A stream of ants go and return via same path from point $A\to B\to A$ with constant velocity, FROM ANY GIVEN POINT TWO ANTS PASS BY PER SECOND IN ONE WAY.It takes one minute to go $A\to B$ for any ant. how many returning ants will meet an ant which is going from $A\to B$?
Honestly I am not understanding the problem, mainly I do not understand the situation I have written in block letters. could any one explain me what is happening? may be i am not understanidng this simple problem as I am not good in english language.
The capitalized section seems to mean that if you’re standing at any point $P$ beside the path, in every second two ants travelling from $A$ to $B$ will pass by you. In other words, the ants are half a second apart: if one ant is passing you right now, the next one will pass you half a second from now. Let $d$ be the spacing between one ant and the next in centimetres, say; each ant travels a distance $d$ in half a second, or $2d$ cm/s.
Now consider an ant $P$ leaving $A$. On its way to $B$ this ant will pass a number of ants travelling from $B$ to $A$; we want to know how many. There’s no real harm in supposing that an ant arrives at $A$ just as $P$ leaves; call this ant $Q_0$. Let $Q_1$ be the first ant that $P$ passes, $Q_2$ the second, and so on. Start a clock at time $t_0=0$ when $P$ leaves and $Q_0$ arrives at $A$, and let $t_k$ be the time in seconds when $P$ meets $Q_k$.
$P$ and $Q_1$ meet at time $t_1$. At time $t_0$ the ant $P$ is moving at $2d$ cm/s in one direction, and $Q_1$ is moving at $2d$ cm/s in the opposite direction towards him, so they are approaching each other at a speed of $4d$ cm/s. $Q_1$ is $d$ cm behind $Q_0$ on the path from $B$ to $A$, so at time $t_0=0$ he’s $d$ cm away from $P$. $P$ and $Q_1$ are approaching each other at a relative speed of $4d$ cm/s, so they will meet in $\frac14$ s, at time $t_1=t_0+\frac14=\frac14$.