How to calculate minimum constant speed to catch up with constantly accelerated moving point My homework wants me to calculate the minimal velocity needed to catch up with a train moving past me.
The train is accelerating with $a_z = const.$. After it drives past me with $v_{0}$, I take 2 seconds to react, after which I sprint after it with $v_p$.
What is the minimum $v_p$ with which I can manage to catch the train?
My attempt so far is this:
I set $x_z = x_p$.
$x_z = \int_{0}^{t} \int_{0}^{t} a_z dt dt$
$    = \int_{0}^{t} a_z t + v_0 dt $
$    = \frac{1}{2} a_z t^2 + v_0 t + x_0 $
$    = \frac{1}{2} a_z t^2 + v_0 t $
and
$x_p = v_p t$
and so I solve the following for t
$x_z = x_p$.
$\frac{1}{2}a_z t^2 + v_0 t = v_p t$
$\frac{1}{2}a_z t + v_0 = v_p$
$\frac{1}{2}a_z t = v_p - v_0$
$t = \frac{2(v_p - v_0)}{a_z}$ 
And so I am looking for $\min_{v_p} t = \frac{2(v_p - v_0)}{a_z} $ 
As I have two variables I dont know what to do or minimize. Clearly there is a space of solutions, but I dont know how to find the minimum.
Is my approach correct until here? Can you give me any tips?

Since you react after 2 seconds
$$
x_z = {1 \over 2}a_z(t+2)^2 +v_0(t+2)
$$
and Since 
$$
x_z=x_p
$$
$$
v_pt={1 \over 2}a_z(t+2)^2 +v_0(t+2)
$$
$$
v_p={1 \over 2t}a_z(t+2)^2 +{v_0(t+2) \over t}
$$
$$
v_p={1 \over 2t}a_z(t^2+2t+4) +{v_0(t+2) \over t}
$$
$$
v_p={1 \over 2}a_z(t+2+4/t) +v_0(1+2/t)
$$
$$v_p=a_z t/2+2a_z/t +2v_0/t +a_z+v_0$$
to find the minimum value of function we equate the slope to zero
$$d v_d/dt  = a_z/2 - 2(a_z+v_0)/t^2 = 0$$
$$ => t= \sqrt{4(1+v_0/a_z)}$$
so if you substitute in above equation you will get minimum constant speed
