Without differentials and integration edit

Begin with the definition of acceleration:

${\displaystyle a={\frac {v_{f}-v_{i}}{\Delta t}}}$ 

where ${\textstyle \Delta t}$  is the time interval. This is true because the acceleration is constant. The left hand side is this constant value of the acceleration and the right hand side is the average acceleration. Since the average of a constant must be equal to the constant value, we have this equality. If the acceleration was not constant, this would not be true.

Now solve for the final velocity:

${\displaystyle v_{f}=v_{i}+a\Delta t\,\!}$ 

Square both sides to get:

${\displaystyle v_{f}^{2}=(v_{i}+a\Delta t)^{2}=v_{i}^{2}+2av_{i}\Delta t+a^{2}(\Delta t)^{2}\,\!}$

(1)

The term ${\displaystyle (\Delta t)^{2}\,\!}$  also appears in another equation that is valid for motion with constant acceleration: the equation for the final position of an object moving with constant acceleration, and can be isolated:

${\displaystyle x_{f}=x_{i}+v_{i}\Delta t+a{\frac {(\Delta t)^{2}}{2}}}$ 

${\displaystyle x_{f}-x_{i}-v_{i}\Delta t=a{\frac {(\Delta t)^{2}}{2}}}$ 

${\displaystyle (\Delta t)^{2}=2{\frac {x_{f}-x_{i}-v_{i}\Delta t}{a}}=2{\frac {\Delta x-v_{i}\Delta t}{a}}}$

(2)

Substituting (2) into the original equation (1) yields:

${\displaystyle v_{f}^{2}=v_{i}^{2}+2av_{i}\Delta t+a^{2}\left(2{\frac {\Delta x-v_{i}\Delta t}{a}}\right)}$ 

${\displaystyle v_{f}^{2}=v_{i}^{2}+2av_{i}\Delta t+2a(\Delta x-v_{i}\Delta t)}$ 

${\displaystyle v_{f}^{2}=v_{i}^{2}+2av_{i}\Delta t+2a\Delta x-2av_{i}\Delta t\,\!}$ 

${\displaystyle v_{f}^{2}=v_{i}^{2}+2a\Delta x\,\!}$ 

Using differentials and integration edit

Begin with the definition of acceleration as the derivative of the velocity:

${\displaystyle a={\frac {dv}{dt}}}$ 

Now, we multiply both sides by the velocity ${\textstyle v}$ :

${\displaystyle v\cdot a=v\cdot {\frac {dv}{dt}}}$ 

In the left hand side we can rewrite the velocity as the derivative of the position:

${\displaystyle {\frac {dx}{dt}}\cdot a=v\cdot {\frac {dv}{dt}}}$ 

Multiplying both sides by ${\textstyle dt}$  gets us the following:

${\displaystyle dx\cdot a=v\cdot dv}$ 

Rearranging the terms in a more traditional manner:

${\displaystyle a\,dx=v\,dv}$ 

Integrating both sides from the initial instant with position ${\textstyle x_{i}}$  and velocity ${\textstyle v_{i}}$  to the final instant with position ${\textstyle x_{f}}$  and velocity ${\textstyle v_{f}}$ :

${\displaystyle \int _{x_{i}}^{x_{f}}{a}\,dx=\int _{v_{i}}^{v_{f}}v\,dv}$ 

Since the acceleration is constant, we can factor it out of the integration:

${\displaystyle {a}\int _{x_{i}}^{x_{f}}dx=\int _{v_{i}}^{v_{f}}v\,dv}$ 

Solving the integration:

${\displaystyle {a}{\bigg [}x{\bigg ]}_{x=x_{i}}^{x=x_{f}}=\left[{\frac {v^{2}}{2}}\right]_{v=v_{i}}^{v=v_{f}}}$ 

${\displaystyle {a}\left(x_{f}-x_{i}\right)={\frac {v_{f}^{2}}{2}}-{\frac {v_{i}^{2}}{2}}}$ 

The factor ${\textstyle x_{f}-x_{i}}$  is the displacement ${\textstyle \Delta x}$ :

${\displaystyle a\Delta x={\frac {1}{2}}\left(v_{f}^{2}-v_{i}^{2}\right)}$ 

${\displaystyle 2a\Delta x=v_{f}^{2}-v_{i}^{2}}$ 

${\displaystyle v_{f}^{2}=v_{i}^{2}+2a\Delta x}$ 

From the work-energy theorem edit

The work-energy theorem states that

${\displaystyle \Delta E_{K}=W}$ 

${\displaystyle {\frac {m}{2}}\left(v_{f}^{2}-v_{i}^{2}\right)=F\Delta x}$ 

which, from Newton's second law of motion, becomes

${\displaystyle {\frac {m}{2}}\left(v_{f}^{2}-v_{i}^{2}\right)=ma\Delta x}$ 

${\displaystyle v_{f}^{2}-v_{i}^{2}=2a\Delta x}$ 

${\displaystyle v_{f}^{2}=v_{i}^{2}+2a\Delta x}$ 

• Equation of motion

• Torricelli's theorem