Many kinematics problems are given to students in high school and college physics courses which involve the five following measurable quantities of a body in motion:

• ${\displaystyle a\,}$ - acceleration (most often assumed to be constant)
• ${\displaystyle t\,}$ - time elapsed
• ${\displaystyle v_{f}\,}$ - final velocity (this variable may also be represented by just ${\displaystyle v\,}$)
• ${\displaystyle v_{i}\,}$ - initial velocity (this variable may also be represented by ${\displaystyle u\,}$ or ${\displaystyle v_{0}\,}$)
• ${\displaystyle x\,}$ - distance traveled (this variable may also be represented by ${\displaystyle d\,}$ or ${\displaystyle s\,}$)

If one knows any three of the above quantities for a given situation, then one may solve for any of the other two. The equations below demonstrate this property.

${\displaystyle a={\dfrac {v_{f}^{2}-v_{i}^{2)){2x))\,}$

${\displaystyle a=2{\dfrac {x-v_{i}t}{t^{2))}\,}$

${\displaystyle a=2{\dfrac {v_{f}t-x}{t^{2))}\,}$

${\displaystyle a={\dfrac {v_{f}-v_{i)){t))\,}$

${\displaystyle t={\dfrac {2x}{v_{f}+v_{i))}\,}$

${\displaystyle t={\dfrac {-v_{i}+{\sqrt {v_{i}^{2}+2ax))}{a))\,}$

${\displaystyle t={\dfrac {-v_{f}+{\sqrt {v_{f}^{2}-2ax))}{-a))\,}$

${\displaystyle t={\dfrac {v_{f}-v_{i)){a))\,}$

${\displaystyle v_{f}={\dfrac {2x}{t))-v_{i}\,}$

${\displaystyle v_{f}={\sqrt {v_{i}^{2}+2ax))\,}$

${\displaystyle v_{f}={\dfrac {2x+at^{2)){2t))\,}$

${\displaystyle v_{f}=v_{i}+at\,}$

${\displaystyle v_{i}={\dfrac {2x}{t))-v_{f}\,}$

${\displaystyle v_{i}={\sqrt {v_{f}^{2}-2ax))\,}$

${\displaystyle v_{i}={\dfrac {2x-at^{2)){2t))\,}$

${\displaystyle v_{i}=v_{f}-at\,}$

${\displaystyle x={\dfrac {v_{f}+v_{i)){2))t\,}$

${\displaystyle x={\dfrac {v_{f}^{2}-v_{i}^{2)){2a))\,}$

${\displaystyle x=v_{i}t+{\tfrac {1}{2))at^{2}\,}$

${\displaystyle x=v_{f}t-{\tfrac {1}{2))at^{2}\,}$