Why is the no v kinematic equation helpful for solving kinematics problems?

[jsspoon]

In the no v kinematic equation, $x={x}_{o}+{v}_{o}t+a{t}^{2}/2$, why do you have to solve $a{t}^{2}/2$ first before solving down completely?

 

[MarkFL]

Hello and welcome to MHB! (Wave)

What do you mean by "solve $at^2/2$ first?"

 

[suluclac]

Since t has the power of 2 for the acceleration, perhaps this is why the OP has stated "first" . . .?
Also, jsspoon, is there any specific question regarding the kinematic equation?

 

[HallsofIvy]

If by "solve" you simply mean "evaluate x for a given value of t", you do not need to evaluate \(\displaystyle \frac{1}{2}at^2\). You can do the calculations in any order. If you mean "solve for t for a given value of x", again there is nothing special about the \(\displaystyle \frac{1}{2}at^2\) term- you can use the quadratic formula to solve.

And why was this called "no v kinematic equation"?

 

[Ackbach]

If by "solve" you simply mean "evaluate x for a given value of t", you do not need to evaluate \(\displaystyle \frac{1}{2}at^2\). You can do the calculations in any order. If you mean "solve for t for a given value of x", again there is nothing special about the \(\displaystyle \frac{1}{2}at^2\) term- you can use the quadratic formula to solve.

And why was this called "no v kinematic equation"?

The OP'er was one of my students, and we had labels for the four kinematic equations:
\begin{align*}
y&=y_0+v_{0y}t+a_y t^2/2 \qquad \text{no }v_y \\
\Delta y&=(v_y+v_{0y})t/2 \qquad \text{no }a_y \\
v_y&=v_{0y}+a_y t \qquad \text{no }y \\
v_y^2&=v_{0y}^2+2a_y \Delta y \qquad \text{no }t.
\end{align*}
Since, in kinematics, there are basically four players: $y, v_y, a_y, t$, there's one kinematic equation corresponding to which kinematic variable is missing. Knowing these four equations helps the students with the algebra, because they can just solve the one they need, and not necessarily have to plug one equation into another.