Problem 41 - Precalculus Mathematics for Calculus

Question

Solve the equation for the indicated variable. $$h=\frac{1}{2} g t^{2}+v_{0} t ; \quad \text { for } t$$

Step-by-Step Solution

Answer

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

Step 1: Recognize the Equation Type

This is a quadratic equation in terms of the variable \( t \) because of the \( t^2 \) term. The equation is in the form \( h = \frac{1}{2} g t^2 + v_0 t \), which is similar to the standard quadratic equation \( ax^2 + bx + c = 0 \).

Step 2: Rearrange the Equation

Subtract \( h \) from both sides of the equation to set it to 0: \( \frac{1}{2} g t^2 + v_0 t - h = 0 \).

Step 3: Identify Coefficients

Identify the coefficients for the quadratic formula. Here, \( a = \frac{1}{2}g \), \( b = v_0 \), and \( c = -h \).

Step 4: Apply the Quadratic Formula

Use the quadratic formula, which is \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), to solve for \( t \). Substitute \( a = \frac{1}{2}g \), \( b = v_0 \), and \( c = -h \).

Step 5: Simplify the Expression

Substitute the known values into the formula: \[ t = \frac{-v_0 \pm \sqrt{v_0^2 - 4 \left( \frac{1}{2}g \right)(-h)}}{2 \left( \frac{1}{2}g \right)} \]. Simplify further to give: \[ t = \frac{-v_0 \pm \sqrt{v_0^2 + 2gh}}{g} \].

Step 6: Determine the Viable Solution

The equation for \( t \) gives two solutions due to the \( \pm \) sign. In physical situations, time is usually positive, so typically only the positive solution is meaningful: \[ t = \frac{-v_0 + \sqrt{v_0^2 + 2gh}}{g} \].

Key Concepts

Solving for a variable

In many mathematical problems, solving for a variable means isolating one variable on one side of an equation. In the given exercise, we aim to solve the equation for the variable \( t \). This involves rearranging terms and applying appropriate strategies to isolate \( t \) in the equation. To solve for \( t \) in the quadratic equation \( h = \frac{1}{2} g t^2 + v_0 t \), we perform several key steps:

• Recognize the type of equation: Identify that it is a quadratic equation due to the \( t^2 \) term.
• Rearrange the equation: Bring all terms to one side so that it can be set to zero, resembling the standard form \( ax^2 + bx + c = 0 \).
• Identify coefficients: Extract the coefficients \( a \), \( b \), and \( c \) that define the equation, which are critical for further steps.

Rearranging equations and identifying coefficients are foundational skills in solving for variables, which improve with practice and comprehension.

Quadratic formula

The quadratic formula is a universal tool used to find the solutions of any quadratic equation of the form \( ax^2 + bx + c = 0 \). It is expressed as:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula allows us to directly compute the values of \( t \) by substituting the values of the coefficients \( a \), \( b \), and \( c \) from our quadratic equation.

• Calculation of discriminant: The expression under the square root sign, \( b^2 - 4ac \), is called the discriminant, and it determines the nature of the solutions.
• Solving: Substitute the coefficients into the formula. In our example with \( a = \frac{1}{2}g \), \( b = v_0 \), and \( c = -h \), result in two potential solutions.
• Interpretation: These solutions characterize the values where the quadratic equation is satisfied.

In practice, understanding when and how to apply the quadratic formula is a critical step in solving quadratic equations.

Physics application

The equation in question isn't just a mathematical exercise; it has practical implications in physics. Specifically, it is related to motion under constant acceleration, a common topic in physics, particularly when analyzing projectile motion or free-fall situations.

• Understanding Variables: In physics, \( h \) can represent height, \( g \) the acceleration due to gravity, usually approximated as 9.81 m/s², and \( v_0 \) the initial velocity of the object.
• Significance: Solving for \( t \) using these parameters helps predict when an object, subject to gravity and initial velocity, reaches a certain height \( h \).
• Practical Use: Often used in determining the time an object takes to reach a certain point during its motion, which is crucial in fields like engineering and space sciences.

By understanding this application, students gain insight into the real-world relevance of quadratic equations and their role in modeling dynamic systems in physics.