Finding a time-dependent vector

[PFuser1232]

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

[/B]
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.

 

[spl-083902]

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

 

[Chestermiller]

Think of this as a linear interpolation problem.

A = r₁ at t = t₁

A = r₂ at t = t₁ + T

Chet

 

[PFuser1232]

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?

 

[Chestermiller]

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.