Problem 75 - Introductory Algebra for College Students

Question

The distance, \(d,\) in feet, that an object falls in \(t\) seconds is modeled by the formula \(d=16 t^{2} .\) Use this formula to solve Exercises \(75-76\) If you drop a rock from a cliff 400 feet above the water how long will it take for the rock to hit the water?

Step-by-Step Solution

Answer

The rock will hit the water in 5 seconds.

Step 1: Express the Given Problem as an Equation

Write the given formula \(d = 16t^2\) and substitute the given distance (d = 400 feet) to get \(400 = 16t^2\)

Step 2: Re-arrange the Equation to Solve for \(t\)

To isolate \(t\), divide both sides of the equation by 16 to get \(t^2 = \frac{400}{16} = 25\)

Step 3: Solve for \(t\)

Take the square root of both sides. This gives \(t = \sqrt{25} = 5\) seconds. Therefore, it will take 5 seconds for the rock to hit the water.

Key Concepts

Quadratic Equations

Quadratic equations are a crucial part of algebra and arise whenever we deal with equations involving the square of a variable (like the equation provided in our exercise). A standard quadratic equation is expressed in the form \(ax^2 + bx + c = 0\). These equations have distinctive features due to the squared term, which affects both the shape of their graph—typically a parabola—and the nature of their solutions.

The equation \(16t^2 = 400\) from our exercise is a pure quadratic equation because there is no linear term (\(bx\)) or constant term (\(c\)) other than what we solve for. To find \(t\), we rearrange the equation to express \(t\) in terms of known values and simplify using algebraic operations.

• The first step is to get the \(t^2\) term alone by dividing both sides by 16.
• Then, solve for \(t\) by taking the square root of both sides.

In this straightforward case, it results in \(t = 5\). Understanding these principles is essential for solving more complex scenarios where quadratic equations may involve more coefficients or require different techniques such as factoring or using the quadratic formula.

Distance Formula

The distance formula used here, \(d = 16t^2\), is a specific application of a physics concept where distance is determined based on constant acceleration. This formula is derived from the principles governing the motion of objects under gravity without air resistance. The 16 in the formula represents half of the gravitational acceleration constant when working in feet and seconds units.

This formula tells us how distance (\(d\)) is related to time (\(t\)) under specific conditions. It shows us that the distance an object falls is proportional to the square of the time it has been falling. This relationship is fundamental in physics and helps us compute how quickly an object will reach a certain distance.

• Replace \(d\) with the known distance to solve for time \(t\).
• Understanding this formula helps in various applications, including calculating fall time for objects from different heights.

Using such a formula simplifies finding how long it takes for an object to reach the ground, assuming no other forces act upon it.

Problem Solving

Problem-solving in algebra often involves methodically applying known formulas and rearranging expressions to find unknown values. The problem given here is a classic example where we are asked to determine time using a simple algebraic restructuring of an equation.

The process starts by correctly setting up the equation using familiar formulas, like the distance formula. Then, by isolating the variable of interest (in this case \(t\)), we apply operations to simplify and solve the equation:

• Identify what the problem is asking you to find (time \(t\)).
• Use logical steps to isolate this variable by performing algebraic operations like division and square roots.

This systematic approach not only provides the solution \(t = 5\) seconds but also ingrains an important method for tackling similar problems. By practicing these steps, solving algebra problems becomes an approachable task with a clear path to finding answers.