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Chapter 2: Problem 31

Suppose a rocket ship in deep space moves with constant acceleration equal to \(9.8 \mathrm{~m} / \mathrm{s}^{2}\), which gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s} ?(\mathrm{~b})\) How far will it travel in so doing?

Short Answer

Expert verified
(a) Approximately 3.06 million seconds, (b) about 4.58 trillion meters.

Step by step solution

01

Understanding the Problem

We need to determine two things about a rocket ship accelerating at \(9.8 \, \text{m/s}^2\) from rest: (a) the time it takes to reach a speed of \( \frac{1}{10} \) of the speed of light; (b) the distance it covers during this acceleration.
02

Calculate Target Speed

First, determine the target speed, which is \( \frac{1}{10} \) of the speed of light: \[ v = \frac{1}{10} \times 3.0 \times 10^8 \, \text{m/s} = 3.0 \times 10^7 \, \text{m/s}. \]
03

Use Kinematic Equation for Time

We can find the time \( t \) using the equation \( v = at \). Rearrange to solve for \( t \):\[ t = \frac{v}{a} = \frac{3.0 \times 10^7 \, \text{m/s}}{9.8 \, \text{m/s}^2}. \]
04

Calculate Time

Substitute the known values:\[ t = \frac{3.0 \times 10^7}{9.8} \approx 3.06 \times 10^6 \, \text{seconds}. \]
05

Use Kinematic Equation for Distance

To find how far the rocket travels, use the equation for distance under constant acceleration: \[ d = \frac{1}{2} a t^2. \]
06

Calculate Distance

Substitute the known values to find the distance:\[ d = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (3.06 \times 10^6)^2 \, \text{s}^2 \approx 4.58 \times 10^{13} \, \text{meters}. \]

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Key Concepts

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

Constant Acceleration
In physics, constant acceleration refers to a scenario where an object's velocity changes at a consistent rate over time. This means the speed increases (or decreases) uniformly as time progresses. In our exercise, the rocket is accelerating at a constant rate of \(9.8 \, \text{m/s}^2\). This mirrors Earth's gravity, which is why it simulates normal gravitational force. Consistent acceleration simplifies calculations because there's a reliable change in velocity over each time increment.
  • Helps predict future velocity and position
  • Simplifies motion equations
Understanding constant acceleration is crucial when dealing with motion in physics, as it affects how we calculate time and distance.
Speed of Light
The speed of light is one of the most fundamental constants in the universe, traveling at approximately \(3.0 \times 10^8 \, \text{m/s}\). It represents the maximum speed at which all energy, matter, and information in the universe can travel. In our exercise, the rocket aims to reach a speed of one-tenth of the speed of light.
  • Speed of light is denoted as \(c\)
  • Used as a constant in many physics equations
  • Approximately \(299,792,458 \, \text{m/s}\)
This constant plays a pivotal role in relativity, where traveling at or near the speed of light significantly affects the laws of physics as we know them.
Uniformly Accelerated Motion
Uniformly accelerated motion refers to motion in which the acceleration of an object is constant in both magnitude and direction. The rocket in our exercise exemplifies this by accelerating at \(9.8 \, \text{m/s}^2\) from rest. The equations of motion for such scenarios are straightforward, allowing calculations of speed and distance significantly easier. Key equations involve relationships like:
  • Velocity: \(v = u + at\)
  • Displacement: \(d = ut + \frac{1}{2} a t^2\)
In scenarios of uniformly accelerated motion, objects adhere strictly to these mathematical principles, making predictions and measurements very reliable.
Kinematic Equations
Kinematic equations describe the motion of objects without considering external forces causing the motion. They are crucial in calculating different aspects of an object's movement.
  • \(v = u + at\)
  • \(d = ut + \frac{1}{2} a t^2\)
  • \(v^2 = u^2 + 2ad\)
In our exercise, these equations help us determine both the time required to reach a certain speed and the distance traveled during this time. By rearranging and substituting known values into the kinematic formulas, we can solve for unknowns efficiently. Understanding how to manipulate these equations is a cornerstone in solving problems involving motion, ensuring we can seamlessly calculate the parameters involved.

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Most popular questions from this chapter

A rocket-driven sled running on a straight, level track is used to investigate the effects of large accelerations on humans. One such sled can attain a speed of \(1600 \mathrm{~km} / \mathrm{h}\) in \(1.8 \mathrm{~s}\), starting from rest. Find (a) the acceleration (assumed constant) in terms of \(g\) and (b) the distance traveled.

To set a speed record in a measured (straight-line) distance \(d\), a race car must be driven first in one direction (in time \(t_{1}\) ) and then in the opposite direction (in time \(t_{2}\) ). (a) To eliminate the effects of the wind and obtain the car's speed \(v_{c}\) in a windless situation, should we find the average of \(d / t_{1}\) and \(d / t_{2}\) (method 1) or should we divide \(d\) by the average of \(t_{1}\) and \(t_{2} ?\) (b) What is the fractional difference in the two methods when a steady wind blows along the car's route and the ratio of the wind speed \(v_{w}\) to the car's speed \(v_{c}\) is \(0.0240 ?\)

A car moves along an \(x\) axis through a distance of \(900 \mathrm{~m}\), starting at rest \((\) at \(x=0)\) and ending at rest \((\) at \(x=900 \mathrm{~m})\). Through the first \(\frac{1}{4}\) of that distance, its acceleration is \(+2.25 \mathrm{~m} / \mathrm{s}^{2}\). Through the rest of that distance, its acceleration is \(-0.750 \mathrm{~m} / \mathrm{s}^{2}\). What are (a) its travel time through the \(900 \mathrm{~m}\) and \((\mathrm{b})\) its maximum speed? (c) Graph position \(x\), velocity \(v\), and acceleration \(a\) versus time \(t\) for the trip.

From January 26,1977, to September 18,1983, George Meegan of Great Britain walked from Ushuaia, at the southern tip of South America, to Prudhoe Bay in Alaska, covering \(30600 \mathrm{~km}\). In meters per second, what was the magnitude of his average velocity during that time period?

The sport with the fastest moving ball is jai alai, where measured speeds have reached \(303 \mathrm{~km} / \mathrm{h}\). If a professional jai alai player faces a ball at that speed and involuntarily blinks, he blacks out the scene for \(100 \mathrm{~ms}\). How far does the ball move during the blackout?
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