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Various position functions from dx/dt

  1. Oct 27, 2013 #1
    Hi,I am just starting to solve calculus problems but,how is it possible,that from v(t) we can get so many differently position function.I mean,if we have v(t)=dx/dt we should always get just v=at+v0,and from this to get s(t)=1/2at^2+v0t+v0.But I have found many motion problems in which position functions have different forms,like for example:s(t)=t^2+4t^2 or s(t)=-t-t^2.And many others like if acceleration is not constant and so on.

    So my question.If we solve some problem,we always start with v=dx/dt and therefore,we should get always just one,and still the same form of position function,should not we?

    thanks for answers
  2. jcsd
  3. Oct 27, 2013 #2


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    Why should we always get that?
  4. Oct 27, 2013 #3
    I will ask otherwise.How do I get v(t) by derivation of s(t) if I do not know function s(t)?If I have some motion problem where we have function s(t) like known stuff,from what is this function determined?I mean,how I can know s(t) if I do not know v(t).And how I could know v(t) if I do not know s(t)?
    If we are trying to solve some motion problem and we have just known variables and parameters,but we have no function,how we can solve the problem?I mean,it is determined,this function,which is neccessery only experimentally?
    Last edited: Oct 27, 2013
  5. Oct 28, 2013 #4
    OK,well when we get other form?Not at+v0?If we still do dx/dy?I know,we must know x.What is hidden in ,,x".And that is my question.How we know,or where or from we can find out what is ,,x"?
  6. Oct 28, 2013 #5


    Staff: Mentor

    If you don't know the formula for s(t), you can still get v(t) (or at least an approximation of it) by using numerical techniques of differentiation. For example, if you have these data points - (0, 2), (1, 5), (2, 10) - with the first coordinate being the time t and the second coordinate being position s, you can choose a polynomial or other function to fit the data to, and then differentiate. Alternatively, you can find the slope between successive pairs of points to get approximate values for v(t).
    Certain functions are reasonable models for projectile motion. The simpler functions such as s(t) = 16t2 give OK results and are easy to work with, but only take into account the gravitational force. For better results that take air resistance into account, the functions are more complicated, but give results that are more true to life. In general the equations of motion are arrived at by experimentation over hundreds of years (going back to Galileo).
  7. Oct 29, 2013 #6
    Thank you very much for answers. :)
  8. Nov 6, 2013 #7
    Hi,I just want to ensure.If I have for example velocity function:v(t)=2t-t^2,then t^2 expression is only time over 2 or some constant is next to t^2,but this constant=1?Or how is it possible,that we have only t^2,it seems,like we could have function that for example is s(t)=t^2 and then meter=second??? Dimensionally seems to be incorrect....
  9. Nov 6, 2013 #8


    Staff: Mentor

    The constants wouldn't be dimensionless. In your formula, v(t) = 2t - t2, the coefficient of t would have to be <distance>/<time2> and the coefficient of t2 would have to be <distance>/<time3>. That way the right side would be in units of <distance>/<time> to match the left side.

    On the other hand, maybe the problem isn't concerned with units, in which case all they're interested in is the numeric value of the velocity.
  10. Nov 6, 2013 #9
    Yes,but what is the value of coefficient???For example the value of the coeffcient of t2 is 1?
    Last edited: Nov 6, 2013
  11. Nov 6, 2013 #10


    Staff: Mentor

    Yes. If the coefficient of some variable is 1, we usually don't bother to write it. For example x and 1x mean the same thing.
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