Question

consider the equation $v = \sqrt{\frac{2e_k}{m}}$, where $m = 125$. complete the statement below about the relationship between $e_k$ and $v$. as $e_k$ increases, $v$ increases at dropdown with options: the same rate as, a faster rate than, a slower rate than $e_k$.

Explanation:

Step1: Analyze the Function Type

The equation is \( v=\sqrt{\frac{2E_k}{m}} \) with \( m = 125 \), so it simplifies to \( v=\sqrt{\frac{2}{125}E_k}=\sqrt{\frac{2}{125}}\cdot\sqrt{E_k} \). This is a square - root function of the form \( y = a\sqrt{x} \) where \( y = v \), \( x = E_k \), and \( a=\sqrt{\frac{2}{125}}>0 \).

Step2: Compare Rates of Change

For a linear function \( y = mx + b \), the rate of change is constant. For a square - root function \( y=\sqrt{x} \), the derivative (rate of change) \( y^\prime=\frac{1}{2\sqrt{x}} \), which is a decreasing function of \( x \).

Let's consider the rate of change of \( v \) with respect to \( E_k \). We can rewrite \( v = k\sqrt{E_k} \) where \( k=\sqrt{\frac{2}{125}} \). The derivative of \( v \) with respect to \( E_k \) is \( \frac{dv}{dE_k}=\frac{k}{2\sqrt{E_k}} \).

For the function \( E_k\) (considered as a function of itself, with a rate of change of 1), as \( E_k \) increases, \( \frac{dv}{dE_k}=\frac{k}{2\sqrt{E_k}} \) decreases.

We can also think about the relationship between the growth of \( v \) and \( E_k \). If \( E_k \) doubles from \( E_{k1}\) to \( E_{k2} = 2E_{k1}\), then \( v_1=\sqrt{\frac{2}{125}E_{k1}} \) and \( v_2=\sqrt{\frac{2}{125}\times2E_{k1}}=\sqrt{2}\times\sqrt{\frac{2}{125}E_{k1}}=\sqrt{2}v_1\approx1.414v_1 \). So when \( E_k \) is multiplied by 2, \( v \) is multiplied by \( \sqrt{2}\approx1.414 \), which is less than 2.

This means that as \( E_k \) increases, \( v \) increases, but the factor by which \( v \) increases is less than the factor by which \( E_k \) increases. So \( v \) increases at a slower rate than \( E_k \).

Answer:

a slower rate than