/*! 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 70 In 1998 , NASA launched Deep Spa... [FREE SOLUTION] | 91Ó°ÊÓ

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In 1998 , NASA launched Deep Space \(I\) (DS-1), a spacecraft that successfully flew by the asteroid named 1992 KD (which orbits the sun millions of miles from the earth). The propulsion system of DS-1 worked by ejecting high-speed argon ions out the rear of the engine. The engine slowly increased the velocity of DS-1 by about \(+9.0 \mathrm{~m} / \mathrm{s}\) per day. (a) How much time (in days) would it take to increase the velocity of DS-1 by \(+2700 \mathrm{~m} / \mathrm{s}\) ? (b) What was the acceleration of \(\mathrm{DS}-1\left(\mathrm{in} \mathrm{m} / \mathrm{s}^{2}\right) ?\)

Short Answer

Expert verified
(a) 300 days; (b) \(1.04 \times 10^{-4} \mathrm{~m/s^2}\).

Step by step solution

01

Identify Known Variables for Part (a)

In part (a), we need to find the time it takes to increase the velocity by \(+2700 \mathrm{~m/s}\). We know that DS-1's velocity increases by \(+9.0 \mathrm{~m/s}\) per day.
02

Calculate Time for Velocity Change

To find the time in days, divide the total increase in velocity \(+2700 \mathrm{~m/s}\) by the daily increase in velocity \(+9.0 \mathrm{~m/s/day}\):\[\text{Time} = \frac{2700 \mathrm{~m/s}}{9.0 \mathrm{~m/s/day}} = 300 \text{ days}\]
03

Identify Known Variables for Part (b)

Now for part (b), we calculate the acceleration. The change in velocity \(+9.0 \mathrm{~m/s}\) occurs over a period of 1 day. We need acceleration in \(\mathrm{m/s^2}\).
04

Convert Time for Acceleration Calculation

Convert the period of 1 day to seconds, since acceleration \(a\) has units \(\mathrm{m/s^2}\). Since 1 day has 86400 seconds (\(24 \times 60 \times 60\)), use this for conversion.
05

Calculate Acceleration

Calculate acceleration using \\[a = \frac{\Delta v}{\Delta t}\]where \(\Delta v = 9.0 \mathrm{~m/s}\) and \(\Delta t = 86400 \mathrm{~s}\).\[a = \frac{9.0}{86400} \approx 1.04 \times 10^{-4} \mathrm{~m/s^2}\]

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Key Concepts

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

Kinematics in Space Missions
Kinematics deals with the motion of objects without considering the forces causing the motion. It's essential for understanding how spacecraft move through space.
When NASA's Deep Space I (DS-1) was launched, it followed principles of kinematics by determining how its speed changed over time. The spacecraft could adjust its velocity each day using a special propulsion system. This constant change in speed can be calculated by comparing the desired change in velocity to the change it achieves daily.
  • This exercise involves calculating the time it would take for DS-1's velocity to increase from zero to a specific value.
  • It's a practical example of how kinematic equations can be used in real-world scenarios, such as space exploration missions.
  • These principles help scientists predict how much time and energy they'll need for a spacecraft to reach its destination or to change its trajectory.
Understanding Acceleration Calculation
Acceleration is a measure of how quickly an object's velocity changes. It is one of the fundamental aspects of motion in physics. In the case of DS-1, understanding its acceleration was crucial to ensure it could reach its target destination efficiently. To calculate acceleration, you need to know two things:
1. The change in velocity, known as \( \Delta v \)2. The time over which this change occurs, known as \( \Delta t \)
The formula for acceleration is:\[a = \frac{\Delta v}{\Delta t}\] In the DS-1 system, \( \Delta v \) is the velocity change per day, and it's vital to convert the time from days to seconds for standard units of acceleration (meters per second squared).
The steps involved are:
  • Ensure you convert days into seconds since there are 86400 seconds in a day.
  • Apply the formula to find acceleration: divide the change in velocity by time in seconds.
  • For DS-1, this calculated to approximately \(1.04 \times 10^{-4}\; \mathrm{m/s^2}\).
The Significance of NASA Space Missions
NASA's space missions, like the one involving Deep Space I, demonstrate the powerful application of physics in real-world scenarios. These missions are critical for advancing our understanding of the universe and testing new technologies.
The propulsion system used in DS-1 was a groundbreaking ion propulsion system, which worked by ejecting high-speed ions to produce thrust. This is different from conventional propulsion methods and showcases the innovative spirit of NASA's research and technology development.
Key points about NASA space missions include:
  • Testing new technologies to enhance space exploration capabilities.
  • Understanding trajectories, speeds, and acceleration to ensure accurate mission planning.
  • Gaining data that assists in future exploration, such as missions to asteroids or even further celestial bodies.
These missions rely heavily on physics and mathematics to solve challenges such as how to efficiently propel spacecraft over vast distances. Science fiction becomes science reality through such explorations.

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Most popular questions from this chapter

Two runners start one hundred meters apart and run toward each other. Each runs ten meters during the first second. During each second thereafter, each runner runs ninety percent of the distance he ran in the previous second. Thus, the velocity of each person changes from second to second. However, during any one second, the velocity remains constant. Make a position-time graph for one of the runners. From this graph, determine (a) how much time passes before the runners collide and (b) the speed with which each is running at the moment of collision.

Review Conceptual Example 7 as background for this problem. A car is traveling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is \(27.0 \mathrm{~m} / \mathrm{s},\) and during a 5.0 -s interval, it changes to a final speed of (a) \(29.0 \mathrm{~m} / \mathrm{s}\) and ( b) \(23.0 \mathrm{~m} / \mathrm{s}\). In each case, find the acceleration (magnitude and algebraic sign) and state whether or not the car is decelerating.

Review Interactive LearningWare 2.2 at in preparation for this problem. A race driver has made a pit stop to refuel. After refueling, he leaves the pit area with an acceleration whose magnitude is \(6.0 \mathrm{~m} / \mathrm{s}^{2}\); after \(4.0 \mathrm{~s}\) he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant speed of \(70.0 \mathrm{~m} / \mathrm{s}\) overtakes and passes the entering car. If the entering car maintains its acceleration, how much time is required for it to catch the other car?

Review Interactive Solution 2.49 at before beginning this problem. A woman on a bridge \(75.0 \mathrm{~m}\) high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 7.00 \(\mathrm{m}\) more to travel before passing under the bridge. The stone hits the water \(4.00 \mathrm{~m}\) in front of the raft. Find the speed of the raft.

A speed ramp at an airport is basically a large conveyor belt on which you can stand and be moved along. The belt of one ramp moves at a constant speed such that a person who stands still on it leaves the ramp 64 s after getting on. Clifford is in a real hurry, however, and skips the speed ramp. Starting from rest with an acceleration of \(0.37\) \(\mathrm{m} / \mathrm{s}^{2}\), he covers the same distance as the ramp does, but in one-fourth the time. What is the speed at which the belt of the ramp is moving?

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