Problem 62 - Precalculus: Functions and Graphs

Question

The formula occurs in the indicated application. Solve for the specified variable. \(s=\frac{1}{2} g t^{2}+v_{0} t\) for \(t \quad\) (distance an object falls)

Step-by-Step Solution

Answer

The formula for \(t\) is \(t = \frac{-v_0 + \sqrt{v_0^2 + 2gs}}{g}\).

Step 1: Identify the Known Equation

We have the equation for the distance an object falls: \[s = \frac{1}{2} g t^2 + v_0 t\]We need to solve this equation for the variable \(t\).

Step 2: Rearrange Equation as a Quadratic

Reorder the equation in the form of a quadratic equation: \[\frac{1}{2} g t^2 + v_0 t - s = 0\]This can be expressed as: \[a t^2 + b t + c = 0\]where \(a = \frac{1}{2} g\), \(b = v_0\), and \(c = -s\).

Step 3: Apply the Quadratic Formula

We solve the quadratic equation for \(t\) using the quadratic formula: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substitute the values of \(a\), \(b\), and \(c\): \[t = \frac{-v_0 \pm \sqrt{v_0^2 - 2gs}}{g}\]

Step 4: Simplify the Expression

This gives the solutions for \(t\): \[t = \frac{-v_0 + \sqrt{v_0^2 + 2gs}}{g} \quad \text{and} \quad t = \frac{-v_0 - \sqrt{v_0^2 + 2gs}}{g}\]Since time cannot be negative, we choose the positive root:\[t = \frac{-v_0 + \sqrt{v_0^2 + 2gs}}{g}\]

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the solutions of quadratic equations. A quadratic equation is typically expressed in the form \(ax^2 + bx + c = 0\) where \(a\), \(b\), and \(c\) are coefficients. The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), allows us to solve for the variable \(x\).
- **Understanding the components**: - The term \(b^2 - 4ac\) is known as the discriminant. It indicates the nature of the roots: - If it is positive, there are two distinct real roots. - If it is zero, there is exactly one real root (the roots are repeated). - If it is negative, the solution involves complex numbers.- **Application to falling objects**: - In the context of kinematic equations, the variables \(a\), \(b\), and \(c\) correlate with physical constants or measured quantities. For example, when solving for time \(t\) in a motion equation like \(s = \frac{1}{2} g t^2 + v_0 t\), the quadratic formula helps find when exactly an object reaches a certain position \(s\).

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. They help in calculating displacement, velocity, time, and acceleration, crucial aspects of any motion scenario. One common kinematic equation is \(s = \frac{1}{2} g t^2 + v_0 t\), used to describe an object's vertical motion under gravity. - **Key elements in the equation**: - \(s\) denotes the distance covered. - \(g\) represents the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\) on Earth). - \(v_0\) symbolizes the initial velocity. - \(t\) is the time of travel. - **Relevance**: - These equations are fundamental in physics, providing the basis for understanding free-falling objects and projectile motion. - By manipulating and solving these equations, we can predict the future positions and velocities of moving objects, whether it's a car accelerating down a highway or an apple falling from a tree. In scenarios like the given exercise, kinematic equations assist in finding how long an object takes to reach a certain height, given its starting speed and the force of gravity.

Solving for Variables

Solving for a specific variable means isolating that variable on one side of the equation to find its value. This is a vital skill in algebra and is used across different mathematical disciplines, including calculus and physics. To solve for a variable, consider the steps taken in rearranging and simplifying equations. - **Transformation into solvable forms**: - Identify all terms containing the desired variable. - Use algebraic operations like addition, subtraction, multiplication, or division to separate the variable. - Rearrange the equation to form a standard quadratic equation if it involves a square term.- **Application in real-world problems**: - Solving \(s = \frac{1}{2} g t^2 + v_0 t\) for \(t\) involves recognizing that it's a quadratic equation in \(t\).
- Applying algebraic rules to transform the equation simplifies the process of solving for \(t\), making complex physics problems manageably concise.
- This practice becomes essential in fields like engineering and physics, where precise values of variables lead to accurate predictions and decisions.