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

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SUMMARY

The discussion centers on the "no v kinematic equation," specifically the equation $x={x}_{o}+{v}_{o}t+\frac{1}{2}at^{2}$. Participants clarify that the term $\frac{1}{2}at^{2}$ does not need to be solved first when evaluating $x$ for a given value of $t$. The equation is part of a set of four kinematic equations used in physics to solve problems involving motion, each corresponding to a missing variable: position, velocity, acceleration, or time. Understanding these equations simplifies the algebra involved in solving kinematics problems.

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  • Understanding of basic kinematics concepts
  • Familiarity with the four kinematic equations
  • Knowledge of algebra and the quadratic formula
  • Basic grasp of motion variables: position, velocity, acceleration, time
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  • Study the derivation and applications of the four kinematic equations
  • Learn how to apply the quadratic formula to solve kinematics problems
  • Explore examples of motion problems using the "no v" kinematic equation
  • Investigate the relationship between acceleration and time in motion equations
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Students of physics, educators teaching kinematics, and anyone looking to deepen their understanding of motion equations and problem-solving techniques in physics.

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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?
 
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Hello and welcome to MHB! (Wave)

What do you mean by "solve $at^2/2$ first?"
 
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?
 
If by "solve" you simply mean "evaluate x for a given value of t", you do not need to evaluate [math]\frac{1}{2}at^2[/math]. 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 [math]\frac{1}{2}at^2[/math] term- you can use the quadratic formula to solve.

And why was this called "no v kinematic equation"?
 
Last edited by a moderator:
HallsofIvy said:
If by "solve" you simply mean "evaluate x for a given value of t", you do not need to evaluate [math]\frac{1}{2}at^2[/math]. 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 [math]\frac{1}{2}at^2[/math] 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.
 

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