Question

The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the time elapsed until a sound is heard. If \(t_{1}\) is the time (measured in seconds) that it takes for the object to strike the water, then \(t_{1}\) will obey the equation \(s=16 t_{1}^{2}\), where \(s\) is the distance (measured in feet). It follows that \(t_{1}=\frac{\sqrt{s}}{4}\). Suppose that \(t_{2}\) is the time that it takes for the sound of the impact to reach your ears. Because sound waves are known to travel at a speed of approximately 1100 feet per second, the time \(t_{2}\) to travel the distance \(s\) will be \(t_{2}=\frac{s}{1100} .\) See the illustration. Now \(t_{1}+t_{2}\) is the total time that elapses from the moment that the object is dropped to the moment that a sound is heard. We have the equation $$ \text { Total time elapsed }=\frac{\sqrt{s}}{4}+\frac{s}{1100} $$ Find the distance to the water's surface if the total time elapsed from dropping a rock to hearing it hit water is 4 seconds.

Short answer

The distance to the water's surface is 225 feet.

Step-by-step solution

Define the given total time

The total time elapsed from dropping a rock to hearing it hit the water is given as 4 seconds. Therefore, we have: \[\frac{\sqrt{s}}{4} + \frac{s}{1100} = 4\]

Introduce a substitution for simplicity

Let's denote \(x = \sqrt{s}\). This implies that \(s = x^2\). Substituting these into the equation, we get: \[\frac{x}{4} + \frac{x^2}{1100} = 4\]

Clear the fraction by multiplying through by 1100

To eliminate the fractions, multiply every term by 1100: \[1100 \cdot \frac{x}{4} + 1100 \cdot \frac{x^2}{1100} = 4 \cdot 1100\]Simplify to get: \[275x + x^2 = 4400\]

Form a quadratic equation

Rearrange the equation to form a standard quadratic equation: \[x^2 + 275x - 4400 = 0\]

Solve the quadratic equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 275\), and \(c = -4400\):\[x = \frac{-275 \pm \sqrt{275^2 - 4(1)(-4400)}}{2(1)}\]Solve the discriminant:\[x = \frac{-275 \pm \sqrt{75625 + 17600}}{2}\]\[x = \frac{-275 \pm \sqrt{93225}}{2}\]\[x = \frac{-275 \pm 305}{2}\]This gives two solutions: \(x = 15\) and \(x = -580\). Since \(x\) represents \(\sqrt{s}\) and must be positive, we take \(x = 15\).

Find the distance to the water surface

Now, recall that \(s = x^2\). Thus, \(s = 15^2 = 225\). The distance to the water's surface is 225 feet.

Key concepts

quadratic equations

Quadratic equations are a fundamental part of algebra and appear frequently in physics problems. They have the standard form \( ax^2 + bx + c = 0 \). In our exercise, we transformed a real-world problem into a quadratic equation during the solving process. Given the equation \(\frac{x}{4} + \frac{x^2}{1100} = 4 \), we multiplied through by 1100 to eliminate fractions and obtained the quadratic form \( x^2 + 275x - 4400 = 0 \). The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) allows us to find the roots of the equation, where 'a', 'b', and 'c' are coefficients. This method is efficient and provides a systematic approach to solve quadratic equations, guaranteeing that you can find all possible solutions.

distance-speed-time relationship

Understanding the distance-speed-time relationship is a cornerstone of physics. The relationship is given by the equation \( s = vt \), where 's' is the distance, 'v' is the speed, and 't' is the time. In our well problem, there are two parts to consider. First, the time \( t_1 \) it takes for the rock to fall, and second, the time \( t_2 \) it takes for the sound to travel back up. For the falling object, the equation \( s = 16t_1^2 \) describes the distance fallen due to gravity. The sound wave, traveling at 1100 feet per second, follows \( t_2 = \frac{s}{1100} \). These equations help us understand that the total time is the sum of the time for both events to occur.

problem-solving steps

Breaking down a physics problem into clear, manageable steps can simplify even the most complex problems. Here’s how we approached the well exercise:

• First, we stated the total time equation \( \frac{\sqrt{s}}{4} + \frac{s}{1100} = 4 \).
• Next, we introduced a substitution \( x = \sqrt{s} \) to simplify our equation.
• We rearranged to get a quadratic form \( x^2 + 275x - 4400 = 0 \).
• Using the quadratic formula, we solved for 'x'.
• Finally, we squared 'x' to find the distance 's' to the water surface.

By following such a systematic approach, students can avoid becoming overwhelmed and ensure they don’t miss any steps.

sound wave speed

The speed of sound is a critical element in many physics problems, especially those involving waves. Sound travels through air at roughly 1100 feet per second depending on factors such as temperature and humidity. In our scenario, after the rock hits the water, the sound wave travels back up the well. We calculate this time using \( t_2 = \frac{s}{1100} \). Understanding the properties and speed of waves is essential since they affect how we perceive distances and time in real-life applications. Knowledge of wave speed helps us accurately determine how long it takes for sound to travel over a given distance, making this concept invaluable in problem-solving.