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I guess that it would be when it is near B, because it is under constant acceleration.I think you can make a stab at that.
I guess that it would be when it is near B, because it is under constant acceleration.I think you can make a stab at that.
When what is near B? When A is near B? A and B are given, you can't move them.I guess that it would be when it is near B, because it is under constant acceleration.
When the object moving from A to B is near B, because it is under a constant acceleration.When what is near B? When A is near B? A and B are given, you can't move them.
It is a very simple question - if you want to get there ASAP, and your speed is limited only by the speed of light, what speed do you have to go at (as near as possible)?
Sorry, but I have no idea what you are trying to say.When the object moving from A to B is near B, because it is under a constant acceleration.
Yes.To answer your question, it would be as near as [possible to] the speed of light.
I'll try to explain myself better. The objecc is moving from A to B, starting at rest with a constant acceleration a until it reaches B. So the fastest the object will be travelling would be when it gets to point B.Sorry, but I have no idea what you are trying to say.
My comment about relativity and infinite speed was in the context of the constant speed case.The objecc is moving from A to B, starting at rest with a constant acceleration a
So, if in the case of constant speed, the object speed must be near the speed of light.My comment about relativity and infinite speed was in the context of the constant speed case.
For constant acceleration it gets a bit tricky. Not sure what that means within relativity.
Yes, depending on what exactly the question was intended to say.So, if in the case of constant speed, the object speed must be near the speed of light.
I really don't know that the questio was trying to say. So well, I guess that there arte those two cases.Yes, depending on what exactly the question was intended to say.
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?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.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.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.1) constant velocity gets there in the least amount of time
Effectively turning it into this model: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.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##?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?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?Ok, but how are you using it, i.e. what is s here?
Yes.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?Yes.
No, you must answer using the given variables L and a.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?The only two cases I can think of are..
1) Accelerate, constant velocity, decelerate
2) Accelerate, decelerate with no constant velocity phase.