A runs 7/4 times as fast as B. If A gives B a start of 84m, how far must the winning post be…?

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.

• Please attach proper tags with this and the other question; I don't know which tags are suitable. – user103816 Jun 2 '15 at 15:21
• Which text is this? I'd like to make sure not to use it, if it's saying things like "$7/4$ times faster" when it means "$7/4$ times as fast". – Cameron Buie Jun 2 '15 at 16:27
• @CameronBuie It is 7/4 times as fast I intentionally changed the text to make the title small but I think I changed the meaning -- I'm non-native English speaker. – user103816 Jun 2 '15 at 16:31
• I commend you on your excellent translation! That subtle point was the only confusing part of the post. – Cameron Buie Jun 2 '15 at 16:45

Once you see that $A$ gains $3$ metres over $B$ in a $7$-metre race, the question becomes: by how much must we scale the $7$-metre race to compensate for the $84$-metre head start?
If the race is $7k$ metres long, $A$ will gain $3k$ metres. We want to find $7k$, given that $3k = 84$. Thus $7k = 7\big(\dfrac{84}3\big) = 196$.
• How did you deduce that if the race is 7k metres long then $A$ will gain $3k$ metres? – user103816 Jun 2 '15 at 16:27
• In a $7$-metre race, $A$ gains $3$ metres. That relationship is linear: it can be scaled to any length, which is to say that for every $7$ metres of race, $A$ gains $3$ metres. In a $700$-metre race, for example, $A$ would gain $300$ metres. Linear relationships appear in many different situations; for instance, a store that offers "buy $10$, get $1$ free" might as well be offering "buy $10k$, get $k$ free", or if you make $20\$$an hour, then you make 20k\$$ in$k\$ hours, etc. – Théophile Jun 2 '15 at 17:57