So I was solving a problem and in that there was this statement written "3 men $A, B$ and $C$ respectively leave $P$ for $Q$ after equal time intervals such that they reach $Q$ simultaneously." and below is how I interpreted it mathematically :-
Let the distance be $d$ units.
Let speed of $A, B$ and $C$ be $V_A,V_B$ and $V_C$ units respectively.
Let the equal time interval be $x$ units.
Let $t$ be the time taken by $A$ units.
Hence mathematically it should be :-
$d=V_A \times t = V_B \times (t-x)=V_C \times (t-2x)$ and this shows that $V_A<V_B<V_C$.
But in the solution hint the below relation was written and I am not able to understand how this relation has come up ;
$\frac{d}{V_A}-\frac{d}{V_B}=\frac{d}{V_B}-\frac{d}{V_C}$
How can we get to this relation? Please help me on this !!!
Thanks in advance !!!
Time = Distance/Speed
The distance is same for each person ($d$). The times they take to get from P to Q are $T$, $T-x$, and $T-2x$, respectively.
So you have,
$T = d/V_A$
$T-x = d/V_B$
$T-2x = d/V_C$
You can see that $d/V_A - d/V_B = x$ and $d/V_B - d/V_C = x$.
It may be more intuitive to think about it this way: B took $x$ seconds fewer than A to get from P to Q (given). A took $d/V_A$ seconds to get from P to Q. B took $d/V_B$ seconds, which is $x$ seconds fewer than $d/V_A$. This means $d/V_B = d/V_A -x $, and so on.