# Torricelli's Equation

I'm Wishing To Make A Few Notes About Some Things About Physics , I So Thought in Equation Torricelli, Is An Equation that All Beginner Will See !

The Equation Is An Equation Torricelli kinematic was discovered by Evangelista Torricelli, one whose function is the ability to calculate the final speed of a body in uniform rectilinear motion varied (accelerated motion) without having to know the time interval in which this remained movimento. the great advantage of this equation is that the time factor does not existe.

The equation has the form:

Where $v_f$ and $v_o$ Represent the initial and final velocities of the body, respectively, $\Delta s$ represents the distance traveled and $a$ is the acceleration.

This equation can be deduced from the following equations:

Isolating $t$ in the second equation :

And replacing it with the first, We have to :

# Examples

I ) A motorcycle has an initial speed of $20 m/s$ and acquires a constant acceleration equal to $2m / s^2$ . Calculate your speed in $m / s$ to cover $100 m$

Solution :

$v^2 = v_o^2 + 2a \Delta s$

$v^2= 20^2+2.2.100$

$v^2 = 400+400$

$v^2 = 800$

$v = \sqrt{800}$

$\boxed{v = 28,28 m/s }$

II ) A particle initially at rest shall be accelerated continuously at a rate of $3.0 m/s^2$ in the direction of the trajectory. Having driven $24 m$ , its speed is :

Solution :

Removing the text data, We have :

$v_o = 0$

$a = 3 m/s^2$

$\Delta s = 24 m$

As we do not know the time of movement applies to equation Torricelli :

$v^2 = v_o^2 + 2a \Delta s$

$v^2 = 0 \cdot 2 + 2 \cdot 3 \cdot 24$

$v^2 = 144$

$\boxed{v = 12 m/s}$

III ) A driver is traveling by car on a highway at a constant speed of $90 km / h$ , when he notices a horse in front of him and decides to stop printing a constant deceleration of $18 km / h$ per second. Calculate the minimum stopping distance in meters.

Solution :

First we transform the initial speed of $km / h$ to $m / s$, then we will analyze the acceleration as it slows down when the driver indicates that he steps on the brakes, ie, have a negative acceleration that is in $km / h$ and needs to be transformed in $m / s$ by $s$ will be the same as $m / s^2$.

$v_o = 90km / h = 25 m / s$

$a = 18 km / h = - 5m / s^2$

$v = 0$

$v^2 = v_o^2 + 2a \Delta s$

$0^2 = 25^2 +2.(-5). \Delta s$

$0 = 625 - 10 \Delta s$

$-625 = -10 \Delta s$

$- \Delta s = \frac{625}{-10}$ (Isolating the variable)

$- \Delta s = - 62.5$

$\boxed{\Delta s = 62,5m}$

**If You Know Any Problem About This , Comment Here For Others Make This Note Be Important !

*Thanks ! :D

Note by Gabriel Merces
7 years, 4 months ago

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If a mass m moves a distance s with uniform acceleration a, looses its velocity from Vo to V, equating work and kinetic energy, we get WD = loss in KE..>$\\ m * a * s=1/2mVo^2-1/2*m*V^2 \implies$ your equation directly. It is from the energy equation.

- 7 years ago

this equation can also be derived from calculus

- 7 years, 4 months ago

Hey Gabriel, can you please make a note on electricity and magnetism related physics?

- 7 years, 4 months ago

@Kartik Sharma Do You Want What Topic Specific ?

Have Much Topics , I'll Try Make A Set About It !

- 7 years, 4 months ago

I want a set on both these topics explaining almost everything from basics to intermediate(advanced is your choice, actually I am not so advanced, that's why)

Again, thank you in anticipation!

- 7 years, 4 months ago

@Gabriel Merces Hi Gabriel can you do a note on rotational mechanics please

- 7 years, 4 months ago

There is also an equation for finding displacement in a certain second.

$s_n=u+\frac{1}{2}a(2n-1)$, where $s_n$ is the displacement in the $n^{th}$ second, $u$ is the initial velocity at $t=0$, and $a$ is the constant acceleration.

- 7 years, 4 months ago

can you please derive the equation? it works,, but with respect to dimensions i found it a little weird.

- 7 years, 4 months ago

The derivation is very simple by calculus.

- 7 years, 4 months ago

Sure. Here is the derivation.

Say a particle is at point $O$ at $t=0$. The particle moves in a straight line under constant acceleration to a point $P$ in time $t=n-1$, and then to a point $Q$ in time $t=n$. Thus, displacement in the $n^{th}$ second is just $OQ-OP$.

Using the formula for displacement, $s=ut+\frac12(a)(t^2)$

$OQ=un+\frac12(a)(n^2)$

$OP=u(n-1)+\frac12(a)((n-1)^2)$

Thus, $OP-OQ=\boxed{s_n= u + \frac12(a)(2n-1)}$

I realise that the dimensions do not add up, but you have to consider $t=1$ and the formula is derived, and is true for all forms of straight line uniform motion.

- 7 years, 4 months ago

Dimensions do add up. It is not simply "u" but .." u * 1sec."

- 7 years ago

using calculus is much more easy

- 7 years, 4 months ago

good one, and it`s very convincing , thanks!!!

- 7 years, 4 months ago

@Gabriel Merces Hi Gabriel can you do a note on waves too please.

- 7 years, 4 months ago

Here Have A Little About Waves : Waves

- 7 years, 4 months ago