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How to find r(t) by given:v(t) r(2) ?

  1. Nov 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the position vector at any given moment ( r(t) )

    v(t)=(6t^2+4t)i+(3t-2)j+(12t^3-t^3)k m/s
    r(t=2)= (3,-0.5,-2)


    2. Relevant equations

    r(t) = r0+ integral(v(t))

    3. The attempt at a solution

    I tried to integrate v(t) and get r(t) but I don't know how to calculate r0.
    what am I supposed to do with t=2 I was given?
    if I do integral of v(t) and calculate the it by t=2 - t=0, what would be the meaning of this?
    t=0 doesn;t have to give r(0)=(0,0,0)

    We learned to get the vectors with known starting points so I am puzzled about finding them myself and unfortunately all the questions deal with finding them -.-
    I seek to learn the meaning of the whole picture.
    then I can solve the rest.
     
  2. jcsd
  3. Nov 10, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Lenjaku! Welcome to PF! :smile:

    In the formula r(t) = r(0) + ∫ (v(t)) dt,

    the limits of integration have to be 0 and t …

    r(t) = r(0) + ∫0t (v(t)) dt​

    But you can substitute any value a of t …

    r(t) = r(a) + ∫at (v(t)) dt :wink:
     
  4. Nov 10, 2012 #3
    Unfortunately I am still trying to figure out things we only had 1 lecture about this.
    If I take the integral of t=2 and reduce t=0 it woudlnt give me anything (I think).

    Just to see what happens I integrated v(t) and placed t=2. the numbers are nothing near r(2).

    it turned out to be:
    ∫v(t)=24i+(-1)j+12k
    r(2)=3i-0.5j-2k

    What would be the meaning of this difference?

    Both is the same time,but apparently the starting point affects the result....did I calculate ∫v(2) as r(0,0,0) starting point? if so why it turned to be so if at all?
     
  5. Nov 10, 2012 #4

    tiny-tim

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    Hi Lenjaku! :smile:
    not following you :confused:

    v(t)=(6t2+4t)i+(3t-2)j+(12t3-t3)k m/s

    (hmm … that last bracket looks wrong :redface:)

    can you show all the steps for the way you calculated ∫ v(t) dt?
     
  6. Nov 10, 2012 #5
    I integrated v(t) as:
    (6t3)/3 + (4t2)/2 , (3t2)/2 - 2t,(12t4)/4 - (9t4)/4

    And then placed t=2.

    r(2) was given.

    I simply wanted to see what it turns to be to understand what is going on there...

    I still don't understand how to find r(t) at any given time when I am only given r(t=2) and no r(t=0) how do I find r(t=0) (do I need it?, if not why)?

    I think I need it to put it in the equation:

    r(t=0) + ∫v(t)
    I have v(t) I can get its integral but what do I do with the given r(t=2)?
    Why was I given this for?
     
  7. Nov 10, 2012 #6

    tiny-tim

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    v(t2) - v(t1) = [(6t3)/3 + (4t2)/2 , (3t2)/2 - 2t,(12t4)/4 - (9t4)/4]t1t2

    you subtract the initial value from the final value
     
  8. Nov 10, 2012 #7
    I am sorry I don;t get you.
    The only t I was given is t=2.
    And I was give the components of r(t=2) so I assume the answer should do something with it.

    I don;t know what r(t=0) is.

    What do I do with the r(t=2) I was given?

    Doing as u suggested was my initiate approach with no any further thinking it seems logical but it leaves me with too many questions.

    I am thinking of x=x0+v0t
    taking v as constant. So v0=v.
     
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