# About the discovery of v^2 = u^2+2as...

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v2 = u2+2as

For the above equation, it was found by eliminating t from one of the below simultaneous equations, and substitute it into another euqation.....

1) v = at + u

2) s = ((v+u)/2)t

where v = finally velocity

a = acceleration

t = time

u = initial velocity

s = displacement

but who started this kind of equation? What kind of problem did this equation's inventor face upon which he forms this equation?

I know it might be, because this equation contains much many information like acceleration, displacement, final/initial velocity, etc..., so it could much handy to use for solving problems, but however, if I see the equation alone, it makes no sense, since it is just another way of expressing v2 = (at + u)2........

I am just curious about it, because in one way or another, I can solve any physics problem till now without using that equation although my teacher told me it is much more handy since a long time ago...

Any body?

Albert

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its just a consequence of a bit of algebraic manipulation. nothing special there.

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I know bloodhound, but, as I have come across physics for several years, I have never face a problem to manipulate those algebraic symbols in this way... the subject v2 does not make sense to physics, why would I find something like the square of final velocity?

so, there must be something more in its background, since it is not just algebraic manipulation that you just come accross naturally, but it's an equation that many textbooks taught!

Any body?

Albert

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I know bloodhound' date=' but, as I have come across physics for several years, I have never face a problem to manipulate those algebraic symbols in this way... the subject v[sup']2[/sup] does not make sense to physics, why would I find something like the square of final velocity?

so, there must be something more in its background, since it is not just algebraic manipulation that you just come accross naturally, but it's an equation that many textbooks taught!

Any body?

Albert

Why can't the square of the final velocity exist?

As a corrolary, what happens when you isolate for final velocity?

If you like these sorts of things, then look up kinematics. There's much more where that came from. More specifically, dynamics and such.

It gets even more interesting when you use seperation, integration and such.

Enjoy!

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1) and 2) only work with problems involving constant acceleration.

1) are a direct result of solving the second order differential equation

x''=a , or (v'=a) where a is constant.

so u get v = x' = at + c (for some constant c = v(t0)=u = initial velocity)

and then x = (at^2)/2 + ut + d (for some constant d = x(t0) . usually taken to be 0 .i.e the origin for convinience)

with a bit of manipulation

$x=ut+\frac{at^2}{2}$

$=ut+\frac{(v-u)t^2}{2t}$

$=\frac{2ut+vt-ut}{2}$

$=\frac{(v+u)t}{2}$

What i was doing there was basically finding the area under the velocity time graph. As you must know the area is equal to the displacement.... not the same as distance travelled.

But i guess you will have to wait a little while , to understand these results properly

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bloodhound, I know much about how you manipulate algebra and such..... but why did the oringinal person (who found out this equation) take equation 1) and 2) and eliminate t from from? What is he/she trying to find?

It is like when you find the area under the velocity time graph, and the area determines the distance

but how about finding the area under distance time graph? It makes no sense, since distance times time doesnot make anything.....

Just like the equation, what is he/she trying to find?

Albert

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There is more than one way to arrive at the result.

W= KE2 - KE1 for a system without potential energy.

For a constant force (acceleration), W=Fs=mas=1/2mv22 - 1/2mv12

m drops out. Multiply through by 2.

No time terms at all to have worried about.

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swansont, can you tell me what are the algebraic symbols:

W, K, E2, E1, s, m, v2, v1?

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Work

Kinetic Energy @ 2 & @1

Displacement

Mass

Velocity @ 2 & @ 1

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Swansont, I know there is more than one way/solution to solve any particular physics problem, and I know you are giving another example for this. But, what I am asking is, before v2 = u2 + 2as has been used widely, what did the original maker of this equation come across or, what is he/she thinking to approach the solution of problem?

Albert

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He/she was probably trying to find out the speed of an object, when he/she knew only the initial speed, the disctance travelled and the acceleration. In that instance, one would naturally use the expressions for distance and velocity in terms of time, which were derived above, and remove the dependence on time. One could in principle work out the time from the distance equation and stick it into the speed equation, but if you have multiple cases to do, this is very inefficient. It is better to substiture an equation for time in terms of distance into the equation for speed.

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Swansont' date=' I know there is more than one way/solution to solve any particular physics problem, and I know you are giving another example for this. But, what I am asking is, before v[sup']2[/sup] = u2 + 2as has been used widely, what did the original maker of this equation come across or, what is he/she thinking to approach the solution of problem?

Albert

The derivation I gave is from the idea that work changes kinetic energy. So I imagine that the idea came from work and energy.

I think that the equation became "widely used" almost immediately. It's not like physicists would have labored for years using just one or two kinematics equations, and had to wait for the others to be "discovered."

All of the kinematics equations stem from the definitions of the terms and some simple calculus and algebra. They were probably all originally derived at about the same time.

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He/she was probably trying to find out the speed of an object, when he/she knew only the initial speed, the disctance travelled and the acceleration. In that instance, one would naturally use the expressions for distance and velocity in terms of time, which were derived above, and remove the dependence on time. One could in principle work out the time from the distance equation and stick it into the speed equation, but if you have multiple cases to do, this is very inefficient. It is better to substiture an equation for time in terms of distance into the equation for speed.

thx Severian, and Swansont

To me, I never remember those formula by heart, only the physics principles, because we could solve problems in many ways......

The reason I asked is this is a formula given importantly from my textbook, and people in my class, just remember it by heart to deal with all kinds of physics tests by substituing the algebraic symbols by numbers....

My textbook is from Oxford Uni Press. I think all the physics "teaching" book should not include equation like this, because people just remember it by heart without knowing any physics principles behind it

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Its funny that you show say that, because I gave a tutorial the other day where someone used exactly that formula. I took the opposite attitude to your books - I made him derive it. It is certainly not an equation that I remember, but it is so trivial to derive that you don't need to.

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Its funny that you show say that, because I gave a tutorial the other day where someone used exactly that formula. I took the opposite attitude to your books - I made him derive it. It is certainly not an equation that I remember, but it is so trivial to derive that you don't need to.

That seems to be the dividing line between the serious college science students and the ones who are taking the class because they have to. I've done the same thing, and just said "If you can't remember it, the derivation is trivial" and watch 90% of the students blanch in horror at the thought.

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Galileo was the first person to solve those equations. Look up his "Discourses Concerning Two New Sciences", written in 1636. First he finds the average speed ((v+u)/2) then finds v^2 = 2as (supposing u=0, starting from rest). It's all done geometrically; this was before they invented modern algebraic methods. You may find the geometric reasoning helps make the physics behind the equations more intuitive. One reason for eliminating t from the equation was that his method of measuring time was very crude (water clock) so he could get more accurate answers using distance (rulers are much more accurate and convenient than water clocks). The main reason for these investigations was simple curiosity, but more specifically, they wanted to know how hard weights would hit, when dropped from a height, to help design weapons like catapults. I hope this answers your question.

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it all started with definition of velocity as: v = s / t (change of the position (displacemet) in certin time)

s represent displacement and

t change in time

then the acceleration as: a = v / t (change of the velocity in certin time) v represent change in velocity and t change in time

after some time "they" came to function (displacement) of s at some time t:

s(t) =s(0) + v*t + (1/2)*(at^2)

after that comes mathemacial manipulations....

somehow i belive it was well known before gallileo

albertlee about your question...niether there is such thing as regular velocity(v) in nature, its just how we call displacement in some time ...its just mathematics... but!! theres no such thing as mathematics its just human way of thinking .. it all started with counting sheeps and look where we have finished... im trying to say, dont try to picture everything because picturing is nothing but masking reality , hope you got it

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