I have a problem to solve and I got stack. The question is: Having a vehicle that weighs $m$ that can move at variable speed but with maximum acceleration of $a_1$ and minimum deceleration of $a_2$ calculate the minimum time to cover distance $\ell$. The starting and ending speed should be zero.

I need some pointers or a link, please and thank you.

  • $\begingroup$ Does it need to stop at the destination? If no, I think you need to accelerate the vehicle as much as you can until the end. If yes, you need to accelerate the vehicle as much as you can until it reaches a prechosen speed V, then at a later time, start to decelerate it as much as you can until it stops exactly at the destination. Since there is only one single variable V to deal with, you can easily optimize time with the best choice of V (apparently V will be the V with which you have to immediately start to decelerate). $\endgroup$ – verdelite Nov 23 '19 at 22:27
  • $\begingroup$ @verdelite Yes it does need to stop at the end. I updated the post $\endgroup$ – Dimitris Karagiannis Nov 23 '19 at 22:29

We accelerate and then when there is no choice anymore, we decelerate. Indeed, if your minimal deceleration decelaration was infinite, you would have been able to decelerate completely brutaly at the end and that would have been the best. Now you have to tend to that limit. The idea to prove it is to look at how to optimize the distance because it's more simple to do it.

Let's write $v_M$ the maximal velocity. Deceleration is from $v_M$. Then the time to decelerate with the minimal decelaration $a_m$ is $t_d=v_M/a_m$ on a distance $l_d=\int_0^{t_d}(v_M-a_m t)dt=v_M t-a_m t^2/2 $. Now the maximal acceleration is $a_M$. Then $v(t)=\int^t_0 a_M t'dt'=a t$. And $x(t)=at^2/2$. At $t_M$ we got to the maximal velocity : $v_M=a t_a$ and the distance $l_a=at_a^2/2$. We have $L=l_a+l_d=a_M t_a^2/2+(a_M t_a) (a_M t_a /a_m)-a_m (a_M t_a/a_m)^2/2=t_a^2(a_M/2-a_M^2/a_m/2)$.

Ok now you can finish. You have $t_a$, so you have $l_a$. But then you have also $l_d=L-l_a$. And from this you get $t_d$, and then $T=t_a+t_d$.

I hope I didn't do too much mistakes :)

I didn't write the whole answer with a rigourous proof to my answer. But that's a sketch of how to tackle the issue.


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