Answered

The process for rationalizing a denominator in a variable expression is the same as in a numeric expression. Here's a real-world example.

The kinetic energy of a rollercoaster car is given by the formula [tex]$k = \frac{1}{2} m v^2$[/tex], where [tex]$k$[/tex] is the kinetic energy, [tex][tex]$m$[/tex][/tex] is the mass of the car, and [tex]$v$[/tex] is the velocity of the car. Solving this formula for [tex]$v$[/tex], we get [tex]$v = \sqrt{\frac{2k}{m}}$[/tex].

Which formula gives the velocity of the car in simplest form?

A. [tex][tex]$v = \frac{\sqrt{2km}}{2m}$[/tex][/tex]
B. [tex]$v = \frac{\sqrt{2km}}{m}$[/tex]
C. [tex]$v = \frac{\sqrt{2k}}{m}$[/tex]



Answer :

GinnyAnswer
Let's start by examining the given kinetic energy formula and the derived formula for velocity.

The kinetic energy [tex]\( k \)[/tex] of the car is given by:
[tex]\[ k = \frac{1}{2}mv^2 \][/tex]

We need to solve this formula for [tex]\( v \)[/tex] (velocity). Here are the steps:

1. Isolate [tex]\( v^2 \)[/tex]:
\begin{align}
k &= \frac{1}{2}mv^2 \\
2k &= mv^2
\end{align}

2. Solve for [tex]\( v \)[/tex]:
\begin{align}
v^2 &= \frac{2k}{m} \\
v &= \sqrt{\frac{2k}{m}}
\end{align}

Now, we have the derived formula for velocity:
[tex]\[ v = \sqrt{\frac{2k}{m}} \][/tex]

Let's compare this with the given answer choices:

1. Option 1: [tex]\( v = \frac{\sqrt{2km}}{2m} \)[/tex]
- Simplify the expression:
\begin{align}
v &= \frac{\sqrt{2km}}{2m} \\
&= \frac{\sqrt{2} \cdot \sqrt{km}}{2m} \\
&= \frac{\sqrt{2} \cdot \sqrt{k} \cdot \sqrt{m}}{2m} \\
&= \frac{\sqrt{2k}}{2\sqrt{m}} \\
&= \frac{\sqrt{2k}}{\sqrt{m}\sqrt{m}} \\
&= \frac{\sqrt{2k}}{m \sqrt{m}} \\
\end{align}
- This expression does not match the simplified derived formula [tex]\(\sqrt{\frac{2k}{m}}\)[/tex].

2. Option 2: [tex]\( v = \frac{\sqrt{2km}}{m} \)[/tex]
- Simplify the expression:
\begin{align}
v &= \frac{\sqrt{2km}}{m} \\
&= \frac{\sqrt{2} \cdot \sqrt{km}}{m} \\
&= \frac{\sqrt{2} \cdot \sqrt{k} \cdot \sqrt{m}}{m} \\
&= \frac{\sqrt{2k} \cdot \sqrt{m}}{m} \\
&= \frac{\sqrt{2k}}{\sqrt{m}}
\end{align}
- This expression is equivalent to [tex]\(\sqrt{\frac{2k}{m}}\)[/tex].

3. Option 3: [tex]\( v = \frac{\sqrt{2k}}{m} \)[/tex]
- Simplify the expression:
\begin{align}
v &= \frac{\sqrt{2k}}{m} \\
\end{align}
- This expression does not include [tex]\(\sqrt{m}\)[/tex] in the correct position compared to the simplified derived formula.

So, among the given options, the correct and simplest form of the formula for velocity [tex]\( v \)[/tex] is:
[tex]\[ v = \frac{\sqrt{2km}}{m} \][/tex]

Therefore, the correct option is:
[tex]\[ 2 \][/tex]