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Finding a time-dependent vector

  1. Feb 27, 2015 #1
    1. The problem statement, all variables and given/known data

    Consider two points located at ##\vec{r}_1## and ##\vec{r}_2## separated by a distance ##r##. 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

    $$\vec{A}(t) = \vec{A}(t_1) + \int_{t_1}^t \frac{d\vec{A}}{dt'} dt'$$
    $$\vec{A}(t) = \vec{A}(t_2) + \int_{t_2}^t \frac{d\vec{A}}{dt'} dt'$$

    3. The attempt at a solution

    Since the time derivative of ##\vec{A}## is constant, I have pulled it out of the integral, eliminated it from both equations, and solved for ##\vec{A}(t)##. I ended up with an expression that looks too messy, though, so I don't know whether my approach (eliminating the derivative from both equations) is correct.
     
  2. jcsd
  3. Feb 27, 2015 #2
    How are you eliminating the constant from the equations?
    and why do you think you need 2 equations to solve it? I think you can do it with just 1
     
  4. Feb 27, 2015 #3
    Think of this as a linear interpolation problem.

    A = r1 at t = t1

    A = r2 at t = t1 + T

    Chet
     
  5. Feb 28, 2015 #4
    Since the derivative of ##\vec{A}(t)## is constant, we can equate it to the average rate of change ##\frac{\vec{r}_2 - \vec{r}_1}{T}##
    So we have:

    $$\vec{A}(t) = \vec{r}_1 + \frac{\vec{r}_2 - \vec{r}_1}{t_2 - t_1} (t - t_1)$$

    Is this correct?
     
  6. Feb 28, 2015 #5
    Sure.
     
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