Definition 6.1.1.
AlgebraAlgebra is a set of tools for representing relationships between quantities as equations, and for manipulating those equations to solve for unknown quantities.
Section 6.1 Prep
This preparatory section will equip you with the knowledge and skills to wield algebra with confidence and precision. Through a combination of instructional video, explanatory text, and practice questions, you will become adept at all things algebra.
Subsection 6.1.1 Prep Videos: Units Conversions
Watch the following video(s) to get an introduction to unit conversions.
Subsection 6.1.2 Prep Text: Algebra
Read the following text to learn about algebra. Don’t forget to try the checkpoint questions along the way!
In physics, equations describe relationships between measurable quantities such as time, distance, velocity, and force. While physics helps you build these relationships, algebra allows you to rearrange them to answer specific questions.
Note that the land of algebra is vast! This section will cover a few of the most common algebra ideas that appear early on in our quest through the mechanics of the universe. More algebra topics will be covered as the need arises in later sections.
An important guiding principle in algebra is: whatever you do to one side of an equation, you must also do to the other. This preserves equality and ensures the relationship remains valid.
IMPORTANT: The solutions in this Algebra Chapter 6 show step-by-step methods to solve of unkonwn quantities.
- The order of algebraic steps shown in the solutions may differ from your own approach, which is completely acceptable.
- In later chapters and sections, solutions may not include every algebraic step, instead focusing on the physics and science in the solutions.
Subsubsection 6.1.2.1 Solving a Single Equation for a Single Unknown
Definition 6.1.2.
One Equation, One UnknownA "one equation, one unknown" situation in algebra involves a single equation with one unknown quantity. The unknown quantity is typically found by rearranging the equation using the rules of algebra so that the unknown appears by itself.
Example 6.1.3.
Consider the follwoing equation:
\(v = v_0 + at \)
Where \(v\text{,}\) \(v_0\text{,}\) \(a\) are known quantities, and \(t\) is the unknown quantitiy. By rearranging the equation, we can isolate (i.e. solve for) \(t\text{.}\)
First subtract \(v_0\) from both sides of the equation:
\((v)-v_0 = (v_0 + at)-v_0\)
which results in:
\(v-v_0 = at\)
Note that in the above step we "subtracted \(v_0\) from both sides". Another commom way to say this is: "subtract \(v_0\) to the other side", since the end result is "moving" \(v_0\) from one side to the other.
Then divide by sides by \(a\text{:}\)
\(\frac{v-v_0}{a} = \frac{at}{a}\)
which results in:
\(\frac{v-v_0}{a} = t\)
Now that \(t\) is all by itself, we are done. However, folks seem to like our unknown quantities that have been solved for to be on the left side of the quation instead of the right side. So if you want, you could also write the above solution
\(t = \frac{v-v_0}{a}\)
Checkpoint 6.1.4.
Consider the follwoing equation:
\(x = x_0 + vt + \frac{1}{2}at^2\)
Where \(x\text{,}\) \(x_0\text{,}\) \(v\text{,}\) \(t\) are known quantities, and \(a\) is the unknown quantitiy.
Solve for \(a\text{.}\)
Hint.
Start by subtracting both \(x_0\text{,}\) and \(vt\) from both sides. Then divide both sides to isolate \(a\text{.}\)
Answer.
\(a = \frac{2(x-x_0-vt)}{t^2}\)
Solution.
Our original equation is:
\(x = x_0 + vt + \frac{1}{2}at^2\)
Subtracting both \(x_0\text{,}\) and \(vt\) from both sides.:
\((x)-x_0-vt = (x_0 + vt + \frac{1}{2}at^2)-x_0-vt\)
which simplifies to:
\(x-x_0-vt = \frac{1}{2}at^2\)
Next you can divide by sides by \(\frac{1}{2}\)
\(\frac{x-x_0-vt}{\frac{1}{2}} = \frac{(\frac{1}{2}at^2)}{\frac{1}{2}}\)
which simplifies to:
\(2(x-x_0-vt) = at^2\)
Now divide both sides by \(t^2\)
\(\frac{2(x-x_0-vt)}{t^2} = \frac{at^2}{t^2}\)
which simplifies to:
\(\frac{2(x-x_0-vt)}{t^2} = a\)
So using the commom convention of our unknown on the left hand side, the answer is:
\(a= \frac{2(x-x_0-vt)}{t^2}\)
Subsubsection 6.1.2.2 Quadratic Equations
Definition 6.1.5.
Quadratic EquationA quadratic equation is a special case of a "one equation, one unknown" situation. The standard form used to represent a quadratic equation is: \(ax^2 + bx + c = 0 \text{,}\) where \(a\text{,}\) \(b\text{,}\) and \(c\) are known quantities, \(x\) is the unknown quantitiy, and \(a \neq 0\text{.}\)
Example 6.1.6.
Consider the follwoing equation:
\(v = v_0 + at \)
Where \(v\text{,}\) \(v_0\text{,}\) \(a\) are known quantities, and \(t\) is the unknown quantitiy. By rearranging the equation, we can isolate (i.e. solve for) \(t\text{.}\)
First subtract \(v_0\) from both sides of the equation:
\((v)-v_0 = (v_0 + at)-v_0\)
which results in:
\(v-v_0 = at\)
Note that in the above step we "subtracted \(v_0\) from both sides". Another commom way to say this is: "subtract \(v_0\) to the other side", since the end result is "moving" \(v_0\) from one side to the other.
Then divide by sides by \(a\text{:}\)
\(\frac{v-v_0}{a} = \frac{at}{a}\)
which results in:
\(\frac{v-v_0}{a} = t\)
Now that \(t\) is all by itself, we are done. However, folks seem to like our unknown quantities that have been solved for to be on the left side of the quation instead of the right side. So if you want, you could also write the above solution
\(t = \frac{v-v_0}{a}\)
Checkpoint 6.1.7.
Consider the follwoing equation:
\(x = x_0 + vt + \frac{1}{2}at^2\)
Where \(x\text{,}\) \(x_0\text{,}\) \(v\text{,}\) \(t\) are known quantities, and \(a\) is the unknown quantitiy.
Solve for \(a\text{.}\)
Hint.
Start by subtracting both \(x_0\text{,}\) and \(vt\) from both sides. Then divide both sides to isolate \(a\text{.}\)
Answer.
\(a = \frac{2(x-x_0-vt)}{t^2}\)
Solution.
Our original equation is:
\(x = x_0 + vt + \frac{1}{2}at^2\)
Subtracting both \(x_0\text{,}\) and \(vt\) from both sides.:
\((x)-x_0-vt = (x_0 + vt + \frac{1}{2}at^2)-x_0-vt\)
which simplifies to:
\(x-x_0-vt = \frac{1}{2}at^2\)
Next you can divide by sides by \(\frac{1}{2}\)
\(\frac{x-x_0-vt}{\frac{1}{2}} = \frac{(\frac{1}{2}at^2)}{\frac{1}{2}}\)
which simplifies to:
\(2(x-x_0-vt) = at^2\)
Now divide both sides by \(t^2\)
\(\frac{2(x-x_0-vt)}{t^2} = \frac{at^2}{t^2}\)
which simplifies to:
\(\frac{2(x-x_0-vt)}{t^2} = a\)
So using the commom convention of our unknown on the left hand side, the answer is:
\(a= \frac{2(x-x_0-vt)}{t^2}\)
Subsubsection 6.1.2.3 Connecting Equations (Substitution & Systems)
Subsubsection 6.1.2.4 Covariational Reasoning
Example 6.1.8.
This is a the first example problem example. This is a good spot for a question.
This is the begining of the solution.
Include as many paragraphs as needed. You can include math stuff too.
Maybe a short debreif after the examples?
I also like to have 1-3 check point questions after the examples if any. The check point questions have a dropdown that shows the question, then there are three more drop downs for hints, answer, and solution.
Checkpoint 6.1.9.
Problem statement goes here.
Hint.
Be nice, include a hint.
Answer.
Answer goes here.
Solution.
A detailed solution goes here.
Can always have multiple paragraphs in any of these sections.
Checkpoint 6.1.10.
Problem statement goes here.
Hint.
Be nice, include a hint.
Answer.
Answer goes here.
Solution.
A detailed solution goes here.
Can always have multiple paragraphs in any of these sections.
Why Use Them?
This is a why section. Not sure if I will always have this since I try to motivate the topic throughout. But keep this short, very short if included.
Bonus Trick: something fun
Is this topic related to something else fun, if so mention it here.
Keep it short.
If the bonus trick needs a longer example, include it after here.
Example 6.1.11.
This is a the first example problem example. This is a good spot for a question.
This is the begining of the solution.
Include as many paragraphs as needed. You can include math stuff too.
Very short parting word. Keep it whimiscal.
Quick review questions after this section. These questions have a drop down for the problem statement, then have multiple choice options. Each option has a hint after clicked and checked by user.
Subsection 6.1.3 Prep Questions: Algebra
Algebra
Checkpoint 6.1.12.
- Correct answer.
- Correct!
- Incorrect answer 1
- Your targeting system glitched, try again. Give a hint.
- Incorrect answer 2
- Your enchanted click missed the mark, try again. Give a hint.
- Incorrect answer 3
- Your digital blade missed the mark, try again. Give a hint.
Problem statment goes here.
