# What is the maximum speed that an object must reach to satisfy minimum transit time?

#### Davidllerenav

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.

#### haruspex

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I guess that it would be when it is near B, because it is under 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)?

#### Davidllerenav

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)?
When the object moving from A to B is near B, because it is under a constant acceleration.
To answer your question, it would be as near as the speed of light.

#### haruspex

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When the object moving from A to B is near B, because it is under a constant acceleration.
Sorry, but I have no idea what you are trying to say.
To answer your question, it would be as near as [possible to] the speed of light.
Yes.

#### Davidllerenav

Sorry, but I have no idea what you are trying to say.
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.

#### haruspex

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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

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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

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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

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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

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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

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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

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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

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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

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

#### haruspex

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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

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The only two cases I can think of are..

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

#### haruspex

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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

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

"What is the maximum speed that an object must reach to satisfy minimum transit time?"

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