
Formula help.
If two persons A and B start at the same time in opposite directions from two points and arrive at two points in 'a' and 'b' hr, respectively after having met, then , A's speed/B's speed = root b/root a.
Can you help how this formula came up? Appreciate your help.

Re: Formula help.
Is it a physics problem? Is it an accelerated movement? I'm not sure what you meant. I guess you meant A starts from a and gets to b and B starts from b and gets to a but in that case I get to V_{b}/V_{a} = 1. Also, by root did you mean square root? Please, explain a little bit more or write the exercise here because it seems to me that something is missing or something... (Wondering)

Re: Formula help.
No this is a basic math problem...Yes, the root refers to square root.

Re: Formula help.
I've been trying to find a way, still seems very confusing. Do you have any more information about the starting and ending points? Sorry to keep asking but up to now I keep thinking that something is missing.
However, I tell you what have I thought so far and maybe it can help:
A and B go in opposite direction and they're going to meet. When they meet, they keep moving and reach two different points at two different times. Let's say that 0 is our meeting point, A moves to +x and B to x. I define two points that are going to be reached by each person: P_{a} and P_{b}. I know that a is the time A takes to reach P_{a} and same thing with b. Finally, I get two equations:
$\displaystyle a \bullet V\sub{a} = P\sub{a}$
$\displaystyle b \bullet V\sub{b} = P\sub{b}$
If I sum both of them:
$\displaystyle a \bullet V \sub{a}  b \bullet V \sub{b} = P\sub{a}  P\sub{b}$
Then I got stuck.

Re: Formula help.
I have an example problem using this formula....
A man starts from B to K, another from K to B at the same time. After passign each other they complete their journey in 3 1/3 and 4 4/5 hours, respectively. Find the speed of the second man if the speed of the first is 12 km/hr.
1'st man's speed/2nd man's speed = root 4 4/5 / root 3 1/3
=> 6/5
=>2nd man's speed =10 km/hr

Re: Formula help.
Finally I think I got it.
I'll take your first example: person A goes +x and met B at certain time. B goes x. That time would be the same for both because they started at the same time.
$\displaystyle A \rightarrow M \leftarrow B$ where M is the meeting point.
The distance between A and M and B and M is unknown since we don't know their velocities but we can obtain them using the second part of the problem where A gets to where B was.
A: $\displaystyle M \rightarrow B$
B: $\displaystyle A \leftarrow M$
The problem says that after they meet (or after they get to M), after "a" hours A arrives to where B was and after "b" hours B arrives where A was. Now we can find an expression for each distance:
$\displaystyle \overline{AM} = b \bullet v_B$
$\displaystyle \overline{BM} = a \bullet v_A$
So, what we've done here is get the segment's width by multiplying the velocity times time. Notice I'm using the second part of the problem, after the meeting because I have enough information there. Person A goes from M to B in a hours.
Now we need to find the time that takes A and B to get to M which, I repeat, it's the same for both. Let's call it $\displaystyle t_M$.
By crossmultiplying:
M to A in b hours:
$\displaystyle \overline{AM} \longrightarrow b$
$\displaystyle \overline{BM} \longrightarrow t_M = \frac{\overline{BM} \bullet b}{\overline{AM}} = a \bullet b \bullet \frac{v_A}{b \bullet v_B} = a \bullet \frac{v_A}{v_B}$
M to B in a hours:
$\displaystyle \overline{BM} \longrightarrow a$
$\displaystyle \overline{AM} \longrightarrow t_M = \frac{\overline{AM} \bullet a}{\overline{BM}} = a \bullet b \bullet \frac{v_B}{a \bullet v_A} = b \bullet \frac{v_B}{v_A}$
Finally, both results are equal so:
$\displaystyle a \bullet \frac{v_A}{v_B} = b \bullet \frac{v_B}{v_A}$
$\displaystyle \frac{v_A^2}{v_B^2} = \frac{b}{a} $
$\displaystyle \frac{v_A}{v_B} = \sqrt{\frac{b}{a}}$
That's it! I tried to be as clear as possible but, please, if you didn't understand anything ask me and I'll try to make it clearer.
EDIT: maybe it's not quite relevant to point this since it's a math problem but the crossmultiplying would be impossible if the velocity changes through time. In this case, it was a constant.

Re: Formula help.
@ russo! Great help! Thanks!

Re: Formula help.
@ Russo: I have a question, sorry for asking this very late...how do we conclude that the time taken by A and B to reach M is the same when we dont know their speeds till the meeting point M?

Re: Formula help.
Because if two things encounter at the same point, it SHOULD be at the same time. Let me tell you an example. Imagine you are walking down the street and you meet someone you know. By the time, you start talking both can look at their watches and expect to have the same time. That's why, when the problem says that both start at the same time (let's say t = 0), the precise moment they meet they're in t = t_{e
Also, when you're talking about encounters you can assume that they're in the same time and position (unless they're are in different reference systems) }

Re: Formula help.
Thanks, that is nice to learn!