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How Calculus works?

  1. Sep 23, 2013 #1
    Dear Sirs,

    I am wondering what is the difference between v=s/t & dv = ds/dt where v-velocity,s-displacement,t-time.

    Consider Gallelio's equations->
    v=u+at--(equation-1) &
    s=ut+1/2at^2--(equation-2)
    where u-initial velocity & a-acceleration

    My doubt is on following point;

    If we do,v = s/t in equation-2,we get->v=u+1/2at not v=u+at
    but if we do,dv=ds/dt in equation-2,we get=>v=u+at ,exactly the equation we want.

    why only differentiation give the result,not just mere divison?????
     
  2. jcsd
  3. Sep 23, 2013 #2

    arildno

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    v=s/t is the AVERAGE velocity over the total time INTERVAL "t".
    Thus, it does NOT give the velocity at the INSTANT "t"
     
  4. Sep 23, 2013 #3
     
    Last edited: Sep 23, 2013
  5. Sep 23, 2013 #4
    dv/dt is at an instantaneous time... Not division.
     
  6. Sep 24, 2013 #5

    Mark44

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    It's the acceleration at an arbitrary time.
     
  7. Sep 24, 2013 #6

    UltrafastPED

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    Galileo's equations work only for constant accelerations; the theory was worked out by the "Oxford Calculators":
    http://en.wikipedia.org/wiki/Oxford_Calculators

    Galilleo's experiments showed that gravity was constant (at the surface of the earth), and was thus able to apply these equations.

    Calculus allows you to work with arbitrary accelerations; instead of working with algebraic averages it makes use of limits. Between the time of Galileo and Newton, Rene Descartes invented analytic geometry ... this is the tool required to move from the geometric analysis of Galileo to the calculus of Newton and Leibniz.
     
  8. Sep 25, 2013 #7
    equation [itex]v=u+at[/itex] is merely rearranged from the average acceleration formula [itex]\displaystyle\frac{v-u}{t}=a[/itex].

    In calculus we get the instantaneous versions of these, [itex]\displaystyle\frac{dv}{dt} = a[/itex], and [itex]v(t)=\displaystyle\int a(t) dt[/itex]
     
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