The problem statement in the book is:
$A$ runs $7/4$ times as fast as $B$. If $A$ gives $B$ a start of $84$m, how far must the winning post be so that $A$ and $B$ might reach it at the same time?
The solution given in the book is:
Ratio of the rates of $A$ and $B = 7/4 : 1 = 7:4$. So, in race of $7$m, A gains $3$m over $B$.
∴ $3$m are gained by $A$ in a race of $7$m.
∴ $84$m are gained by $A$ in a race of $\left( \dfrac{7}{3} \times 84 \right) m = 196m$.
∴ Winning post must be 196m away from the starting point.
Question: I don't understand how we can use the unitary method after the second step, with 3m gain and 7m distance. Please explain using correct method why is unitary method applicable here.
My Understanding: Let $A$'s speed be $a$ and $B$'s speed be $b$. $a:b=7:4$, so $a=7x$ and $b=4x$. Initially $A$ and $B$ are $84m$ apart such that $B$ is ahead of $A$. Now relative speed of $A$ w.r.t. $B$ is $a-b$ in magnitude or $7x-4x=3x$.
In time $t_0$ the separation between $A$ and $B$ is reduced by $3xt_0$ and $A$ travels a net distance w.r.t ground equal to $7xt_0$. Now in time $t_0= \frac 1x$ $A$ moves $7m$ w.r.t. ground and the separation reduced is $3m$. In time $t_0=\frac {1}{3x}$, $A$ travels $\dfrac 73$m and the separation reduced is $1$m. In time $t_0= \dfrac{84}{3x}$ A travels $(7*84)/3$m and the separation reduced is $84$m, that is both $A$ and $B$ coincide, so $(7*84)/3$m must be the winning post's distance.
My understanding is quite equivalent to unitary method but I am having difficulty in converting my method into unitary method.
Basically I want a proof/derivation of the fact that we can directly use unitary method there.
Edit: I realize that I need to understand unitary method, please give me some references for unitary method explanation.
