a constant acceleration. Both cars cover a distance of \(460 \mathrm{m}\) in \(210 \mathrm{s}\). Assume that they are moving in the \(+x\) direction. Determine (a) the constant velocity of car A, (b) the final velocity of car \(\mathrm{B},\) and (c) the acceleration of car B.

A car makes a trip due north for three-fourths of the time and due south one- fourth of the time. The average northward velocity has a magnitude of \(27 \mathrm{m} / \mathrm{s},\) and the average southward velocity has a magnitude of \(17 \mathrm{m} / \mathrm{s} .\) What is the average velocity (magnitude and direction) for the entire trip?

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