Answer :
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]
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]