Max Speed to Move from Point A to Point B

[haruspex]

The objecc is moving from A to B, starting at rest with a constant acceleration a

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.

 

[Davidllerenav]

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.

So, if in the case of constant speed, the object speed must be near the speed of light.

 

[haruspex]

So, if in the case of constant speed, the object speed must be near the speed of light.

Yes, depending on what exactly the question was intended to say.

 

[Davidllerenav]

Yes, depending on what exactly the question was intended to say.

I really don't know that the questio was trying to say. So well, I guess that there arte those two cases.

 

[CWatters]

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.

 

[Davidllerenav]

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.

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?

 

[haruspex]

it start ar rest and accelerates at a and at somep point it decelerate at a in order to be at rest at B

Pity that wasn't stated in the first place.
So turn that into some equations.

Which are the two cases?

Unless there is also a max allowed speed, I can only think of one case.

 

[hmmm27]

Not trying to be funny, but are you and the instructor both native speakers of the same language ?

The "two cases" are:

1) constant velocity gets there in the least amount of time
2) symmetrical acc/deceleration gets there in the least amount of time.

Pick one (or both) and state the condition - which includes the term "maximum velocity" - that makes it true.

 

[haruspex]

1) constant velocity gets there in the least amount of time

Given the clarification that it starts and finishes at rest, I don't see how that can be a sensible case.
Also, we've seen no mention of a maximum velocity, so dropping the start and finish at rest constraint for the constant speed case doesn't help.
If we were given a max speed the answer would depend on the relationship between that, the acceleration a and the distance.

This leaves the view that
- it starts and finishes at rest
- in between, any mix of accelerating at a, decelerating at a, and constant speed
But I count that as one case, not two.

 

[hmmm27]

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.

 

[haruspex]

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.

Effectively turning it into this model:

If we were given a max speed the answer would depend on the relationship between that, the acceleration a and the distance.

...maybe.

 

[Davidllerenav]

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, 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]

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 $v_0=0$?

Ok, but how are you using it, i.e. what is s here?

 

[Davidllerenav]

Ok, but how are you using it, i.e. what is s here?

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]

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?

Yes.

 

[Davidllerenav]

Yes.

Ok. But what should I replace on the variables? Or should I just leave those two equations?

 

[haruspex]

Ok. But what should I replace on the variables? Or should I just leave those two equations?

No, you must answer using the given variables L and a.
Use your two equations to find out when you have to switch to decelerating.

 

[CWatters]

The only two cases I can think of are..

1) Accelerate, constant velocity, decelerate
2) Accelerate, decelerate with no constant velocity phase.

 

[haruspex]

The only two cases I can think of are..

1) Accelerate, constant velocity, decelerate
2) Accelerate, decelerate with no constant velocity phase.

But isn't the second just a special case of the first?

 

[CWatters]

Indeed. But it's the best I can come up with for "two cases".

 

[Davidllerenav]

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.

 

[haruspex]

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.

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.

 

[Davidllerenav]

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.

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]

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?

Yes, but the task it minimise the total time of travel. How should you apportion the time to the three stages?