I'm having some difficulties dealing with this problem: A train starts travelling from A towards B. It's velocity is v. Simultaneously train starts travelling at the same velocity from B to A. The distance from A to B is d. A fly starts to move the same moment the trains do. It starts from A and moves towards B with valocity u > v. whan it encounters the second train it reverses direction until in encounters the first train. It then changes direction again until it encounters the second train again and so on until the two trains crash.
Now, I'm aware there are many people all over the internet discussing and explaining this problem. Although I'm not really understanding it.
The easy way of soliving it is:
$$ \frac{d}{2} = v\times t_{tot} $$ $$ t_{tot}=\frac{d}{2 \times v}$$
$$d_{tot}=u \times t_{tot}=\frac{d\times u}{2v}$$
this is correct, isn't?
I then tried to build the sereis. I called $d_0$ the first leg (before the fly encounters the second train for the first time). $$ d_0 + v\times t_0 = d $$ $$ u\times t_0 + v\times t_0=d $$ $$ t_0 = \frac{d}{u+v} $$ therefore $ d_0 = d-v\times t_0=d-v \frac{d}{u+v}= d\frac{u}{u+v} $
the other steps are analogus: $ d_1 +v\times t_1 = d_0 = d\times \frac{u}{u+v} $
$t_1=\frac{d}{\left(u+v \right )^2} $
$d_1 = d\times \frac{u}{u+v}-v\frac{d}{(u+v)^2}=d\times \frac{u}{u+v}\left(1-\frac{v}{v+u}\right)=d\times \frac{u^2}{(u+v)^2} $
therefore $$ d_n = d \times \left(\frac{u}{u+v}\right)^{n+1} $$ Is it correct so far?
I then calculated the partial sum: $$ S_n = d\times \frac{u}{u+v}\frac{1-\left(\frac{u}{u+v}\right)^{n+1}}{1-\frac{u}{u+v}}$$
I then calculated the limit: $$ \lim_{n\rightarrow +\infty}{Sn}=d\times \frac{u}{u+v}\frac{1}{1-\frac{u}{u+v}}=d\times \frac{u}{u+v}\frac{1}{\frac{u+v-u}{u+v}}=\frac{u}{v}d $$
which is double the result I got before. Where did I go wrong?
You can see that $d_1+v\times t_1=d_0$ is wrong since:
You see, $d_1+v\times t_1<d_0$.
Note: black line: time-space line of train, and red line is ones of the fly. The gray lines are just alignments.