Torricelli's equation

Begin with the definition of acceleration:

${\displaystyle a_{x}={\frac {{v_{f}}_{x}-{v_{i}}_{x}}{\Delta t}}}$

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:

${\displaystyle {v_{f}}_{x}={v_{i}}_{x}+a_{x}\Delta t\,\!}$

Square both sides to get:

${\displaystyle {v_{f}}_{x}^{2}=({v_{i}}_{x}+a_{x}\Delta t)^{2}={v_{i}}_{x}^{2}+2a_{x}{v_{i}}_{x}\Delta t+a_{x}^{2}(\Delta t)^{2}\,\!}$

(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:

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

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

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

(2)

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

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

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

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

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

• Equation of motion

• Torricelli's theorem