Problem 51
Question
The velocity \(v\) of an object dropped from a tall building is given by the formula \(v=\sqrt{64 d},\) where \(d\) is the distance the object has dropped. Solve the formula for \(d\)
Step-by-Step Solution
Verified Answer
If \(v\) is the velocity of an object dropped from a tall building, the distance \(d\) that the object has dropped can be calculated using the formula \(d = \frac{v^2}{64}\).
1Step 1: Isolate the square root
To isolate \(d\), first we need to isolate the square root, by dividing both sides of the equation by \(\sqrt{64}\). This gives us \(\frac{v}{\sqrt{64}} = \sqrt{d}\).
2Step 2: Eliminate the square root
Now, we want to get rid of the square root on the right side. To do this, we square both sides of the equation. This gives us \((\frac{v}{\sqrt{64}})^2 = d\).
3Step 3: Simplify the expression
Finally, we simplify the left side of the equation to get the final expression for \(d\). As \(\sqrt{64} = 8\), our equation becomes \(\frac{v^2}{64} = d\).
Key Concepts
Radical EquationsAlgebraic ManipulationMathematical Formulas
Radical Equations
Radical equations involve equations where variables are under a radical sign, which often signifies a square root or higher-order roots. These equations require special attention because the presence of a radical can complicate the process of solving for a variable. A key step in dealing with radical equations is understanding that the radical must be removed to isolate the variable and proceed with solving.
To eliminate a square root, you can square both sides of the equation. For instance, in the equation \(v = \sqrt{64d}\), \(d\) is under the square root. By squaring both sides, you effectively remove the square root and allow for simpler algebraic manipulation. Remember to check your solutions in the original equation, as squaring can introduce extraneous solutions that don't satisfy the original radical equation.
The method of dealing with radical equations efficiently needs careful consideration of each step, such as isolating the radical term first and then proceeding to square both sides. This approach prevents errors such as incorrect simplification or misunderstanding of the root properties.
To eliminate a square root, you can square both sides of the equation. For instance, in the equation \(v = \sqrt{64d}\), \(d\) is under the square root. By squaring both sides, you effectively remove the square root and allow for simpler algebraic manipulation. Remember to check your solutions in the original equation, as squaring can introduce extraneous solutions that don't satisfy the original radical equation.
The method of dealing with radical equations efficiently needs careful consideration of each step, such as isolating the radical term first and then proceeding to square both sides. This approach prevents errors such as incorrect simplification or misunderstanding of the root properties.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging and simplifying equations to isolate the variable in question. This skill is essential when solving equations, especially those involving radicals, as it requires precision and understanding of mathematical operations.
Steps in algebraic manipulation include:
Steps in algebraic manipulation include:
- Isolating the term with the variable you need to solve for, such as by dividing or multiplying both sides of an equation.
- Applying inverse operations to simplify each side – these can include squaring, factoring, or canceling terms.
- Simplifying the remaining expression to the most reduced form possible.
Mathematical Formulas
Mathematical formulas serve as tools for expressing relationships between different variables, making them crucial in problem-solving across various mathematical contexts. The formula \(v = \sqrt{64d} \) connects velocity \(v\) with distance \(d\) dropped by an object, implicating a direct relationship through square roots.
Understanding mathematical formulas in-depth involves recognizing the components within such formulas and how altering one can affect others. It's also important to know that formulas can be rearranged. In the original exercise, this rearrangement was done to express \(d\) in terms of \(v\), showing that a formula can have multiple expressions depending on which variable you need to isolate.
Breaking down formulas to identify their core parts and purposes – such as constants, variables, and operations (like squaring) – supports how we can understand and manipulate them to solve equations. This skill is invaluable, as it offers the ability to one confidently solve real-world problems efficiently, by applying the right formulas appropriately.
Understanding mathematical formulas in-depth involves recognizing the components within such formulas and how altering one can affect others. It's also important to know that formulas can be rearranged. In the original exercise, this rearrangement was done to express \(d\) in terms of \(v\), showing that a formula can have multiple expressions depending on which variable you need to isolate.
Breaking down formulas to identify their core parts and purposes – such as constants, variables, and operations (like squaring) – supports how we can understand and manipulate them to solve equations. This skill is invaluable, as it offers the ability to one confidently solve real-world problems efficiently, by applying the right formulas appropriately.
Other exercises in this chapter
Problem 51
Graph. Find the domain and the range of each function. \(y=-3 \sqrt{x-\frac{3}{4}}+7\)
View solution Problem 51
For each function \(f,\) find \(f^{-1},\) the domain and range of \(f\) and \(f^{-1},\) and determine whether \(f^{-1}\) is a function. $$ f(x)=\frac{x^{2}}{2}
View solution Problem 51
Let \(f(x)=3 x^{2}+2 x-8\) and \(g(x)=x+2 .\) Perform each function operation and then find the domain. $$ -f(x)+4 g(x) $$
View solution Problem 51
Add or subtract. $$ \frac{1}{1-\sqrt{5}}+\frac{1}{1+\sqrt{5}} $$
View solution