Torricelli’s equation

• Demonstration of Torricelli’s equation

Starting from the time function of speed, we have:

v = v ₀ + at

Squaring both sides of the equality:

v ² = (v ₀ + at) ²

Now we will develop the remarkable product (v ₀ + at) ² , so that:

v ² = v ₀ ² + 2.v ₀ .at + a ² t ²

For the last two terms of the function, we will isolate the factor 2a:

v ² = v ₀ ² + 2a (v ₀ .t + ½ at ² )
Equation A

With equation A in hand, we will proceed to the clockwise function of the position in uniformly varied motion:

S = S ₀ + v ₀ .t + ½ at ²

S – S ₀ = v ₀ .t + ½ at ²

As S – S ₀ = ΔS, we have:

ΔS = v ₀ .t + ½ at ²
Equation B

Finally, we will substitute equation B into equation A:

v ² = v ₀ ² + 2a (v ₀ .t + ½ at ² )

v ² = v ₀ ² + 2aΔS
Torricelli’s equation

Note that there is no time dependence, since the terms of the equation are the final velocity of the mobile (v), initial speed of the mobile (v ₀ ), acceleration (a) and the space covered (ΔS).

• Torricelli’s name “Equation” is not adequate!

The term equation is not correct because what we have above is a correspondence between elements, and not an equality satisfied by just a few values. Therefore, we should call the above term a Torricelli function , but traditionally, even if it is not the proper form, this expression is recognized as a Torricelli equation.

• Numerical example

A car starting from rest has a constant acceleration equal to 5 m/s ² . Determine the distance covered by the mobile when its speed is equal to 72 km/h.

Resolution:

Taking the data from the question, we have:

a = 5 m/s ²

v = 72 km/h ÷ 3.6 = 20 m/s

v ₀ = 0 (Mobile starting from rest)

ΔS = ?

Substituting the above values ​​into Torricelli’s equation , we have:

v ² = v ₀ ² + 2aΔS

20 ² = 0 + 2.5.ΔS

400 = 10.ΔS

ΔS = 40 m