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Chapter 16: Problem 62

A rocket, starting from rest, travels straight up with an acceleration of \(58.0 \mathrm{m} / \mathrm{s}^{2} .\) When the rocket is at a height of \(562 \mathrm{m},\) it produces sound that eventually reaches a ground-based monitoring station directly below. The sound is emitted uniformly in all directions. The monitoring station measures a sound intensity \(I\). Later, the station measures an intensity \(\frac{1}{3} I .\) There are no reflections. Assuming that the speed of sound is \(343 \mathrm{m} / \mathrm{s}\) find the time that has elapsed between the two measurements.

Short Answer

Expert verified
The time elapsed between the two measurements is 1.20 seconds.

Step by step solution

01

Find the Time Taken to Reach 562 m

We first determine the time it takes for the rocket to reach the height of 562 meters using the equation of motion for uniformly accelerated linear motion: \[ s = ut + \frac{1}{2}at^2 \]where \(s = 562\, \mathrm{m}, u = 0\, \mathrm{m/s}\), and \(a = 58.0\, \mathrm{m/s}^2\). Substituting these values, we get:\[562 = 0 * t + \frac{1}{2} * 58.0 * t^2 \]Solving for \(t\):\[562 = 29 t^2 \ t^2 = \frac{562}{29} \ t = \sqrt{19.379} \ t \approx 4.40 \text{ seconds}\]
02

Calculate the Initial Intensity and Its Distance

The sound intensity at 562 meters is \(I\). The intensity is inversely proportional to the square of the radius (distance from the source), so use the relationship\[I \propto \frac{1}{r^2},\]where \(r\) is the distance from the sound source to the monitoring station. Initially, \(r = 562\) m.
03

Determine the New Distance for Intensity \(\frac{1}{3}I\)

If the intensity diminishes to \(\frac{1}{3}I\), the new distance \(r_2\) from the source follows:\[\frac{1}{3}I \propto \frac{1}{r_2^2}\]Equating the intensities:\[I \times r^2 = \frac{1}{3}I \times r_2^2 \Rightarrow 3r^2 = r_2^2 \Rightarrow r_2 = \sqrt{3}r = \sqrt{3} \times 562 \approx 973.34 \text{ m}\]
04

Calculate the Time for Sound to Reach Both Distances

Calculate time for sound to travel to each distance:For the initial distance (562 m), time \( t_1 = \frac{562}{343} \approx 1.64 \) seconds.For the new distance (973.34 m), time \( t_2 = \frac{973.34}{343} \approx 2.84 \) seconds.
05

Find the Time Elapsed Between Measurements

Calculate the time interval between the two measurements:\[ t_{\text{elapsed}} = (t_2 - t_1) = (2.84 - 1.64) \text{ seconds} = 1.20 \text{ seconds}.\]

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Key Concepts

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

Understanding Acceleration
Acceleration is a key concept in physics, especially in kinematics. It describes how the velocity of an object changes over time. When you see a value like \(58.0\ \text{m/s}^2\), it means for every second, the speed of the rocket increases by 58 meters per second.
There are some important points to remember:
  • Uniform acceleration: In this exercise, acceleration is constant, meaning it doesn't change as time progresses. This simplifies calculations significantly.
  • Units of acceleration: The units \(\text{m/s}^2\) tell us how speed in meters per second changes every second.
  • Starting from rest: The rocket begins at a velocity of 0\ \text{m/s}, making initial calculations straightforward as the initial velocity \(u\) becomes zero in equations.
Using these points, you can predict how long it takes for the rocket to reach a certain height using the kinematic equation:
\[s = ut + \frac{1}{2}at^2\]where \(s\) is the displacement, \(u\) is the initial velocity, and \(a\) is the acceleration.
Always substitute known values carefully to avoid errors in calculations.
Equations of Motion
The equations of motion are essential for solving problems involving kinematics in physics. They provide relationships between displacement, velocity, acceleration, and time.
In this scenario, the rocket's motion is governed by the basic equation:
\[s = ut + \frac{1}{2}at^2\]
  • Displacement \(s\): It's the height the rocket achieves, here 562 meters.
  • Initial Velocity \(u\): Since the rocket starts from rest, \(u = 0\).
  • Time \(t\): This is what we solve for using the equation with given values of \(s\) and \(a\).
  • Constant Acceleration \(a\): Provided as \(58.0\ \text{m/s}^2\), it remains unchanged throughout the motion.
Understanding these relationships is crucial as they allow you to calculate missing quantities as long as other variables are known, which makes analysis of motion predictable and methodical.
Deciphering Sound Intensity
Sound intensity is crucial in understanding how sound spreads through space from a source and how this affects the strength or loudness of the sound.
Some key factors:
  • Inverse Square Law: As sound travels, its intensity decreases with the square of the distance from its source. Thus, if you move twice as far, the intensity drops by a factor of four.
  • Initial Intensity \(I\): At 562 meters, this is the reference intensity sound emits uniformly in all directions.
  • Change in Intensity: A reduction to \(\frac{1}{3}\ I\) indicates an increase in distance from the source.
  • Distance Calculations: By understanding the inverse square law, we can find the new distance when intensity changes, using formulas like\[I \propto \frac{1}{r^2}\]This means specific calculations connect distance changes to sound intensity transformations.
By meeting these calculations with the given speed of sound, we find out how timing relates to varying sound intensities, helping you calculate elapsed time based on these auditory measurements.

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Most popular questions from this chapter

A speedboat, starting from rest, moves along a straight line away from a dock. The boat has a constant acceleration of \(+3.00 \mathrm{m} / \mathrm{s}^{2}\) (see the figure). Attached to the dock is a siren that is producing a \(755-\mathrm{Hz}\) tone. If the air temperature is \(20^{\circ} \mathrm{C},\) what is the frequency of the sound heard by a person on the boat when the boat's displacement from the dock is \(+45.0 \mathrm{m} ?\)

Light is an electromagnetic wave and travels at a speed of \(3.00 \times 10^{8} \mathrm{m} / \mathrm{s} .\) The human eye is most sensitive to yellow-green light, which has a wavelength of \(5.45 \times 10^{-7} \mathrm{m} .\) What is the frequency of this light?

When an earthquake occurs, two types of sound waves are generated and travel through the earth. The primary, or \(P\), wave has a speed of about \(8.0 \mathrm{km} / \mathrm{s}\) and the secondary, or \(\mathrm{S}\), wave has a speed of about \(4.5 \mathrm{km} / \mathrm{s}\). A seismograph, located some distance away, records the arrival of the P wave and then, 78 s later, records the arrival of the \(S\) wave. Assuming that the waves travel in a straight line, how far is the seismograph from the earthquake?

Have you ever listened for an approaching train by kneeling next to a railroad track and putting your ear to the rail? Young's modulus for steel is \(Y=2.0 \times 10^{11} \mathrm{N} / \mathrm{m}^{2},\) and the density of steel is \(\rho=7860 \mathrm{kg} / \mathrm{m}^{3} .\) On a day when the temperature is \(20^{\circ} \mathrm{C},\) how many times greater is the speed of sound in the rail than in the air?

A loudspeaker in a parked car is producing sound whose frequency is \(20510 \mathrm{Hz} .\) A healthy young person with normal hearing is standing nearby on the sidewalk but cannot hear the sound because the frequency is too high. When the car is moving, however, this person can hear the sound. (a) Is the car moving toward or away from the person? Why? (b) If the speed of sound is \(343 \mathrm{m} / \mathrm{s},\) what is the minimum speed of the moving car?
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