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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)?

Seeker
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Creedance
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    Please edit to include your efforts. And for clarity: is $v_2=v$? Or was $v2$ meant to denote $v^2$? – lulu Aug 31 '22 at 11:28
  • İt is v^2 sory. – Creedance Aug 31 '22 at 11:32
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    I don't think you can work this problem without knowing the length of a truck. Or making some assumptions. – B. Goddard Aug 31 '22 at 11:38
  • Actually its basically asking the min of save driving distance. and the lenght of the road is infinite or not mentioned so you dont need the lenght of the truck – Creedance Aug 31 '22 at 11:40
  • Think that i didnt understand is what is it mean with the "within a time unit" – Creedance Aug 31 '22 at 11:41
  • İs it askin the max number of trucks who passes in the 1 unit time.so if the number of the trucks is N each speed of the trucks is 1/N am ı correct – Creedance Aug 31 '22 at 11:44
  • The time unit doesn't matter, nor does the length of a truck, just assign values. – lulu Aug 31 '22 at 11:44
  • If one truck is a mile long, I think your answer will be different than if one truck is an inch long. – B. Goddard Aug 31 '22 at 11:44
  • @B.Goddard As I understand the problem, the length of a truck is irrelevant. The optimal gap should be independent of that. True, the actual number of trucks that pass in a given time interval will depend on the length, but the gap should not. – lulu Aug 31 '22 at 11:45
  • The problem is i it just simply asking the min(v^2/100+v/4)but its can be infinitly small – Creedance Aug 31 '22 at 11:46
  • Small distance more trucks – Creedance Aug 31 '22 at 11:46
  • @Creedance Yes, but if $v=0$ then no trucks at all pass a point in a given interval, so that's not optimal. – lulu Aug 31 '22 at 11:46
  • First do it for points (assuming a truck is a point). At speed $v$, how many points pass in an hour (assuming they all maintain safe distance). Now optimize that number in terms of $v$. Then, to check @B.Goddard's issue, assume that the "points" have length $L$ and repeat the calculations. I think the optimal $v$ should be the same but it is definitely worth checking. – lulu Aug 31 '22 at 11:49
  • So what is the min of the V.actually this question is from the CAS intelligent scale test so the answer may can be the "smallest vlaue of the given sum" – Creedance Aug 31 '22 at 11:49
  • @B.Goddard Counter to my intuition, the length of the truck definitely does matter. – lulu Aug 31 '22 at 12:09

2 Answers2

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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 = $\frac{\sqrt a}{5}+\frac1 4$

Abel Wong
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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.

Caspar Heusinger
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