The equation has a factor of $\dfrac{1}{2}$ on the right. What operation will undo multiplying by $\dfrac{1}{2}$ so that you can get rid of the fraction?

After you clear the fraction, how can you isolate $v^2$? Once you have an equation of the form $v^2 = \text{(something)}$, what operation do you use to solve for $v$?

When you take the square root while solving $v^2 = \text{(something)}$, remember that there are usually two roots. How does the condition $v>0$ affect which root you keep?

In Desmos, type m = 2 and K = 5 (or any positive values you like) to create constants for mass and kinetic energy. Then enter the function y = 0.5*m*x^2 to represent $K = \frac{1}{2} m v^2$ with $x$ standing in for $v$.

Add a horizontal line y = K. The intersection of y = 0.5*m*x^2 and y = K gives the value of $x$ (the speed) that satisfies the original equation for your chosen $m$ and $K$. Use Desmos’s intersection tool to see the positive $x$-coordinate of this point.

For each option (A–D), type its right-hand expression into a new Desmos line using your chosen values of $K$ and $m$ (for example, v_A = 2K/m, v_B = sqrt(m/(2K)), etc.). Compare each numeric result to the positive intersection $x$-value from step 2; the correct choice is the one whose value matches that intersection.

We are given

and we want to solve this equation for $v$ in terms of $K$ and $m$, assuming $v>0$.

First, clear the fraction by multiplying both sides by 2:

Next, divide both sides by $m$ to get $v^2$ alone:

So now we know that $v^2$ equals $2K$ divided by $m$.

To solve for $v$, take the square root of both sides of $v^2 = \dfrac{2K}{m}$:

Because the problem tells us $v>0$, we choose the positive root only:

This matches choice D.