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:[ 1]
v
f
2
=
v
i
2
+
2
a
Δ
x
{\displaystyle v_{f}^{2}=v_{i}^{2}+2a\Delta x\,}
where
v
f
{\displaystyle v_{f}}
velocity along the x axis on which the acceleration is constant.
v
i
{\displaystyle v_{i}}
a
{\displaystyle a}
acceleration along the x axis, which is given as a constant.
Δ
x
{\displaystyle \Delta x\,}
displacement .In this and all subsequent equations in this article, the subscript
x
{\displaystyle x}
v
f
x
{\displaystyle {v_{f}}_{x}}
This equation is valid along any axis on which the acceleration is constant.
Without differentials and integration
edit
Begin with the following relations for the case of uniform acceleration:
x
f
−
x
i
=
v
i
t
+
1
2
a
t
2
{\displaystyle x_{f}-x_{i}=v_{i}t+{\tfrac {1}{2}}at^{2}}
(1 )
v
f
−
v
i
=
a
t
{\displaystyle v_{f}-v_{i}=at}
(2 )
Take (1), and multiply both sides with acceleration
a
{\textstyle a}
a
(
x
f
−
x
i
)
=
a
v
i
t
+
1
2
a
2
t
2
{\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:
a
(
x
f
−
x
i
)
=
v
i
(
a
t
)
+
1
2
(
a
t
)
2
{\displaystyle a(x_{f}-x_{i})=v_{i}(at)+{\tfrac {1}{2}}(at)^{2}}
(4 )
Use (2) to substitute the product
a
t
{\textstyle at}
a
(
x
f
−
x
i
)
=
v
i
(
v
f
−
v
i
)
+
1
2
(
v
f
−
v
i
)
2
{\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:
a
(
x
f
−
x
i
)
=
v
i
v
f
−
v
i
2
+
1
2
v
f
2
−
v
i
v
f
+
1
2
v
i
2
{\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
v
i
v
f
{\textstyle v_{i}v_{f}}
a
(
x
f
−
x
i
)
=
1
2
v
f
2
−
1
2
v
i
2
{\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:
v
f
2
=
v
i
2
+
2
a
Δ
x
{\displaystyle v_{f}^{2}=v_{i}^{2}+2a\Delta x}
(8 )
Using differentials and integration
edit
Begin with the definition of acceleration as the derivative of the velocity:
a
=
d
v
d
t
{\displaystyle a={\frac {dv}{dt}}}
Now, we multiply both sides by the velocity
v
{\textstyle v}
v
⋅
a
=
v
⋅
d
v
d
t
{\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:
d
x
d
t
⋅
a
=
v
⋅
d
v
d
t
{\displaystyle {\frac {dx}{dt}}\cdot a=v\cdot {\frac {dv}{dt}}}
Multiplying both sides by
d
t
{\textstyle dt}
d
x
⋅
a
=
v
⋅
d
v
{\displaystyle dx\cdot a=v\cdot dv}
Rearranging the terms in a more traditional manner:
a
d
x
=
v
d
v
{\displaystyle a\,dx=v\,dv}
Integrating both sides from the initial instant with position
x
i
{\textstyle x_{i}}
v
i
{\textstyle v_{i}}
x
f
{\textstyle x_{f}}
v
f
{\textstyle v_{f}}
∫
x
i
x
f
a
d
x
=
∫
v
i
v
f
v
d
v
{\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:
a
∫
x
i
x
f
d
x
=
∫
v
i
v
f
v
d
v
{\displaystyle {a}\int _{x_{i}}^{x_{f}}dx=\int _{v_{i}}^{v_{f}}v\,dv}
Solving the integration:
a
[
x
]
x
=
x
i
x
=
x
f
=
[
v
2
2
]
v
=
v
i
v
=
v
f
{\displaystyle {a}{\bigg [}x{\bigg ]}_{x=x_{i}}^{x=x_{f}}=\left[{\frac {v^{2}}{2}}\right]_{v=v_{i}}^{v=v_{f}}}
a
(
x
f
−
x
i
)
=
v
f
2
2
−
v
i
2
2
{\displaystyle {a}\left(x_{f}-x_{i}\right)={\frac {v_{f}^{2}}{2}}-{\frac {v_{i}^{2}}{2}}}
The factor
x
f
−
x
i
{\textstyle x_{f}-x_{i}}
Δ
x
{\textstyle \Delta x}
a
Δ
x
=
1
2
(
v
f
2
−
v
i
2
)
{\displaystyle a\Delta x={\frac {1}{2}}\left(v_{f}^{2}-v_{i}^{2}\right)}
2
a
Δ
x
=
v
f
2
−
v
i
2
{\displaystyle 2a\Delta x=v_{f}^{2}-v_{i}^{2}}
v
f
2
=
v
i
2
+
2
a
Δ
x
{\displaystyle v_{f}^{2}=v_{i}^{2}+2a\Delta x}
From the work-energy theorem
edit
The work-energy theorem states that
Δ
E
K
=
W
{\displaystyle \Delta E_{K}=W}
m
2
(
v
f
2
−
v
i
2
)
=
F
Δ
x
{\displaystyle {\frac {m}{2}}\left(v_{f}^{2}-v_{i}^{2}\right)=F\Delta x}
which, from Newton's second law of motion, becomes
m
2
(
v
f
2
−
v
i
2
)
=
m
a
Δ
x
{\displaystyle {\frac {m}{2}}\left(v_{f}^{2}-v_{i}^{2}\right)=ma\Delta x}
v
f
2
−
v
i
2
=
2
a
Δ
x
{\displaystyle v_{f}^{2}-v_{i}^{2}=2a\Delta x}
v
f
2
=
v
i
2
+
2
a
Δ
x
{\displaystyle v_{f}^{2}=v_{i}^{2}+2a\Delta x}
![{\displaystyle {a}{\bigg [}x{\bigg ]}_{x=x_{i}}^{x=x_{f}}=\left[{\frac {v^{2}}{2}}\right]_{v=v_{i}}^{v=v_{f}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/661ee4c57c8dbd012bfef71a6602bcc931f30716)
