Derive vector that moves uniformly from one point to another

[Mr Davis 97]

Homework Statement

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

Homework Equations

None

The Attempt at a Solution

I'm having trouble even getting started with this problem... I really need some help.

 

[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

 

[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.

 

[Mr Davis 97]

A drawing might help.
View attachment 105244
Write vector $\vec r$ in terms of $\vec r_1$ and $\vec r_2$ and the ratio t/T first.

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...

 

[ehild]

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...

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

 

[Mr Davis 97]

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

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

 

[ehild]

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

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

 

[Mr Davis 97]

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

 

[ehild]

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

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

 

[Mr Davis 97]

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

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

 

[ehild]

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

Yes!

 

[Mr Davis 97]

Yes!

Woo! Thanks so much!

 

[ehild]

Woo! Thanks so much!

You are welcome

 

[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

 

[ehild]

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

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).

 

[Mr Davis 97]

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

 

[ehild]

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

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.

 

[Mr Davis 97]

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.

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

 

[ehild]

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

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?

 

[Mr Davis 97]

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?

Ah, I see. Makes sense now.