Distance between two particles

[LCSphysicist]

Homework Statement

All below.

Relevant Equations

All below.

So we have 5 incognits 5 equations, is not hard, but need attention.
What you think about? Right?
This type of question is generally more easy to solve, do you know any trick?

 

[Delta2]

Had to spent a little time to understand your first equation , but i think it is correct.

However i do not agree with your 4th equation. Can you expand a bit more how did you arrive at your 4th equation. In my opinion it should be $$4^2=(X_B')^2+(Y_B')^2$$.

Also there is another problem, the problem says that $f(x)=-\sqrt{3x}$ and $x\leq 0$. This means we are into complex numbers? Cause we are taking the square root of a negative x. Probably this is a typo , the problem probably means $x\geq 0$.

 

[Mark44]

... particle B along the graph of $f(x) = -\sqrt{3x}, x \le 0$.

This makes no sense. The domain of $f(x) = -\sqrt{3x}$ is $x \ge 0$, so for any x < 0, the function f is not defined.

Looks like @Delta2 noticed this as well.

 

[LCSphysicist]

Woll, i didn't notice that

Had to spent a little time to understand your first equation , but i think it is correct.

However i do not agree with your 4th equation. Can you expand a bit more how did you arrive at your 4th equation. In my opinion it should be $$4^2=(X_B')^2+(Y_B')^2$$.

Also there is another problem, the problem says that $f(x)=-\sqrt{3x}$ and $x\leq 0$. This means we are into complex numbers? Cause we are taking the square root of a negative x. Probably this is a typo , the problem probably means $x\geq 0$.

The fourth equation: Distance of B from the origem

db = (xb² + yb²)^(1/2)
db' = 1/2 (xb² + yb²)^(-1/2) (2xb xb' + 2yb yb')
but (xb² + yb²)(1/2) is the distance of b from the origem in this instant
db' = xb xb' + yb yb' * (3^-1)
db' = 4
4 * 3 = xb xb' + yb yb'

But since the question is write wrong, i think it is better we don't go on with it :S

 

[LCSphysicist]

Even so, i think interesting discuss which method is right to get the fourth equation, the mine or yours. or maybe both

 

[Delta2]

Woll, i didn't notice thatThe fourth equation: Distance of B from the origem

db = (xb² + yb²)^(1/2)
db' = 1/2 (xb² + yb²)^(-1/2) (2xb xb' + 2yb yb')
but (xb² + yb²)(1/2) is the distance of b from the origem in this instant
db' = xb xb' + yb yb' * (3^-1)
db' = 4
4 * 3 = xb xb' + yb yb'

But since the question is write wrong, i think it is better we don't go on with it :S

What you calculating here is the speed at which B approaches the origin. That's not the same thing as to what we mean simply by its speed. When you write db'=4 that's simply wrong cause you equate B's speed with the Speed that B is approaching the origin. B's speed is simply the magnitude of its velocity vector. Its velocity vector is $$\frac{d\vec{r}}{dt}=X_B'\hat x+Y_B'\hat y$$ and the magnitude of this is $$\|X_B'\hat x+Y_B'\hat y\|=\sqrt{(X_B')^2+(Y_B')^2}$$

 

[Delta2]

In an attempt to make it more clear to you, you are calculating $$\frac{d\|\mathbf{r}\|}{dt}$$ and set it equal to 4, while i believe the correct is $$\left\|\frac{d\mathbf{r}}{dt}\right\|=4$$, because that's the definition of speed, the magnitude of the velocity vector.

 

[etotheipi]

Also note that like @Delta2 said, if $\mathbf{v} = v^i \mathbf{e}_i$, then $\left\| \mathbf{v} \right\| = \sqrt{ \sum (v^i)^2}$, and also $\frac{d\mathbf{v}}{dt} = \frac{dv^i}{dt}\mathbf{e}_i$. Then we have $$\left\| \frac{d\mathbf{v}}{dt} \right\| = \sqrt{\left(\frac{dv^i}{dt}\right)^2}$$whilst$$\frac{d \left\|\mathbf{v}\right\|}{dt} = \frac{v^i \frac{dv^i}{dt}}{\sqrt{ \sum (v^i)^2}}$$and these are not equal. You will have noticed this for uniform circular motion, there the norm of the acceleration vector is non-zero however length of the velocity vector is constant.

 

[Delta2]

I would try to calculate $$\left\|\mathbf{v}_{AB}\right\|=\left\|\mathbf{v}_{A} - \mathbf{v}_{B}\right\|$$since in non-relativistic Physics the relative speed is the rate of separation. You can start off by taking the trajectory $y = -\sqrt{3x}$ and differentiating w.r.t. time, and then using the magnitude constraint, to find the velocity vector of that particle.

I am not so sure the rate of separation as you define it though the magnitude of the difference of velocities, is the same as the rate of change of their distance, doing the math i don't get necessarily equal expressions...

 

[etotheipi]

Yes sorry, that was a big mistake . If $s$ is the separation then $$\begin{align*}\mathbf{r}_{ab} &= \mathbf{r}_a -\mathbf{r}_b \\ \\ s &= \left\| \mathbf{r}_a - \mathbf{r}_b \right\| \\ \\ \frac{ds}{dt} &= \frac{d \left\| \mathbf{r}_a - \mathbf{r}_b \right\|}{dt} \neq \left\| \mathbf{v}_a - \mathbf{v}_b \right\| \end{align*} $$ It seems I made the same error I warned against in #8... many apologies, I need some more caffeine