Two men, M and N started walking towards each other simultaneously from two places F and G, which are 50 km apart.They meet after 5 hrs. After their meeting , M reduced his speed by 1 kmph and N increased his speed by 1 kmph.They arrived at G and F respectively at the same time.Find the initial speed of M?
My Approach:
Let $S_M$ and $S_N$ be the speed of M and N
When the move towards each other Their meeting point will be at d=$50$
Relative speed$= S_M+S_N$
$t=5$
$$50=(S_M+S_N)\cdot5 \quad \implies S_M+S_N=10$$
When they move towards the other destination i.e M moves towards G and N moves towards F. Therefore,at that point their time is equal
So,
$5. S_M/(?)=5 . S_N/(?)$
(?) means I am confused here how to calculate their relative velocity.
Am I doing right till here? Please correct me if I am wrong.
Best Answer
Hint:
Since they arrive at $F$ and $G$, respectively, at the same time we have: $$\frac{5S_M}{S_N+1}=\frac{5S_N}{S_M-1}$$
Since the point where they met is $5S_M$ km apart from $F$ and $5S_N$ km apart from $G$.
Since $S_M+S_N=10$ we have \begin{align} \frac{5S_M}{(10-S_M)+1}&=\frac{5(10-S_M)}{S_M-1} \end{align}