Problem 26 - Beginning and Intermediate Algebra

Question

Solve each equation for the specified variable. (Leave \(\pm\) in the answers.) \(S=v t+\frac{1}{2} g t^{2}\) for \(t\)

Step-by-Step Solution

Answer

t = \frac{-v \pm \sqrt{v^{2} + 2gS}}{g}\.

Step 1 - Recognize the quadratic form

Identify the equation as a quadratic equation in terms of the variable you're solving for, which is \(t\). The given equation is \(S = vt + \frac{1}{2}gt^{2}\).

Step 2 - Rewrite in standard quadratic form

Rewrite the equation in the standard quadratic form \(at^{2} + bt + c = 0\). This form is \(\frac{1}{2}gt^{2} + vt - S = 0\).

Step 3 - Identify coefficients

Identify the coefficients \(a, b, \text{and} c\) in the standard quadratic form. Here, \(a = \frac{1}{2}g, b = v, \text{and} c = -S\).

Step 4 - Apply the quadratic formula

Apply the quadratic formula \(t = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). Substituting the coefficients gives \(t = \frac{-v \pm \sqrt{v^{2} - 4 \frac{1}{2}g (-S)}}{2 \frac{1}{2}g}\).

Step 5 - Simplify the quadratic formula

Simplify the expression inside the square root and the denominator. This results in \(t = \frac{-v \pm \sqrt{v^{2} + 2gS}}{g}\).

Key Concepts

solving quadratic equations

Quadratic equations often appear in physics, especially in problems dealing with motion. These equations are known as quadratics because their highest degree (or power) of the variable is 2. The general form is: \[ax^2 + bx + c = 0\] where:

• \(a\), \(b\), and \(c\) are constants
• \(x\) is the variable we're solving for

Solving these equations involves finding the values of \(x\) that make the equation true. There are various methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula.
To apply these methods successfully, it's crucial to correctly identify the quadratic nature of the problem and rewrite the equation in its standard form.

standard quadratic form

The standard quadratic form is a way of arranging a quadratic equation so that it's easier to solve. The standard form is: \[ax^2 + bx + c = 0\]

• \(a\) is the coefficient of \(x^2\)
• \(b\) is the coefficient of \(x\)
• \(c\) is the constant term

By comparing any given quadratic equation to this form, you can easily determine the coefficients \(a\), \(b\), and \(c\). For example, if we start with the physics equation \[S = vt + \frac{1}{2}gt^2\] and rewrite it in standard form as \[\frac{1}{2}gt^2 + vt - S = 0\] we identify the coefficients as:

• \(a = \frac{1}{2}g\)
• \(b = v\)
• \(c = -S\)

quadratic formula

The quadratic formula is a powerful tool for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is: \[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\]
This formula gives the solutions for \(x\) directly by plugging in the coefficients \(a\), \(b\), and \(c\).

• First, calculate the discriminant: \(b^2 - 4ac\).
• Next, take the square root of the discriminant.
• Finally, use the values of \(a\) and \(b\) to find \(x\) with the formula.

Using our physics example, substitute \(a = \frac{1}{2}g\), \(b = v\), and \(c = -S\) into the formula: \(t = \frac{-v \pm \sqrt{v^{2} - 4 \frac{1}{2}g (-S)}}{2 \frac{1}{2}g}\)
This simplifies to \(t = \frac{-v \pm \sqrt{v^{2} + 2gS}}{g}\).
The solutions for \(t\) are then obtained as: \(t = \frac{-v + \sqrt{v^{2} + 2gS}}{g}\) and \(t = \frac{-v - \sqrt{v^{2} + 2gS}}{g}\).