Torricelli's equation

For other uses, see Torricelli.

In physics, Torricelli's equation, or Torricelli's formula, is an equation created by Evangelista Torricelli to find the final velocity of a moving object with constant acceleration along an axis (for example, the x axis) without having a known time interval.

The equation itself is:

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

where

• $v_{f}$ is the object's final velocity along the x axis on which the acceleration is constant.
• $v_{i}$ is the object's initial velocity along the x axis.
• $a$ is the object's acceleration along the x axis, which is given as a constant.
• $\Delta x\,$ is the object's change in position along the x axis, also called displacement.

In this and all subsequent equations in this article, the subscript $x$ (as in ${v_{f}}_{x}$) is implied, but is not expressed explicitly for clarity in presenting the equations.

This equation is valid along any axis on which the acceleration is constant.

Derivation

Without differentials and integration

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

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

$v_{f}-v_{i}=at$ (2)

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

$a(x_{f}-x_{i})=av_{i}t+{\tfrac {1}{2}}a^{2}t^{2}$ (3)

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

$a(x_{f}-x_{i})=v_{i}(at)+{\tfrac {1}{2}}(at)^{2}$ (4)

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

$a(x_{f}-x_{i})=v_{i}(v_{f}-v_{i})+{\tfrac {1}{2}}(v_{f}-v_{i})^{2}$ (5)

Work out the multiplications:

$a(x_{f}-x_{i})=v_{i}v_{f}-v_{i}^{2}+{\tfrac {1}{2}}v_{f}^{2}-v_{i}v_{f}+{\tfrac {1}{2}}v_{i}^{2}$ (6)

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

$a(x_{f}-x_{i})={\tfrac {1}{2}}v_{f}^{2}-{\tfrac {1}{2}}v_{i}^{2}$ (7)

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

$v_{f}^{2}=v_{i}^{2}+2a\Delta x$ (8)

Using differentials and integration

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

$a={\frac {dv}{dt}}$

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

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

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

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

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

$dx\cdot a=v\cdot dv$

Rearranging the terms in a more traditional manner:

$a\,dx=v\,dv$

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 ${\textstyle v_{f}}$:

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

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

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

Solving the integration:

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

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

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

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

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

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

From the work-energy theorem

The work-energy theorem states that

$\Delta E_{K}=W$

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

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

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

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

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

See also

• Equation of motion

External links

• Torricelli's theorem