Sheldon and his toy train set.

Using $v^2=u^2+2as$ , we get

$y^2=x^2+2as$

Now we need the distance to be half with the acceleration constant and therefore the $2as$ term becomes $as$ which is

$\dfrac{y^2-x^2}{2}$ , this gives the equation.

Let the velocity when the middle of the train crosses the pole be $k ms^{-1}$

$k^2 = x^2 + \dfrac{y^2-x^2}{2}$

This gives $k^2 = \dfrac{y^2+x^2}{2}\implies k = \sqrt{\dfrac{x^2+y^2}{2}}$

Using $v^2=u^2+2as$ we get $y^2=x^2+2aL$ where $L$ is the length of the train. This implies that: $2aL=y^2-x^2$ .
Similarly, if the midpoint of the train passes the pole with a velocity of $z$ , then: $2a(\frac{L}{2})=z^2-x^2$ .
This implies that: $2(z^2-x^2)=y^2-x^2$ $\Rightarrow 2z^2-2x^2=y^2-x^2$ $\Rightarrow 2z^2=y^2+x^2$ $\Rightarrow z=\sqrt{\frac{x^2+y^2}{2}}$

y^2-x^2=2as hence a=(y^2-x^2)/2s now m^2-x^2=2(y^2-x^2/2s)(s/2) therefore m^2=x^2+(y^2-x^2/2) hence m^2=x^2+y^2/2 hence m=sqrt(x^2+y^2/2)

as we all are aware from the formula v.^2-u>^2=2as; where v=final velocity; u=initial velocity; a=constant acceleration; s=displacement; ...so applying this equation for first case when the train has passed whole pole...; 2ad=y.^2-x.^2; d= Lenght of train; then a=(y.^2-x.^2)/2d; ..... now for the scnd case 2a (d/2)=v.^2-x.^2; where v is velocity to be find; putting the value of "a" here; 2 {(y.^2-x.^2)/2d}*d/2=v.^2-x.^2); then we get the value of v as sqrt((y.^2+x.^2)/2); answer

Let the velocity of the middle part be Z, the acceleration of the train be 'a' and length of train be 'l'. So by third eqn of motion, y^2=x^2 + 2al.... this gives us al=(y^2-x^2)/2 We also get an eqn y^2=Z^2+al (as the length here will be l/2) putting the value of al...we get y^2=Z^2+(y^2-x^2)/2 by solving this we get Z=[(y^2+x^2)/2]^1/2