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Derive V2=u2+2as

  1. Sep 5, 2012 #1
    So i am having a slight problem deriving this equation.
    there is a simple way which is
    v=u+at -> t = v-u/a

    s = (v+u)t/2
    so s = (v+u)*(v-u)/2a
    2as = (v+u)(v-u)
    2as = v2-u2
    v2=u2+2as

    but i am having trouble understanding this way

    V2=(u+at)2
    V2=(u+at)(u+at)
    V2=u2+a2t2+2uat - i cant understand how you get from this to;
    V2=u2+2a(ut+0.5at2)

    i know that s=u+u+at/2*t i.e s=u+v/2*t

    so with V2=u2+a2t2+2uat you have a*a t*t u+u+a+a+t+t
    you can remove 2a as V2=u2+2a(ut+0.5at2)
    this leaves a*a t*t u+u+t+t
    then remove u+u+a+t+t for s=u+u+at/2*t
    then i am left with a and t*t and i dont know what to do with it.

    please help this is really grinding on me that i cant derive this, very annoying!
     
    Last edited: Sep 5, 2012
  2. jcsd
  3. Sep 5, 2012 #2
    Hello Johnahh, can you do simple calculus?

    Also please note that if you use the full reply box, not the quick one you can get superscript (X2) and subscript X2 from the symbols on the top line of the frame.
    Note also the useful symbols on the right.
     
  4. Sep 5, 2012 #3
    hi studiot,

    im afraid i am yet to look at calculus, i am just about to start AS levels on monday so i should be learning some this year.
    is calculus needed to solve this problem?
     
  5. Sep 5, 2012 #4

    Simon Bridge

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    No you don't need calculus to solve the problem:
    Starting with the expansion [tex]v^2=u^2+2uat+a^2t^2[/tex]... take out the 2a as a common factor: [tex]v^2=u^2+2a\frac{2uat+a^2t^2}{2a}[/tex] then all that is needed is to show that [tex]s=ut+\frac{at^2}{2}[/tex]... see?

    (Hint: if you click the "quote" button at the bottom of this post, you'll be able to see how I got the equations to come out all nice like that ... though the advanced editing tools will work too.)
     
  6. Sep 5, 2012 #5
    Pity about the calculus as it provides a simple and easy route to building up the full set of equations from the simple definitons of speed, time, distance and acceleration.

    As far as I'm aware the non calculus route requires the assumption of averaging to obtain an initial equation and then it is a matter of crunching through the algebra to obtain the rest.

    Maybe someone else has a better idea, but I hope the other informatiuon was useful.

    go well
     
  7. Sep 5, 2012 #6

    Simon Bridge

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    You can go the calculus method - but simplified by exploiting the geometry of kinematics.
    Students need only to know about the analytic equation of a line, and how to find the areas of simple shapes.

    Once a v-t graph is drawn, the rules are:
    acceleration is the slope and displacement is the area.

    That's all you need to know.

    If the graph depict constant acceleration from u to v in time T then:
    equation of the line is: v(t) = at+u
    slope of the graph: a = (v-u)/T
    area:
    s = (u+v)T/2 (treated as a trapezium)
    s = (v-u)T/2 + uT (treated as a triangle + rectangle: it's the same though)

    Most situations are simpler than this.
    Constant deceleration from velocity v to rest in time T give a triangle so:
    equation of the line is: v(t) = v-at
    a=v/T and s=vT/2 read right off the graph.

    You get the rest of the standard set of kinematic equations by eliminating each variable in turn as in simultanious equations. Five variables is the usual full set: gets you five equations. You can have six if the initial time is not t=0.

    Typically, students are then asked to memorize the standard five equations... and they forget where the equations come from in favor of a "list what you know - pick pre-memorized equation" style of problem solving.

    Just using the rules of calculus really means that students have already done the "find area/find slope" problems that motivate it. Otherwise calculus is just another magic wand.
     
  8. Sep 5, 2012 #7
    I don't know what they do today.

    Coordinate geometry was not part of my early 1960s 'O' level.
    This was introduced at the same time as elementary calculus.

    So if y = mx + c, as an equation, is now part of pre A level maths perhaps some things have improved.
     
  9. Sep 5, 2012 #8

    Simon Bridge

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    Year 10 maths, NZ students are expected to be able to write and use the analytic and standard forms of the equations for a line, and convert between them. They can solve linear simultaneous equations, manipulate and find the roots of quadratic equations and generalize qualitatively to polynomials of any order. They usually just lack confidence.

    We've sacked a lot of geometry though... there used to be a lot of stuff about geometric objects drawn inside circles for example.

    What I related above was standard fare in 5th form physics classes in the 80's. The area/slope form of the definitions set students up for calculus quite nicely so there's this extra payoff for emphasizing the graphical approaches.

    Basically, priorities have shifted around a lot ... looks similar for the UK (I am applying for positions there ...) I don't think actual "coordinate geometry" is done at all - not formally. Graphs are part of number theory in math.

    Calculus used to be taught by rule, and some people still do that, but now there is more emphasis on geometric methods to motivate the concepts.

    One of the massive improvements is that all the bog-standard courses are now online for the students to discover with a bit of prompting ... leaving me free to use the less conventional approaches in the classroom. The wider mix of methods gets more students learning ;)
     
  10. Sep 5, 2012 #9
    Just has a look at GCSE 'standard' and 'higher' maths.

    They certainly do more graphical work than I did.

    But I couldn't find any mention of the formal equations of motion under constant aceleration.

    I don't have any comparable physics texts.

    I wouldn't expect too much freedom if you come to the UK.
    Teaching is more regimented than ever and extremely target lead.
     
  11. Sep 5, 2012 #10
    thanks for the replies

    So i shouldnt be asked to derive these equations? i believe i know the graphs as they are fairly straightforward and you can work them out by knowing the formula.
    i went to school 8 years ago and cant remember much maths from there but i retook GCSE maths last year, it was self taught and i didnt cover polynominals, is this something that i will need for A level maths and physics?

    Thanks
     
  12. Sep 5, 2012 #11
    I can't imagine anyone asking for a derivation of the equations at higher GCSE.

    At A and AS level you can never be quite sure as they are always mucking around with the syllabus.

    However the syllabus is online and for instance the AQA one lists all the equations and states whether you should be able to recall them from memory, or just use themif given and which ones you need to be able to derive, explain or prove.

    You should study the syllabus as well as the subject.

    Finally, and most important,

    Read the question twice and answer the question they have asked, not what you think they have asked.

    As the old carpenter put it ' Measure twice , cut once '

    It is really surprising how many candidates fall down on this.

    You will find lots of really friendly help here for those wanting to make progress. The helpers are really patient.

    Good luck with the next stage of your studies.
     
  13. Sep 5, 2012 #12
    Thanks studio, it seems strange to me people being so nice on the internet lol!
     
  14. Sep 5, 2012 #13
    Yeah must speak to my analyst about that.

    :rofl:
     
  15. Oct 16, 2012 #14
    1st step:
    velocity=distance/time traveled
    v=s/t

    2nd step:
    acceleration=rate of Δv/t, u=initial velocity
    a=Δv/t
    a=(v-u)/t
    at=v-u
    at+u=v
    v=u+at⇔eq'n 1

    3rd step:
    average velocity =(v+u)/2
    apply v=s/t
    s=vt
    subtitute the average velocity
    s=[(v+u)/2]t
    subtitute eq'n 1
    s=[(at+u+u)/2]t
    s=[(at+2u)/2]t
    substitute
    s=[at/2+u]t
    s=at²/2+tu⇔eq'n 2

    4th step:
    eq'n 1;
    (v=u+at) = (v=u+at)²
    v²=(u+at)²⇔expand
    v²=u²+2atu+a²t²⇔factor 2a
    v²=u²+2a(tu+at²/2)⇔eq'n 2
    v²=u²+2as
     
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