/*! 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 18 At a time when mining asteroids ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 18

At a time when mining asteroids has become feasible, astronauts have connected a line between their 3500-kg space tug and a 6200-kg asteroid. Using their tug鈥檚 engine, they pull on the asteroid with a force of 490 N. Initially the tug and the asteroid are at rest, 450 m apart. How much time does it take for the tug and the asteroid to meet?

Short Answer

Expert verified
It takes approximately 64 seconds for the tug and the asteroid to meet.

Step by step solution

01

Identify the Forces and Masses

We have a space tug with mass \( m_1 = 3500 \text{ kg} \) and an asteroid with mass \( m_2 = 6200 \text{ kg} \). The force applied by the tug is \( F = 490 \text{ N} \). The initial distance between them is \( d = 450 \text{ m} \).
02

Calculate Acceleration

The acceleration of both the tug and the asteroid can be determined using Newton's second law, where \( F = m imes a \). First, find the acceleration of the asteroid:\[a_2 = \frac{F}{m_2} = \frac{490}{6200} \approx 0.079 \text{ m/s}^2\].Similarly, find the acceleration of the tug:\[a_1 = \frac{F}{m_1} = \frac{490}{3500} \approx 0.14 \text{ m/s}^2\].These accelerations are directed towards each other.
03

Calculate Total Time to Meet Using Relative Motion

The distance that both the tug and asteroid need to cover in order to meet is the initial distance \( d = 450 \text{ m} \). The effective acceleration \( a \) considering they move towards each other is the sum:\[a = a_1 + a_2 = 0.14 + 0.079 = 0.219 \text{ m/s}^2\].Using the equation of motion for uniformly accelerated motion \( d = \frac{1}{2} a t^2 \), solve for \( t \):\[450 = \frac{1}{2} \times 0.219 \times t^2\]\[t^2 = \frac{450 \times 2}{0.219}\]\[t = \sqrt{\frac{900}{0.219}} \approx 63.97 \text{ seconds}\].

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 91影视!

Key Concepts

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

Newton's Second Law
Newton's second law is fundamental in understanding how forces affect motion. It's expressed as \( F = ma \), where \( F \) is the force applied to an object, \( m \) is the mass, and \( a \) is the acceleration. This law shows that if you apply a force to an object, it will experience acceleration proportional to the force, and inversely proportional to its mass.
This principle is what guides our calculation for the asteroid and tug. By knowing the force applied and the masses involved, we can determine how each object moves.
Acceleration
Acceleration is the rate at which an object's velocity changes over time. In this exercise, acceleration is found using the formula \( a = \frac{F}{m} \).
For both the asteroid and the tug, their respective accelerations were calculated based on their masses and the total force:
  • Asteroid's acceleration: \( a_2 = 0.079 \text{ m/s}^2 \)
  • Tug's acceleration: \( a_1 = 0.14 \text{ m/s}^2 \)
These values tell us how quickly each object starts moving towards each other under the applied force.
Relative Motion
Relative motion involves understanding how two objects move in relation to one another. In our exercise, both the asteroid and the tug start from rest, moving towards each other due to the applied force.
To determine how fast they reach each other, we need to consider their combined accelerations. By adding their individual accelerations, we calculate a total effective acceleration of \( a = 0.219 \text{ m/s}^2 \). This combined acceleration helps us see their interaction better and predicts when they'll meet.
Equation of Motion
The equation of motion is crucial in calculating the time it takes for objects to meet. It's given by the formula \( d = \frac{1}{2} a t^2 \), where \( d \) is the distance covered, \( a \) is the acceleration, and \( t \) is the time.
In the problem, the distance \( d \) was 450 meters and the effective acceleration was \( a = 0.219 \text{ m/s}^2 \). By rearranging the equation and solving for \( t \), we found: \[t = \sqrt{\frac{900}{0.219}} \approx 63.97 \text{ seconds}\]
This equation helps us understand how acceleration and time work together to cover a specific distance.

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 cup of coffee is on a table in an airplane flying at a constant altitude and a constant velocity. The coefficient of static friction between the cup and the table is 0.30. Suddenly, the plane accelerates forward, its altitude remaining constant. What is the maximum acceleration that the plane can have without the cup sliding backward on the table?

A boat has a mass of 6800 kg. Its engines generate a drive force of 4100 N due west, while the wind exerts a force of 800 N due east and the water exerts a resistive force of 1200 N due east. What are the magnitude and direction of the boat鈥檚 acceleration?

A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the earth and the moon. The distance between the earth and the moon is \(3.85 \times 10^{8} \mathrm{m},\) and the mass of the earth is 81.4 times as great as that of the moon.

On earth a block has a weight of 88 N. This block is sliding on a horizontal surface on the moon, where the acceleration due to gravity is \(1.60 \mathrm{m} / \mathrm{s}^{2} .\) As the figure shows, the block is being pulled by a horizontal rope in which the tension is \(T=24 \mathrm{N}\). The coefficient of kinetic friction between the block and the surface is \(\mu_{k}=0.20 .\) Concepts: (i) Which of Newton's laws of motion provides the best way to determine the acceleration of the block? (ii) Does the net force in the \(x\) direction equal the tension \(T ?\) Calculations: Determine the acceleration of the block.

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 915 N and the drag force has a magnitude of 1027 N. The mass of the sky diver is 93.4 kg. What are the magnitude and direction of his acceleration?
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Physical Quantities And Units

Read Explanation

Fluids

Read Explanation

Wave Optics

Read Explanation

Torque and Rotational Motion

Read Explanation
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.