Arithmetic progression with deceleration A train is travelling at $180 \text { km/h }$, $500\text { m }$ away from a train station, what is the constant deceleration needed to get to a complete stop at the station.
A continued question regarding the use of series and sequences to work this out instead of physics. This was in my exam and my teacher said that "Physics was unnecessary for the question." How would you do this question using simple maths?
 A: Speed 50 m/s.
According to galileo's  law of odd numbers, in an de-accelerated motion distance covered in every following second is proportional to the decreasing ratio of odd numbers.
Let it reaches the station in $t$ seconds.
So, distance covered in first second is 50 metre. using the law the distance covered in the $r^{th}$ second will be, $\frac{2(t-r)+1}{2t+1}50$.
Total distance covered will be, $500= \sum_{r=1}^t(\frac{2(t-r)+1}{2t+1})50$ 
which gives $t=10+\sqrt{110} \approx 20$, 
$a=\frac{\Delta v}{t}=\frac{-50}{20}=\frac{-5}{2}$, so the acceleration is $-2.5\frac{m}{s^2}$.
A: If you know that the velocity is zero at the end, you know what the average velocity is.  Using that and the distance, you can compute the total time.  Then with the time and change in velocity, you can compute the acceleration(negative).
Not sure if this is what the teacher had in mind.  Probably not if it involved series and sequences.
ETA: Limit method wasn't as simple as I thought it would be.  Hopefully somebody smarter than me will take it up, but here's what I had.  Also it involves Distance=Velocity$\times$Time.  Not sure if that is "physics"
Imagine breaking up the trip into $x$ segments and each segment, we reduce the velocity by $\frac{180}{x}$.  We can then setup an equation like $$0.5=\lim_{x\to\infty}\sum_{i=0}^x(180-\frac{i}{x}180)(\frac{t}{x})\\0.5=\lim_{x\to\infty}\sum_{i=0}^x\frac{180t(x-i)}{x^2}\\0.5=90t\\t=\frac{1}{180}hrs$$
But given the time and change in velocity, I'm still not sure how you would get the deceleration without using a physics formula. 
A: Let $s(t)$ be the position in km, $v(t)$ the speed in km/hr, $a$ the acceleration in km/hr^2.  We are given $s(0)=0, v(0)=180, s(t)=\frac 12, v(t)=0$  Then $$s(t)=\frac 12at^2+v(0)t+s(0)\\ v=v(0)+at\\a=-\frac {180}t\\\frac 12=-\frac 12\cdot 180t+180t=90t\\
t=\frac 1{180}\\a=-32400$$
