Math puzzle not that hard but i cant find what am i missing

Question

The braking distance for a truck with the speed measured in km/h is $\frac{v^2} {100}$ meters, the “reaction distance” (distance driven during the reaction time) is about $\frac v 4$. For save driving, the distance between one truck driving behind another truck should be at least the sum of the braking distance and the reaction distance. At which speed will a convoy of trucks have the highest number of trucks passing a point along the road (within a time unit)?

Answer 1

The formula for the safe driving distance should be $d_s(v)=\frac{v^2}{100}+\frac{v}{4}$.
Then the distance between the front of the trucks should be $d =d_v+l$ with $l$ being the length of one truck. Then the formula of trucks passing per second (if you convert the velocity to m/s and the distance to m) is $\frac{d}{v}=\frac{\frac{v^2}{100}+\frac{v}{4}+l}{v}$ and then you need to find the minimum of this function.

Answer 2

Min Distance between truck d = $\frac{v^2}{100}+\frac v 4$

Min Time between two trucks t = $\frac{v}{100}+\frac 1 4$

Then min time will achieve with v=0 and every $\frac 1 4$h (reaction distance $\frac v 4$ km? The coefficient may need to be adjusted).

This solution v=0 is stupid. That means infinity trucks are packed together in unit distance, which is impossible. Why?

So we need a truck length term. Min Distance between truck = $\frac{v^2}{100}+\frac v 4+a$

Min Time between two trucks t = $\frac{v}{100}+\frac 1 4+ \frac a v$

Then find $\frac{dt}{dv}=0$, we have $v=10\sqrt a$ and Min Time = $\frac{\sqrt a}{5}+\frac1 4$