Question about moving person and train Dear  friends below you can see question in georgian language and i will translate it into english :(Nino is  simple name of the women in georga)
so Nino is starting her move at the begining of the train(where  driver is sitting) and in direction of train move starts her walking, She passed 45  meter when last wagon of the train passed her, if she would start walking in opposite direction of the train, then she will go 30 meter : question is what is the length of the train(speed of  nino is same in both  case)
let us suppose that length of the train is x, when nino passed  45  meter  and  train's last  wagon passed her , it means that from the starting point train passed 45 meter till nino and  remainder distance will be x-45, when  nino passed 30  meter in opposite direction, it means that   train distance  form ninos last position to  initial position  is 30 and remaining length is x-30,i am   a bit confused  about  equality,if  we introduce speeds, then maybe train's speed and nino's speed is different  and how  we can measure  time periods? can you help  me please
 A: Let say the train is traveling at $x$ meters per second and Nino walks at $y$ meters per second. Let us also say the train has length $L$ meters.
So first, we see that it takes Nino $45/y$ seconds before the train passes her. In that time the last wagon in the train has traveled $L+45$ meters. But using the speed of the train, we have that it has also traveled $45x/y$ meters. So we have
$$ 45\frac{x}{y} = L + 45.$$
Similarly in the second scenario, it takes $30/y$ seconds. But now the wagon has traveled $L-30$. So we have
$$ 30\frac{x}{y} = L - 30. $$
Using both we get
$$\frac{L+45}{45} = \frac{L-30}{30} \Rightarrow 15L = 2*45*30 \Rightarrow L = 180m.$$
Whats interesting about this is that you don't actually need the speeds!
A: Alternative (very similar) approach:
Let:

*

*$D = ~$ length of train.

*$r = ~$ speed of train.

*$s = ~$ speed of Nino.

*$t_1 = ~$ time taken when Nino is traveling in same direction as train.

*$t_2 = ~$ time taken when Nino is traveling in same direction as train.

Then, the problem becomes:

*

*Eq-1: $~D = (r-s)t_1.$

*Eq-2: $~45 = st_1.$

*Eq-3: $~D = (r+s)t_2.$

*Eq-4: $~30 = st_2.$

*$D = \color{red}{?}$
Using Eq-2 and Eq-4, you have that $~\displaystyle t_2 = \frac{2}{3}t_1.$
Using the above result, + Eq-2 and Eq-4, you can refine Eq-1 and Eq-3 to

*

*$D = [rt_1] - 45.$

*$\displaystyle D = \frac{2}{3}\left[r t_1\right] + 30.$
Construing $[rt_1]$ to be a variable, you have two linear equations in the two unknowns $D$ and $[rt_1]$, so you can solve for $D$:

*

*$2D = 2[rt_1] - 90.$

*$3D = 2[rt_1] + 90.$
Subtracting the first equation above from the second gives  $D = 180.$
A: In the time Nino can cross a stationary train,
let the train traverse $k$ times its length
Then $\dfrac{L(k+1)}{L(k-1)} = \dfrac{k+1}{k-1}= \dfrac{45}{30} =\dfrac{3}{2},\;$  which easily yields $k = 5$
and means that length of train $= 6*30$ or $4*45 = 180m$
