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My friend gave me a question today:

We have a point $A$. At a distance of $x_0$ from the point. There is a particle $P_1$. There is another particle, $P_2$, at $A$. $P_1$ moves with velocity $u_1$ towards $A$. At the same instant, $P_2$ moves making an angle $\theta$ with $P_1A$ with a velocity $u_2$. What will be the minimum distance between the two particles when both of them start at the same time?

So can anybody help me with this problem? I tried decomposing it into vectors, used $\tan\theta$, but it did not give me a correct answer.

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Let's assume $P_2$ is in $(0,0)$ at $t=0$. $P_1$ is in $(x_0,0)$ at t=0. (So A-P1 represents the x-axis)

The evolution of $P_2$'s position is given by $\vec{r}_{p2}(t) = u_2\cdot t\cdot(\cos(\theta),\sin(\theta))$ which you can easily get from decomposing.

The evolution of $P_1$'s position is given by $\vec{r}_{p1}(t) = (u_1\cdot t+x_0,0)$

The distance between these particles in function of time is (Pythagoras' Theorem)

$D(t) = \sqrt{ (u_2\cos(\theta)\cdot t-u_1\cdot t-x_0)^2 + u_2^2\sin(\theta)^2\cdot t^2} $

This function obviously won't have a maximum. It is enough to derive it once and set it's derivative equal to 0.

This is the gist of the problem. Feel free to ask if you have any difficulties understanding this. Working your way through the algebra you should find

$t_{min} = \frac{u_2\cdot \cos(\theta)-u_1}{u_1^2+u_2^2-2u_1u_2\cdot\cos(\theta)}x_0 $ as the time at which the distance is minimal

$D_{min} = \frac{u_2x_0\cdot\sin(\theta)}{\sqrt{u_1^2+u_2^2-2u_1u_2\cdot\cos(\theta)}}$ the value of the minimal distance

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  • $\begingroup$ OH! I Got!!, Thank you very much, but I had got it with a very similar method. Thanks, But Although for a clean solution $\endgroup$
    – Dinesh
    Jun 19 '14 at 16:02
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You don't have enough information to properly answer the question.

If $u_1 \le u_2\ and\ \theta \ge \pi/2$, then the shortest distance occurs at the initial conditions.

When $u_1 > u_2$, then the shortest distance occurs when $\angle P_{1-initial}P_1P_2$ is $\pi/2$.

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