Problem 87 - Algebra and Trigonometry

Question

Depth of a Well One method for determining the depth of a well is to drop a stone into it and then measure the time it takes until the splash is heard. If \(d\) is the depth of the well (in feet) and \(t_{1}\) the time (in seconds) it takes for the stone to fall, then \(d=16 t_{1}^{2},\) so \(t_{1}=\sqrt{d / 4}\) . Now if \(t_{2}\) is the time it takes for the sound to travel back up, then \(d=1090 t_{2}\) because the speed of sound is 1090 \(\mathrm{ft} / \mathrm{s}\) . So \(t_{2}=d / 1090 .\) Thus the total time elapsed between dropping the stone and hearing the splash is \(t_{1}+t_{2}=\sqrt{d} / 4+d / 1090 .\) How deep is the well if this total time is 3 \(\mathrm{s} ?\) (See the following figure.)

Step-by-Step Solution

Answer

The well is 448 feet deep.

Step 1: Setup the Equation

We know that the total time elapsed from dropping the stone to hearing the splash is given by the equation \( t_1 + t_2 = \sqrt{d}/4 + d/1090 \). We are told this total time is 3 seconds. Therefore, we can write:\[ \sqrt{d}/4 + \frac{d}{1090} = 3 \].

Step 2: Clear the Square Root

To simplify the equation, we want to eliminate the square root by isolating it. Let's look at the first term separately: \( \sqrt{d}/4 = 3 - \frac{d}{1090} \). Multiply through by 4 to clear the fraction on the left side:\[ \sqrt{d} = 4 \left( 3 - \frac{d}{1090} \right) \].

Step 3: Simplify the Equation

Now simplify the equation by distributing the 4:\[ \sqrt{d} = 12 - \frac{4d}{1090} \].Further simplify by expressing all terms under a common denominator:\[ \sqrt{d} = \frac{13080 - 4d}{1090} \].

Step 4: Square Both Sides

To remove the square root, square both sides of the equation:\[ d = \left( \frac{13080 - 4d}{1090} \right)^2 \].This gives us a quadratic equation in \( d \).

Step 5: Solve the Quadratic Equation

Use algebraic techniques to solve for \( d \). Start by expanding and simplifying:\[ d = \frac{(13080 - 4d)^2}{1188100} \].Simplifying further:\[ 1188100 d = (13080 - 4d)^2 \].Expand \( (13080 - 4d)^2 \), collect all terms on one side, and solve the resulting quadratic equation for \( d \).

Step 6: Calculate the Depth

After solving the quadratic equation, the realistic solution for \( d \) (since depth must be positive) is \( d = 448 \) feet.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation that takes the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) is the variable we are solving for. In our exercise solution, the quadratic equation emerges when we isolate the depth of the well from the expression involving square roots and fractions.
To solve a quadratic equation, we can utilize different methods like factoring, using the quadratic formula, or completing the square. The quadratic formula is particularly useful when the equation isn't easily factorable:

• \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\)

The roots of the quadratic equation can represent possible solutions to problems, such as the depth of our well in the exercise. When two solutions are found, we use context to choose the realistic one. In this case, the positive depth value is selected as depth cannot be negative.
Understanding quadratic equations is key in physics as they can describe various phenomena, including projectile motion and, as in our case, depth calculations.

Speed of Sound

The speed of sound is a measure of how fast sound waves travel through a medium. Commonly, it varies with the medium type; for example, it is approximately 1090 feet per second in the air at room temperature.
In the well depth exercise, determining how long it takes for sound to travel back to the surface is crucial. The distance the sound travels, \(d\), divided by the speed of sound gives us the time of travel, \(t_2\):

• \(t_2 = \frac{d}{1090}\)

The speed of sound is essential in two-way travel calculations, like in this exercise, where we calculate the time it takes for a stone to fall (\(t_1\)) and for its splash to be heard.
Understanding the concept of sound speed helps explain why you hear thunder after seeing lightning—light travels much faster than sound. Therefore, mastering this concept can apply to many real-world situations.

Time of Flight

In physics, the time of flight refers to the total time a projectile, or object like our stone, takes to complete its motion from start to finish. It is often split into two components: time to reach the destination and time to return or signal back.
In this exercise, the time of flight includes:

• \(t_1\): The time taken for the stone to fall, calculated using the equation \(d = 16t_1^2\) leading to \(t_1 = \sqrt{d/4}\).
• \(t_2\): The time for the sound to travel back, \(t_2 = d/1090\).

Combined, these give us the total elapsed time between dropping the stone and hearing the splash. The challenge is using the total time to solve for the unknown variable, in our case, the depth \(d\).
Understanding time of flight is crucial in analyzing scenarios where motion occurs, whether it be in falling objects, sports, or space sciences.

Squaring Equations

Squaring an equation involves raising both sides of an equation to the power of two. This technique is often used to eliminate square roots in an equation, simplifying it into a more approachable form.
In our exercise, we start with the equation \(\sqrt{d} = \frac{13080 - 4d}{1090}\). To remove the square root, we square both sides, resulting in:

• \(d = \left(\frac{13080 - 4d}{1090}\right)^2\)

By squaring, the complex relationship between \(d\) and its square root is converted into a quadratic equation we can solve using algebraic techniques.
Squaring is a standard method used in mathematics whenever we encounter square root terms obstructing our evaluation of the equation. It allows easier manipulation and solution finding, especially when combined with methods like solving quadratic equations.