Torricelli's equation

Begin with the following relations for the case of uniform acceleration:

${\displaystyle x_{f}-x_{i}=v_{i}t+{\tfrac {1}{2}}at^{2}}$

(1)

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

(2)

Take (1), and multiply both sides with acceleration ${\textstyle a}$ 

${\displaystyle a(x_{f}-x_{i})=av_{i}t+{\tfrac {1}{2}}a^{2}t^{2}}$

(3)

The following rearrangement of the right hand side makes it easier to recognize the coming substitution:

${\displaystyle a(x_{f}-x_{i})=v_{i}(at)+{\tfrac {1}{2}}(at)^{2}}$

(4)

Use (2) to substitute the product ${\textstyle at}$ :

${\displaystyle a(x_{f}-x_{i})=v_{i}(v_{f}-v_{i})+{\tfrac {1}{2}}(v_{f}-v_{i})^{2}}$

(5)

Work out the multiplications:

${\displaystyle a(x_{f}-x_{i})=v_{i}v_{f}-v_{i}^{2}+{\tfrac {1}{2}}v_{f}^{2}-v_{i}v_{f}+{\tfrac {1}{2}}v_{i}^{2}}$

(6)

The crossterms ${\textstyle v_{i}v_{f}}$  drop away against each other, leaving only squared terms:

${\displaystyle a(x_{f}-x_{i})={\tfrac {1}{2}}v_{f}^{2}-{\tfrac {1}{2}}v_{i}^{2}}$

(7)

(7) rearranges to the form of Torricelli's equation as presented at the start of the article:

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

(8)

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 ${\textstyle \Delta x}$ :

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

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

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

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