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Particle A moves along the positive x-axis, and particle B along the line $$y=-\sqrt{3}x$$ for $x\in\left(-\infty,0\right]$ where $x$ and $y$ are in meters. At a certain time, $A$ is at the point $\left(5,0\right)$ and moving with speed $3\textit{ ms}^{-1}$, and $B$ is at a distance of $3$ units from the origin and moving with speed $4\textit{ ms}^{-1}$ away from the origin. At what rate is the distance between $A$ and $B$ changing?

My approach: I tried to write relative position/velocity vectors for both $A$ and $B$ at the instant, and got them to be: $$\vec{r_{A,B}}=\frac{13}{2}\hat{i}-\frac{3\sqrt{3}}{2}\hat{j}$$ and $$\vec{v_{A,B}}=5\hat{i}-2\sqrt{3}\hat{j}$$All I know from here is that $v_{A,B}=\dot{r_{A,B}}$ w.r.t. time.

I would appreciate any help. Thanks a lot!

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  • $\begingroup$ Try introduce a var like $t$ be the time from the certain time, what it A and B coordinate about t? $\endgroup$
    – Wuming Liu
    Commented Aug 24, 2023 at 14:24
  • $\begingroup$ @WumingLiu even if I did introduce, it would not be helpful since we are not given that the velocity is constant, neither that it is a function of $t$. Writing $A$ and $B$ as a function of $t$ would be meaningless. $\endgroup$
    – Doge with shades
    Commented Aug 24, 2023 at 14:32
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    $\begingroup$ work out the value of x that puts B at a distance of 3 from the origin, then workout the gradient at that point, which in fact you may already know. Having got the gradient of B's direction, you can get a vector for B's velocity, and also a vector for A's velocity. Then the difference between these two vectors is the relative velocity, from which you can get the relative speed (rate of change in distance at that instant). $\endgroup$
    – Cato
    Commented Aug 24, 2023 at 14:39
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    $\begingroup$ @Cato have I already written the answer in my question? I mean, have I unknowingly glossed over the fact that $\vec{v_{A,B}}=5\hat{i}-2\sqrt{3}\hat{j}$ is the relative velocity, and that $\left|\vec{v_{A,B}}\right|$ is the speed/ROC of distance? $\endgroup$
    – Doge with shades
    Commented Aug 24, 2023 at 15:15
  • $\begingroup$ The position seems to make no difference to me. We know nothing about time or acceleration, which also make no difference to the velocity or relative speed at the time - I make the velocities to be (3, 0) and (2, -2√3) - but where did you get 5 from in the i direction? I was thinking it would be 3-2 =1, although I'm assuming B is moving in the direction of increasing x, but maybe I'm wrong - your reasoning looks correct otherwise $\endgroup$
    – Cato
    Commented Aug 25, 2023 at 7:46

1 Answer 1

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Let $r$ be the distance away from the origin along the line $y=-\sqrt{3}x$ and let $d$ be the distance between the points $A$ and $B$.

By the law of cosines $$d^2=r^2+x^2-2rx\cos{120^{\circ}}=r^2+x^2+rx$$

Plugging in $r=3$ and $x=5$ gives $$d^2=3^2+5^2+3(5)=49\implies d=7$$

Differentiating gives $$2d\frac{dd}{dt}=2r\frac{dr}{dt}+2x\frac{dx}{dt}+r\frac{dx}{dt}+x\frac{dr}{dt}$$ and evaluating at $r=3, x=5, d=7$ and the speeds of $A$ and $B$ gives $$14\frac{dd}{dt}=6(4)+10(3)+3(3)+5(4)=83\implies\frac{dd}{dt}=\frac{83}{14}$$

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