Max Speed to Move from Point A to Point B

AI Thread Summary
The discussion revolves around determining the maximum speed an object must achieve to travel from point A to point B in the least time, given it can accelerate or move at a constant speed. Participants clarify that the object starts from rest, accelerates at a fixed rate, and must decelerate to stop at point B. They debate whether the problem allows for constant speed or if it should only involve acceleration and deceleration phases. The consensus is that basic kinematic equations can be used to solve the problem, focusing on two scenarios: constant velocity and symmetrical acceleration/deceleration. The conversation highlights the need for clear problem wording and assumptions about starting and finishing conditions.
Davidllerenav
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Hi, I need help with this problem:

Homework Statement


Condition: an object has to move from point A to point B in the least time possible. The distance between the points is L. The object can accelerate (decelerate) with a fixed acceleration ##a## or move with a constant speed.

What maximum speed does this object have to reach to satisfy the condition?

Homework Equations


I guess this one:
##v= \frac {ds}{dt}##

The Attempt at a Solution


If the object moves from A to B in the least time possible, that means that ##\Delta t## tends to ##0##. To find the maximum speed I need to find the moment when the object travels more dinstance in the least time. That would be the derivative of the distance ##L## with respect to ##t##, so ##lim_{\Delta t\to 0} \frac{\Delta L} {\Delta t} = \frac {dL} {dt}##. Am I right? The problem is that I don't know hot to find the maximum speed necessary. How do I find the maximum speed?
 
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Davidllerenav said:
The object can accelerate (decelerate) with a fixed acceleration to or move with a constant speed.
The question makes no sense to me. Is this the exact wording? Is it a translation?
 
haruspex said:
The question makes no sense to me. Is this the exact wording? Is it a translation?
It is a translation. It is ##a## instead of "to". So it would be "The object can accelerate (decelerate) with a fixed acceleration ##a## or move with a constant speed."
 
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Davidllerenav said:
It is a translation. It is ##a## instead of "to". So it would be "The object can accelerate (decelerate) with a fixed acceleration ##a## or move with a constant speed."
Any constant speed or a given one? Is relativity to be taken into account?
 
haruspex said:
Any constant speed or a given one? Is relativity to be taken into account?
No costant speed. What do you mean with relativity?
 
Davidllerenav said:
No costant speed. What do you mean with relativity?
Are we limited to the speed of light?
 
haruspex said:
Are we limited to the speed of light?
Relativity is definitely out of the picture here.
 
Davidllerenav said:
No costant speed.
Sorry, I do not understand that answer. We are told we can use the constant acceleration, a, or a constant speed. Or does it mean the object can switch between the two, having started from rest?
 
Davidllerenav said:
Yes. Relativity is definitely out of the picture here.
If relativity is not assumed then the answer is no, we are not limited to the speed of light.
 
  • #10
haruspex said:
Sorry, I do not understand that answer. We are told we can use the constant acceleration, a, or a constant speed. Or does it mean the object can switch between the two, having started from rest?
That's a good question. I guess that it meas that we can use the constant acceleration or the constant speed, since it doesn't say anithing about it starting from rest.
 
  • #11
Davidllerenav said:
That's a good question. I guess that it meas that we can use the constant acceleration or the constant speed, since it doesn't say anithing about it starting from rest.
I'm struggling to find an interpretation that makes for a reasonable question. It seems obvious just to say go at an infinite speed.

If the choice is between
(1) a given acceleration, a, starting from rest and
(2) a given constant speed v
then it will depend on the numeric relationship between a, v and L.

Maybe you have to start and finish at rest, and you can mix accelerating at a, decelerating at a, and moving at constant speed.
 
  • #12
haruspex said:
I'm struggling to find an interpretation that makes for a reasonable question. It seems obvious just to say go at an infinite speed.
Why is it obvious? Wouldn't it be a non-constant speed then?
haruspex said:
If the choice is between
(1) a given acceleration, a, starting from rest and
(2) a given constant speed v
then it will depend on the numeric relationship between a, v and L.

Maybe you have to start and finish at rest, and you can mix accelerating at a, decelerating at a, and moving at constant speed.
I really don't know, I guess that it can start at rest, but why woudl it finish at rest?
 
  • #13
Davidllerenav said:
Why is it obvious? Wouldn't it be a non-constant speed then?
If it does not have to start at rest then you can start at the dedired speed. If you rule out an infinite speed there is no answer, because whatever speed you pick you can do better by going faster.
Davidllerenav said:
why woudl it finish at rest?
I am suggesting that as a requirement in order to make it a reasonable question.
 
  • #14
I agree. Question as written is too obvious. To turn it into a sensible question you must assume it starts or finishes at rest and its most likely they meant you to assume both. So the trip must be divided into three parts...

Acceleration at a
Constant velocity
Deceleration at -aYou need to work out how long each part should be to make the trip in the shortest time, and write an equation for the Max velocity achieved.
 
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  • #15
haruspex said:
If it does not have to start at rest then you can start at the dedired speed. If you rule out an infinite speed there is no answer, because whatever speed you pick you can do better by going faster.

I am suggesting that as a requirement in order to make it a reasonable question.
What about if relativity is considered? If the object can't go with infinite speed.
 
  • #16
I would suggest clarification from the teacher. Like, lots of clarification and highly suggested.

Also, is the answer to be found using calculus? or would simple kinematic equations do.
 
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  • #17
hmmm27 said:
I would suggest clarification from the teacher. Like, lots of clarification and highly suggested.

Also, is the answer to be found using calculus? or would simple kinematic equations do.
I guess kinematics. Since the last topic we saw was Galilean transformation.
 
  • #18
hmmm27 said:
Also, is the answer to be found using calculus? or would simple kinematic equations do.
Whatever the intent, calculus would be overkill.
And you mean kinetics; kinematics is something else.
 
  • #19
haruspex said:
Whatever the intent, calculus would be overkill.
Unless it was a math, not physics, class.
And you mean kinetics; kinematics is something else.
o:) LOL, I was wondering about that the other day : apologies for adding to existing confusion.

Wait, what ?:wideeyed:
 
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  • #20
hmmm27 said:
Unless it was a math, not physics, class.

o:) LOL, I was wondering about that the other day : apologies for adding to existing confusion.

Wait, what ?
Kinematics is not concerned with masses or forces. It is sometimes described as the geometry of motion. E.g. in a mechanical linkage it addresses how the components are constrained to move in relation to each other.
 
  • #21
The OP's question appears to be easily solveable (once the missing bits are found) with "kinematics equations", found in Wikipedia or our own reference post : time, distance, velocity (and acceleration).
 
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  • #22
haruspex said:
Kinematics is not concerned with masses or forces. It is sometimes described as the geometry of motion. E.g. in a mechanical linkage it addresses how the components are constrained to move in relation to each other.
But I think that we need to use kinematics, because we haven't see dynamics yet. Kinectics and dynamics are the same, right?
 
  • #23
Davidllerenav said:
But I think that we need to use kinematics, because we haven't see dynamics yet. Kinectics and dynamics are the same, right?
I erred.
I am used to people referring to kinematics in the context of questions involving accelerations resulting from torques and forces. In the present case, we are unconcerned with such - it is purely a question of the relationships between position, velocity, acceleration and time.
 
  • #24
haruspex said:
I erred.
I am used to people referring to kinematics in the context of questions involving accelerations resulting from torques and forces. In the present case, we are unconcerned with such - it is purely a question of the relationships between position, velocity, acceleration and time.
Exactly. So, If the answer would be infinite speed with no relativity. If now we are limited by speed of light, what would be the answer?
 
  • #25
Davidllerenav said:
Exactly. So, If the answer would be infinite speed with no relativity. If now we are limited by speed of light, what would be the answer?
I think you can make a stab at that.
 
  • #26
haruspex said:
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.
 
  • #27
Davidllerenav said:
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)?
 
  • #28
haruspex said:
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.
 
  • #29
Davidllerenav said:
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.
Davidllerenav said:
To answer your question, it would be as near as [possible to] the speed of light.
Yes.
 
  • #30
haruspex said:
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 traveling would be when it gets to point B.
 
  • #31
Davidllerenav said:
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.
 
  • #32
haruspex said:
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.
 
  • #33
Davidllerenav said:
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.
 
  • #34
haruspex said:
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.
 
  • #35
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.
 
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  • #36
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.
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?
 
  • #37
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
Pity that wasn't stated in the first place.
So turn that into some equations.
Davidllerenav said:
Which are the two cases?
Unless there is also a max allowed speed, I can only think of one case.
 
  • #38
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.
 
  • #39
hmmm27 said:
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.
 
  • #40
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.
 
  • #41
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.
Effectively turning it into this model:
haruspex said:
If we were given a max speed the answer would depend on the relationship between that, the acceleration a and the distance.
...maybe.
 
  • #42
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, 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##?
 
  • #43
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##?
Ok, but how are you using it, i.e. what is s here?
 
  • #44
haruspex said:
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?
 
  • #45
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?
Yes.
 
  • #46
haruspex said:
Yes.
Ok. But what should I replace on the variables? Or should I just leave those two equations?
 
  • #47
Davidllerenav said:
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.
 
  • #48
The only two cases I can think of are..

1) Accelerate, constant velocity, decelerate
2) Accelerate, decelerate with no constant velocity phase.
 
  • #49
CWatters said:
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?
 
  • #50
Indeed. But it's the best I can come up with for "two cases".
 

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