The problem is as follows:
A mass trasit train is slowing down its motion before getting to destination. Sensors in the tracks as read in the control room indicate that the train took $20\,s$ and then $30\,s$ in traveling a quarter of a kilometer successively. Using this information find the length which will be traveled by the train before stopping completely.
The answer from my book is: $602.08\,m$
I don't know exactly how to get to this number. So far the only idea which has had come to my mind is to use the concept of average acceleration:
If I recall correctly this is given by:
$\bar{a}=\frac{v_{2}-v_{1}}{t}$
For this part I'm assuming it is meant the total time: t
Hence:
$\bar{a}=\frac{\frac{250}{30}-\frac{250}{30}}{20+30}=-\frac{1}{12}$
which it makes sense as this object is decreasing its speed. Assuming that the length covered in each portion is equal to $250\,m$ from:
$\frac{1}{4}\cdot 1000= 250\,m$
But the part where I came to use the equation:
$v_f^2=v_i^2-2a\Delta x$
It comes the question, who's $v_{i}$?. So I assumed it is what I assumed is the final:
$v_{i}=\frac{250}{30}$
$\Delta x=\frac{v_f^2-v_i^2}{2a}=\frac{0^2-(\frac{250}{30})^2}{-2\times\left(\frac{1}{12}\right)}=416\frac{2}{3}\,m$
But this isn't close to what my book say the answer is, so what could be wrong here?. Can someone help me?. Could it be that something which I thought is missing or what?. Help please?.
METHOD 1
The average velocities over the 1st and 2nd intervals are $\frac{25}{2}m/s$ and $\frac{25}{3}m/s$ respectively. These are also the velocities at the mid-points in time of these 2 intervals. The changes in velocity and time between the 2 mid-points are $\frac{25}{2}-\frac{25}{3}=\frac{25}{6}m/s$ and $\frac{30}{2}+\frac{20}{2}=25s$ respectively. So the acceleration is $-\frac16m/s^2$.
The velocity at the end of the 2nd interval is $\frac{25}{3}-\frac16 \frac{30}{2}=\frac{35}{6}m/s$
Using the formula $v^2=2as$ the distance travelled from the end of the 2nd interval until coming to rest is $$s=\frac{(\frac{35}{6})^2}{2*\frac16}=\frac{35^2}{12}=102m$$
METHOD 2
You are given distances and times. Presumably the train is undergoing constant (negative) acceleration to the station. The formula which comes to mind is $$s=\frac12 at^2$$
It is slightly easier to think of the equivalent situation in which the train is accelerating out of the station. Measure distances and times from the station to the ends of the two intervals : $$s_1 = \frac12 a t_1^2$$ $$s_2 = \frac12 a (t_1+30)^2$$ $$s_3 = \frac12 a (t_1+50)^2$$
You are given the distances $$250=s_2-s_1 = \frac12 a [(t_1+30)^2-t_1^2]=\frac12 a (30)(2t_1+30)$$ $$250=s_3-s_2 = \frac12 a [(t_1+50)^2-(t_1+30)^2]=\frac12 a (20)(2t_1+80)$$
Comparing the last two equations gives $$30(2t_1+30)=20(2t_1+80)$$ $$3(t_1+15)=2(t_1+40)$$ $$t_1=80-45=35$$ $$250=\frac12 a (30)(2t_1+30)=\frac12 a (30)(100)$$
Compare the last eqn with the 1st : $$\frac{s_1}{250} = \frac{35^2}{30*100}$$ $$s_1 =102 m$$
From the end of the 2nd interval the train travels a further $102m$ before stopping.