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Derive vector that moves uniformly from one point to another

  1. Aug 28, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider two points located at ##\vec{r}_1## and ##\vec{r}_2##, and separated by distance ##r = |\vec{r}_1 - \vec{r}_2|##. Find a time-dependent vector ##\vec{A} (t)## from the origin that is at ##\vec{r}_1## at time ##t_1## and at ##\vec{r}_2## at time ##t_2 = t_1 + T##. Assume that ##\vec{A} (t)## moves uniformly along the straight line between the two points

    2. Relevant equations

    None

    3. The attempt at a solution
    I'm having trouble even getting started with this problem... I really need some help.
     
  2. jcsd
  3. Aug 28, 2016 #2
    I guess you just want to find the equation of a straight line crossing 3 points?

    Edit: as it is stated itlooks like it has to be a straight line just between the two points r1 and r2, so you have quite some freedom of choice between the origin and r1
     
  4. Aug 29, 2016 #3

    ehild

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    A drawing might help.
    upload_2016-8-29_7-24-18.png
    Write vector ##\vec r ## in terms of ##\vec r_1## and ##\vec r_2## and the ratio t/T first.
     
  5. Aug 29, 2016 #4
    I see how we can write ##\vec{r}## in terms of ##\vec{r}_1## and ##\vec{r}_2##, but I don't see where the ration t/T comes in...
     
  6. Aug 29, 2016 #5

    ehild

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    At t=0, ##\vec r =0##, and at t=T, ##\vec r =\vec r_2 -\vec r_1##. So what is ##\vec r(t)##?
     
  7. Aug 29, 2016 #6
    Isn't at ##t = t_1 + T##, ##\vec{r} = \vec{r}_2 - \vec{r}_1##? Sorry, I am just not getting this problem...
     
  8. Aug 29, 2016 #7

    ehild

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    Yes, but what is it at 0<t<T?
     
  9. Aug 29, 2016 #8
    The only thing that I can think of is ##\displaystyle \vec{r}(t) = \frac{t}{T} \vec{r}_2 - \frac{t}{T} \vec{r}_1##
     
  10. Aug 29, 2016 #9

    ehild

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    Correct! It can be written also as ## \vec{r}(t) = \frac{t}{T} (\vec{r}_2 - \vec{r}_1)##. And how do you write ##\vec A ##?
     
  11. Aug 29, 2016 #10
    Is it ##\vec{A} = \vec{r}_1 + \vec{r} (t) = \vec{r}_1 + \frac{t}{T} (\vec{r}_2 - \vec{r}_1)##?
     
  12. Aug 29, 2016 #11

    ehild

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    Yes!
     
  13. Aug 29, 2016 #12
    Woo! Thanks so much!
     
  14. Aug 29, 2016 #13

    ehild

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    You are welcome:smile:
     
  15. Aug 29, 2016 #14
    To make the problem easier did you let ##t_1 = 0##? So that ##t_2 = T##? I just want to make sure I understand the solution
     
  16. Aug 29, 2016 #15

    ehild

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    You are right, t is not the "real" time, but the elapsed time after t1. Better to write it Δt in the formula ##\vec{r}(t) = \frac{Δt}{T} (\vec{r}_2 - \vec{r}_1)##, and Δt=t-t1. You need Δt =t-t1 in the final formula for A(t).
     
  17. Aug 29, 2016 #16
    How can we substitute ##\delta t## for ## t## if that is not how it was when we derived the formula?
     
  18. Aug 29, 2016 #17

    ehild

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    It was assumed in the original formula that t was zero at r1. t was not the time shown by the clock, but the elapsed time, shown by a stopwatch. So it is better to denote it by Δt.
    The problem wants you to give A(t) where t is the time shown by the clock.
     
  19. Aug 29, 2016 #18
    If that is the case, then shouldn't we have started our derivation with the ratio ##\displaystyle \frac{t}{T + t_1}## instead of ##\displaystyle \frac{t}{T}##?
     
  20. Aug 29, 2016 #19

    ehild

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    We should have started it with (t-t1)/T. t-t1 time elapses from t1, You have to write t-t1 instead of t. Look at your final formula.
    ##\vec{A} = \vec{r}_1 + \vec{r} (t) = \vec{r}_1 + \frac{t}{T} (\vec{r}_2 - \vec{r}_1)## It is valid only when t1=0 (and t2=T).
    A(t1) must be r1 and A(t2) must be r2 (t2=t1+T). Is it true?

    Replace t with t-t1.

    ##\vec{A} = \vec{r}_1 + \vec{r} (t) = \vec{r}_1 + \frac{t-t_1}{T} (\vec{r}_2 - \vec{r}_1)##

    Is A(t1) = r1 and A(t2) = r2 (t2=t1+T) true now?
     
  21. Aug 30, 2016 #20
    Ah, I see. Makes sense now.
     
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