On a straight stretch of road of length $L$ there were two stop signs at the two ends. Assume you follow traffic laws and stop at these signs. Your car can accelerate at a constant $a_{1}$ ( and decelerate at another constant $a_{2}$).
Compute the shortest time to travel from one stop sign to another if there is no speed limit in between. Do it again if the speed limit between the signs is $v_{m}$.
What I have tried so far:
Using distance traveled as $s(x)=s_{0}+v_{0}t+a^{2}t$ and $L=s_{1}+s_{2}+v_{max}t$;
- $s_{1}(x)=$$1\over{2}$$a_{1}t^{2}$ (Distance traveled accelerating as $s_{0}$ and $v_{0}$ are both zero)
- $s_{2}(x)=(L-s_{1})+[v_{1}]_{max}t+$$1\over{2}$$a_{2}t^{2}$ (Distance traveled decelerating, with $[v_{1}]_{max}$ being the top speed attained in $s_{1}$)
And so, for part one where we have no speed limit restraint, $v_{max}=0$, then:
$L=$$1\over{2}$$a_{1}t+(L-s_{1})+[v_{1}]_{max}t+$$1\over{2}$$a_{2}t$
$0=[$$1\over{2}$$(a_{1}+a_{2})]t^{2}+[v_{1}]_{max}t-s_{1}$
And I use the quadratic formula, removing the negative value for $t$.
As for part two, I did basically the same but with $v_{max}$ not being zero and $v_{0}$ in $s_{2}(x)$ being equal to $v_{max}$.
While comparing with some classmates, I realized I might not have the correct solution and so I would appreciate if someone could point out any potential flaw in my process.
x(t) will mean position as a function of time. v(t) will mean velocity as a function of time. a(t) will mean acceleration as a function of time. Let’s say T is the time it takes to get to the second stop sign. Then we know, x(0)=0, x(T)=L, v(0)=0, v(T)=0, a(t) will go from a1 and then and t=t2 will change to a2. The lines: v=a1t and v=a2(t-T) would intersect when t=t2, so if a1t2=a2(t2-T), then a1t2/a2=t2-T, then T=t2-a1/a2t2, then T=(1-a1/a2)t2. So if we knew t2, we would know T. 1/2a1(t2)^2+1/2a2(T-t2)^2=L. Try to cancel out the T from those 2 equations to find t2, so you can find T.