Finding a time-dependent vector

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To find a time-dependent vector ##\vec{A}(t)## that moves uniformly between two points ##\vec{r}_1## and ##\vec{r}_2## over a time interval from ##t_1## to ##t_2##, the average rate of change can be expressed as ##\frac{\vec{r}_2 - \vec{r}_1}{T}##. The correct formulation for ##\vec{A}(t)## is given by the equation ##\vec{A}(t) = \vec{r}_1 + \frac{\vec{r}_2 - \vec{r}_1}{t_2 - t_1} (t - t_1)##. This approach simplifies the problem to linear interpolation between the two points. The discussion confirms that using a single equation is sufficient for this linear motion problem.
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Homework Statement



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.

Homework Equations



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

The Attempt at a Solution


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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.
 
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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
 
Think of this as a linear interpolation problem.

A = r1 at t = t1

A = r2 at t = t1 + T

Chet
 
Chestermiller said:
Think of this as a linear interpolation problem.

A = r1 at t = t1

A = r2 at t = t1 + T

Chet

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?
 
MohammedRady97 said:
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?
Sure.
 
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