# Torricelli's equation

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 a{\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 at{\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 v$_i$v$_f${\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 definitions of velocity as the derivative of the position, and acceleration as the derivative of the velocity:
| $v={\frac {dx}{dt}}$ | (9) |
| $a={\frac {dv}{dt}}$ | (10) |
Set up integration from initial position x$_i${\textstyle x_{i}} to final position x$_f${\textstyle x_{f}}
| $\int _{x_{i}}^{x_{f}}a\,dx$ | (11) |
In accordance with (9) we can substitute dx{\textstyle dx} with vdt{\textstyle v\, dt}, with corresponding change of limits.
| $\int _{t_{i}}^{t_{f}}a\,v\,dt$ | (12) |
Here changing the order of a{\textstyle a} and v{\textstyle v} makes it easier to recognize the upcoming substitution.
| $\int _{t_{i}}^{t_{f}}v\,a\,dt$ | (13) |
In accordance with (10) we can substitute adt{\textstyle a\, dt} with dv{\textstyle dv}, with corresponding change of limits.
| $\int _{v_{i}}^{v_{f}}v\,dv$ | (14) |
So we have:
| $\int _{x_{i}}^{x_{f}}a\,dx=\int _{v_{i}}^{v_{f}}v\,dv$ | (15) |
Since the acceleration is constant, we can factor it out of the integration:
| $a\int _{x_{i}}^{x_{f}}dx=\int _{v_{i}}^{v_{f}}v\,dv$ | (16) |
Evaluating the integration:
| $a{\biggx{\bigg ]}_{x=x_{i}}^{x=x_{f}}=\left[{\frac {v^{2}}{2}}\right]_{v=v_{i}}^{v=v_{f}}$ | (17) |
| $a\left(x_{f}-x_{i}\right)={\frac {v_{f}^{2}}{2}}-{\frac {v_{i}^{2}}{2}}$ | (18) |
The factor x$_f$−x$_i${\textstyle x_{f}-x_{i}} is the displacement Δx{\textstyle \Delta x}:
| $2a\Delta x=v_{f}^{2}-v_{i}^{2}$ | (19) |
| $v_{f}^{2}=v_{i}^{2}+2a\Delta x$ | (20) |

## Application

Combining Torricelli's equation with F=ma{\textstyle F=ma} gives the work-energy theorem.
Torricelli's equation and the generalization to non-uniform acceleration have the same form:
Repeat of (16):
| $\int _{x_{i}}^{x_{f}}a\,dx=\int _{v_{i}}^{v_{f}}v\,dv$ | (21) |
Evaluating the right hand side:
| $\int _{x_{i}}^{x_{f}}a\,dx={\tfrac {1}{2}}v_{f}^{2}-{\tfrac {1}{2}}v_{i}^{2}$ | (22) |
To compare with Torricelli's equation: repeat of (7):
| $a(x_{f}-x_{i})={\tfrac {1}{2}}v_{f}^{2}-{\tfrac {1}{2}}v_{i}^{2}$ | (23) |
To derive the work-energy theorem: start with F=ma{\textstyle F=ma} and on both sides state the integral with respect to the position coordinate. If both sides are integrable then the resulting expression is valid:
| $\int _{x_{i}}^{x_{f}}F\,dx=\int _{x_{i}}^{x_{f}}ma\,dx$ | (24) |
Use (22) to process the right hand side:
| $\int _{x_{i}}^{x_{f}}F\,dx={\tfrac {1}{2}}mv_{f}^{2}-{\tfrac {1}{2}}mv_{i}^{2}$ | (25) |
The reason that the right hand sides of (22) and (23) are the same:
First consider the case with two consecutive stages of different uniform acceleration, first from s$_0${\textstyle s_{0}} to s$_1${\textstyle s_{1}}, and then from s$_1${\textstyle s_{1}} to s$_2${\textstyle s_{2}}.
Expressions for each of the two stages:
$a_{1}(s_{1}-s_{0})={\tfrac {1}{2}}v_{1}^{2}-{\tfrac {1}{2}}v_{0}^{2}$ $a_{2}(s_{2}-s_{1})={\tfrac {1}{2}}v_{2}^{2}-{\tfrac {1}{2}}v_{1}^{2}$
Since these expressions are for consecutive intervals they can be added; the result is a valid expression.
Upon addition the intermediate term 1/2 v$_1$$^2${\textstyle {\tfrac {1}{2}}v_{1}^{2}} drops out; only the outer terms 1/2 v$_2$$^2${\textstyle {\tfrac {1}{2}}v_{2}^{2}} and 1/2 v$_0$$^2${\textstyle {\tfrac {1}{2}}v_{0}^{2}} remain:
| $a_{1}(s_{1}-s_{0})+a_{2}(s_{2}-s_{1})={\tfrac {1}{2}}v_{2}^{2}-{\tfrac {1}{2}}v_{0}^{2}$ | (26) |
The above result generalizes: the total distance can be subdivided into any number of subdivisions; after adding everything together only the outer terms remain; all of the intermediate terms drop out.
The generalization of (26) to an arbitrary number of subdivisions of the total interval from s$_0${\textstyle s_{0}} to s$_n${\textstyle s_{n}} can be expressed as a summation:
| $\sum _{i=1}^{n}a_{i}(s_{i}-s_{i-1})={\tfrac {1}{2}}v_{n}^{2}-{\tfrac {1}{2}}v_{i-1}^{2}$ | (27) |