Mathematical Puzzle: A Drag Race of Who Wins I'm having a real difficult time understanding how this problem is solved: 
"Two drivers, Alison and Kevin, are participating in a drag race. Beginning from a standing start, they each proceed with a constant acceleration. Alison covers the last 1/4th of the distance in 3 seconds, whereas Kevin covers the last 1/3rd of the distance in 4 seconds. Who wins and by how much time?"
Distance = $s(t) = \frac 1 2 at^2$.
That just comes with understanding the physics involved. I don't think I can just easily plug numbers into it to create any kind of system of equations.
By solving for t, we get $t = \sqrt{\frac{2s}{a}}$
So far so good, but what follows is where I don't understand conceptually what's really going on.
Alison's acceleration: $3 = \sqrt{\frac{2s}{a}} - \left[ \frac{2\frac{3s}{4}}{a}\right]^{1/2}$
I'm not exactly sure where even that is coming from.
If we solve for a, and my algebra skills are not where they should be since I can't, we should get $a = \frac{s[(\sqrt{2}-\sqrt{3/2}]^2}{9}$ (but how?)
The time for Alison will be this: {(2s)/[(2^(1/2)-(3/2)^(1/2)*s/9)]}^(1/2) = 12 + 6(3)^(1/2)
The same procedure should get Kevin's time. But I don't understand it. Can anybody walk be through this in simple language? Thanks for your time and help!
(Note: I posted this yesterday, but I saw I made a big typo in it. So I deleted it.)
 A: I would do it like this.
First let the whole distance be $12$ units (you can choose the units to fit and $12$ makes everything an integer).
Now at some time $t_1$, Alison has travelled $9$ units so that $9=\frac {at_1^2}2$ and $$t_1^2=\frac {18}a$$
At time $t_2$, Alison has travelled $12$ units, so $$t_2^2=\frac {24}a$$
Now this last quarter of the journey takes $3$ seconds so that $$t_2-t_1=3=\sqrt{\frac {24}a}-\sqrt{\frac {18}a}$$
Squaring this equation and clearing fractions $$9a=24+18-2\sqrt{432}=42-24\sqrt 3$$ and $$a=\frac {14-8\sqrt 3}3$$
Once $a$ is known, $t_2$ is easy to find from the equation for $t_2^2$. And Kevin's journey can be analysed in a similar way.
I'd prefer this systematic approach to trying to write down complicated formulae - there is so much less to go wrong, and if you make a slip it is easier to spot where it is.
A: We have the general equation $s(t) = \frac{1}{2} at^2$, which will we can apply to any time in the race.  If we let $S$ be the end of the race and $T$ be the time it takes for Kevin to reach the end, we will have
$$
T = \sqrt{\frac{2S}{a}}
$$
But, we also know that 3 seconds earlier, he was 1/4 of the total distance away from the finish line, or in other words, he was 3/4 done with the race.  This means at the time $T-3$, he was at the position $\frac{3}{4}S$. So we can write the equation
$$
T-3 = \sqrt{\frac{2(\frac{3}{4}S)}{a}}
$$
subtracting the two equations leads to
$$
3 = \sqrt{\frac{2S}{a}} - \sqrt{\frac{\frac{3}{2}S}{a}}
$$
Notice that the two terms on the right hand side have the factor ${\sqrt{\frac{S}{a}}}$, which we can factor out:
$$
3 = \sqrt{\frac{2S}{a}} - \sqrt{\frac{\frac{3}{2}S}{a}} = \sqrt{\frac{S}{a}}(\sqrt{2} - \sqrt{\frac{3}{2}})
$$
Square both sides to get
$$
9 = \frac{S}{a}(\sqrt{2} - \sqrt{\frac{3}{2}})^2
$$
and now we just solve for $a$, which is just multiplying by $a$ and dividing by $9$:
$$
a = \frac{S}{9}(\sqrt{2} - \sqrt{\frac{3}{2}})^2
$$
which is Kevin's acceleration.  To get Kevin's time, we just plug in the original equation:
$$
T = \sqrt{\frac{2S}{a}} = \sqrt{\frac{2S}{ \frac{S}{9}(\sqrt{2} - \sqrt{\frac{3}{2}})^2}} = \sqrt{\frac{2}{ \frac{(\sqrt{2} - \sqrt{\frac{3}{2}})^2}{9}}} = \sqrt{\frac{18}{ (\sqrt{2} - \sqrt{\frac{3}{2}})^2}}
$$
The exact same strategy works for Alison, with slightly different numbers.
