I asked my teacher today. He said that indeed it start ar rest and accelerates at a and at somep point it decelerate at a in order to be at rest at B. He said that there arte two cases and that I need to choose one and explain why. So yes, I can use SUVAT equations. Which are the two cases?CWatters said:You seem to be making this problem much harder than it is. I very much doubt they intended you to think about limitations due to the speed of light!
It starts from A at rest and accelerates at "a". At some point it must stop accelerating and start to decelerate at "a" in order to be at rest at B.
A few basic equations of motion (eg SUVAT) and you are done.
Pity that wasn't stated in the first place.Davidllerenav said:it start ar rest and accelerates at a and at somep point it decelerate at a in order to be at rest at B
Unless there is also a max allowed speed, I can only think of one case.Davidllerenav said:Which are the two cases?
Given the clarification that it starts and finishes at rest, I don't see how that can be a sensible case.hmmm27 said:1) constant velocity gets there in the least amount of time
Effectively turning it into this model:hmmm27 said:I think choosing a "case" means arbitrarily picking either "cv wins" or "acc/dec wins", then working backwards to state the condition(s) that makes it so.
...maybe.haruspex said:If we were given a max speed the answer would depend on the relationship between that, the acceleration a and the distance.
Ok, can I use ##s=v_0 t+\frac{1}{2}at^2##? And it will end up being ##s=\frac{1}{2}at^2## since ##v_0=0##?haruspex said:Pity that wasn't stated in the first place.
So turn that into some equations.
Unless there is also a max allowed speed, I can only think of one case.
Ok, but how are you using it, i.e. what is s here?Davidllerenav said:Ok, can I use ##s=v_0 t+\frac{1}{2}at^2##? And it will end up being ##s=\frac{1}{2}at^2## since ##v_0=0##?
Well, ##s## can't be ##L##, because it isn't accelerating all way through, it would be the distance before it starts desaccelerating. Whn it starts desaccelerating, it would be ##s=v_0 t-\frac{1}{2}at^2##, right?haruspex said:Ok, but how are you using it, i.e. what is s here?
Yes.Davidllerenav said:Well, ##s## can't be ##L##, because it isn't accelerating all way through, it would be the distance before it starts desaccelerating. Whn it starts desaccelerating, it would be ##s=v_0 t-\frac{1}{2}at^2##, right?
Ok. But what should I replace on the variables? Or should I just leave those two equations?haruspex said:Yes.
No, you must answer using the given variables L and a.Davidllerenav said:Ok. But what should I replace on the variables? Or should I just leave those two equations?
But isn't the second just a special case of the first?CWatters said:The only two cases I can think of are..
1) Accelerate, constant velocity, decelerate
2) Accelerate, decelerate with no constant velocity phase.
That doesn't work if the question intends what I think it does, as laid out in post #39.Davidllerenav said:Thanks, I did it by saying that the max velocity is when the velocity is constant. I showed it with the SUVAT equations, hope it is right.
But if it first goes with a constant acceleratio, then has constant velocity and then desaccelerate with constant acceleration, wouln't it be the max speed when the speed is constant?haruspex said:That doesn't work if the question intends what I think it does, as laid out in post #39.
With no acceleration, it can never start to move.
Yes, but the task it minimise the total time of travel. How should you apportion the time to the three stages?Davidllerenav said:But if it first goes with a constant acceleratio, then has constant velocity and then desaccelerate with constant acceleration, wouln't it be the max speed when the speed is constant?