# Derive vector that moves uniformly from one point to another

1. Aug 28, 2016

### Mr Davis 97

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. Aug 28, 2016

### mastrofoffi

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

3. Aug 29, 2016

### ehild

A drawing might help.

Write vector $\vec r$ in terms of $\vec r_1$ and $\vec r_2$ and the ratio t/T first.

4. Aug 29, 2016

### Mr Davis 97

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...

5. Aug 29, 2016

### ehild

At t=0, $\vec r =0$, and at t=T, $\vec r =\vec r_2 -\vec r_1$. So what is $\vec r(t)$?

6. Aug 29, 2016

### Mr Davis 97

Isn't at $t = t_1 + T$, $\vec{r} = \vec{r}_2 - \vec{r}_1$? Sorry, I am just not getting this problem...

7. Aug 29, 2016

### ehild

Yes, but what is it at 0<t<T?

8. Aug 29, 2016

### Mr Davis 97

The only thing that I can think of is $\displaystyle \vec{r}(t) = \frac{t}{T} \vec{r}_2 - \frac{t}{T} \vec{r}_1$

9. Aug 29, 2016

### ehild

Correct! It can be written also as $\vec{r}(t) = \frac{t}{T} (\vec{r}_2 - \vec{r}_1)$. And how do you write $\vec A$?

10. Aug 29, 2016

### Mr Davis 97

Is it $\vec{A} = \vec{r}_1 + \vec{r} (t) = \vec{r}_1 + \frac{t}{T} (\vec{r}_2 - \vec{r}_1)$?

11. Aug 29, 2016

### ehild

Yes!

12. Aug 29, 2016

### Mr Davis 97

Woo! Thanks so much!

13. Aug 29, 2016

### ehild

You are welcome

14. Aug 29, 2016

### Mr Davis 97

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

15. Aug 29, 2016

### ehild

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).

16. Aug 29, 2016

### Mr Davis 97

How can we substitute $\delta t$ for $t$ if that is not how it was when we derived the formula?

17. Aug 29, 2016

### ehild

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.

18. Aug 29, 2016

### Mr Davis 97

If that is the case, then shouldn't we have started our derivation with the ratio $\displaystyle \frac{t}{T + t_1}$ instead of $\displaystyle \frac{t}{T}$?

19. Aug 29, 2016

### ehild

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?

20. Aug 30, 2016

### Mr Davis 97

Ah, I see. Makes sense now.