/*! 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 75 A spelunker (cave explorer) drop... [FREE SOLUTION] | 91Ó°ÊÓ

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A spelunker (cave explorer) drops a stone from rest into a hole. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\) in air, and the sound of the stone striking the bottom is heard \(1.50 \mathrm{~s}\) after the stone is dropped. How deep is the hole?

Short Answer

Expert verified
The hole is approximately 4.08 meters deep.

Step by step solution

01

Understanding the Problem

The stone is dropped, falls to the bottom of the hole, and the sound travels back up. Total time for the stone to fall and the sound to travel back is 1.50 s.
02

Equations of Motion for the Falling Stone

The stone falls from rest, so its initial velocity \( v_0 = 0 \). Use the equation of motion for the distance traveled by the stone: \[ d = \frac{1}{2}gt^2 \]where \( g = 9.8 \ m/s^2 \) and \( t \) is the time taken to fall.
03

Travel Time for Sound

Let \( t_1 \) be the time for the stone to fall and \( t_2 \) the time for sound to travel back up. The total time is given by:\[ t_1 + t_2 = 1.50 \, s \]
04

Equation for Sound Travel

The distance \( d \) is also covered by sound traveling upward, which takes time \( t_2 \):\[ d = 343 \, m/s \times t_2 \]
05

Substituting and Solving Equations

Substitute \( t_2 = 1.50 - t_1 \) into the sound travel equation:\[ d = 343 \times (1.50 - t_1) \]Set the equations equal:\[ \frac{1}{2}gt_1^2 = 343 \times (1.50 - t_1) \]Substitute \( g = 9.8 \, m/s^2 \) and solve for \( t_1 \). This will yield a quadratic equation.
06

Solving the Quadratic Equation

The quadratic equation is:\[ 4.9t_1^2 + 343t_1 - 514.5 = 0 \]Use the quadratic formula:\[ t_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 4.9 \), \( b = 343 \), \( c = -514.5 \).
07

Calculation

Calculating the above gives:\[ t_1 = \frac{-343 \pm \sqrt{343^2 - 4(4.9)(-514.5)}}{2(4.9)} \]\[ t_1 \approx 0.91 \, s \]
08

Determine Depth Using \( t_1 \)

Using \( t_1 \approx 0.91 \, s \) in the motion equation:\[ d = \frac{1}{2} \times 9.8 \times (0.91)^2 \]\[ d \approx 4.08 \, m \]

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Key Concepts

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

Equations of Motion
Equations of motion are fundamental tools in physics that describe the behavior of objects in motion. They allow us to predict how an object will move under various conditions of velocity, acceleration, and time. The key equation from the step-by-step solution is: - \( d = \frac{1}{2}gt^2 \) - The variable \(d\) represents the distance traveled, \(g\) is the acceleration due to gravity (approximately \(9.8 \, m/s^2\) on Earth), and \(t\) is the time taken. When an object falls freely under gravity, starting from rest, the initial velocity \(v_0\) is zero, simplifying the equation considerably.- By inserting known values and solving for time or distance, students can explore how changes in one parameter affect the others. Equations of motion not only help in solving problems like the spelunker's, they also build a strong foundation for analyzing various dynamic systems.
Speed of Sound
The speed of sound is the rate at which sound waves travel through a medium, which in this problem is air. The given speed of sound is \(343 \, m/s\). - Sound travels relatively fast but is affected by factors such as air temperature and pressure. This constant speed allows us to connect the time it takes for sound to return to the spelunker after the stone hits the bottom. - Using the formula:\[ d = v \times t_2 \]where \(v\) is the speed of sound, and \(t_2\) is the time taken by sound to travel, we can calculate how far the sound must have traveled. This problem exemplifies the way we can use the speed of sound to measure unseen distances.
Quadratic Equation
Quadratic equations are a powerful mathematical tool used when dealing with polynomial expressions of degree two. They often appear in physics when two linear paths or systems intersect or combine. - In our problem, the quadratic equation arises when equating the distances covered during free fall and the sound's return:\[ 4.9t_1^2 + 343t_1 - 514.5 = 0 \]- This equation represents the two unknowns: the time for the stone to fall \(t_1\) and the time for the sound to travel back \(t_2\). - To solve the quadratic equation, we use the quadratic formula:\[ t_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula provides solutions for \(t_1\), allowing us to determine how long it takes the stone to reach the bottom of the hole.
Free Fall
Free fall is the motion of an object where gravity is the only significant force acting upon it. In such a scenario, air resistance is negligible, and the object accelerates downwards at \(9.8 \, m/s^2\). - In the given problem, when the spelunker drops the stone, it enters free fall from rest. This starts a chain of calculations that involves determining how long it takes for the stone to fall based on its acceleration and distance. - Understanding free fall is crucial as it is a basic concept in kinematics and physics, illustrating how objects behave under the influence of Earth's gravity alone. - Observations of free fall help students predict motion characteristics and solve real-world physics problems.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. Kinematics focuses on parameters such as velocity, acceleration, displacement, and time. - In the spelunker's stone scenario, kinematics allows us to separate the motion of the stone into distinct phases: the fall and the sound’s travel back. - By understanding the equations of motion and applying them appropriately, students can tackle complex problems requiring calculations of velocities and positions over time. Kinematics is fundamental in developing an understanding of how objects move, making it an essential component of physics education. - With real-world connections like this, students can appreciate the broad applicability of kinematics.

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Most popular questions from this chapter

A Boeing 747 "Jumbo Jet" has a length of \(59.7 \mathrm{~m}\). The runway on which the plane lands intersects another runway. The width of the intersection is \(25.0 \mathrm{~m} .\) The plane decelerates through the intersection at a rate of \(5.70 \mathrm{~m} / \mathrm{s}^{2}\) and clears it with a final speed of \(45.0 \mathrm{~m} / \mathrm{s}\). How much time is needed for the plane to clear the intersection?

An automobile starts from rest and accelerates to a final velocity in two stages along a straight road. Each stage occupies the same amount of time. In stage 1 , the magnitude of the car's acceleration is \(3.0 \mathrm{~m} / \mathrm{s}^{2}\). The magnitude of the car's velocity at the end of stage 2 is \(2.5\) times greater than it is at the end of stage 1 . Find the magnitude of the acceleration in stage 2 .

(a) In general, does the average acceleration of an object have the same direction as its initial velocity \(v_{0}\), its final velocity \(v,\) or the difference \(v-v_{0}\) between its final and initial velocities? Provide a reason for your answer. (b) The following table lists four pairs of initial and final velocities for a boat traveling along the \(x\) axis. Use the concept of acceleration presented in Section 2.3 to determine the direction (positive or negative) of the average acceleration for each pair of velocities. $$ \begin{array}{|c|c|c|} \hline & \text { Initial velocity } v_{0} & \text { Final velocity } v \\ \hline(\mathrm{a}) & +2.0 \mathrm{~m} / \mathrm{s} & +5.0 \mathrm{~m} / \mathrm{s} \\ \hline(\mathrm{b}) & +5.0 \mathrm{~m} / \mathrm{s} & +2.0 \mathrm{~m} / \mathrm{s} \\ \hline(\mathrm{c}) & -6.0 \mathrm{~m} / \mathrm{s} & -3.0 \mathrm{~m} / \mathrm{s} \\ \hline(\mathrm{d}) & +4.0 \mathrm{~m} / \mathrm{s} & -4.0 \mathrm{~m} / \mathrm{s} \\ \hline \end{array} $$ Problem The elapsed time for each of the four pairs of velocities is \(2.0 \mathrm{~s}\). Find the average acceleration (magnitude and direction) for each of the four pairs. Be sure that your directions agree with those found in the Concept Questions.

Review Interactive Solution 2.29 at in preparation for this problem. Suppose a car is traveling at \(20.0 \mathrm{~m} / \mathrm{s},\) and the driver sees a traffic light turn red. After \(0.530 \mathrm{~s}\) has elapsed (the reaction time), the driver applies the brakes, and the car decelerates at \(7.00 \mathrm{~m} / \mathrm{s}^{2}\). What is the stopping distance of the car, as measured from the point where the driver first notices the red light?

At provides some background for this problem. A ball is thrown vertically upward, which is the positive direction. A little later it returns to its point of release. (a) Does the acceleration of the ball reverse direction when the ball starts its downward trip? (b) What is the displacement of the ball when it returns to its point of release? Explain your answers. Problem If the ball is in the air for a total time of \(8.0 \mathrm{~s}\), what is its initial velocity?

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