Can someone help me out explaining acceleration to a friend?

[RGUN]

You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?

 

[GasX]

Originally posted by: RGUN
Im sorry to disappoint but I indeed am a Mechanical Engineer.

I dont know how else to explain this to you.... Just, look at all the forces in the system, without an impulse you CANNOT just go from accelerating to not, you have to have a linear (or parabolic) transition. There are no impulses!! The removal of the engines force will mean that it is no longer able to sustain its acceleration oppoising drag and the acceleration will begin to decline until it is zero.

This demonstrates the difference between book smarts and street smarts. The slope of the curve you are talking about is, from a practical perspective, vertical. While technically correct, you are actually wrong.

 

[RGUN]

Originally posted by: linen
For me thinking about it this way clears up the argument: acceleration is the second derivative of position, velocity is the first derivative of position, therefore acceleration is the first derivative of velocity....i.e. the rate of change of velocity [(m/s)/s]. When the clutch is engaged, and assuming (rightly) there are forces attempting to slow the bike down (i.e. friction in wheels and air resistance, etc.), the velocity continues to go up for a time, but the rate of change of velocity (i.e. acceleration) goes down, and therefore the bike is decelerating.

Correct, except its not referred to as deceleration until velocity decreases, it is still acceleration but just in a diminishing amount.

 

[jdoggg12]

Don't think of it as letting off the gas, but pulling in the clutch. That will help you get around your questions about what the engine is doing.

My statement to him was this:

"The bike will never go faster than it was going when the clutch was pulled. Think of it like this - does a baseball keep speeding up once it has left the pitchers hand?

(hint: no, it does not)

The INSTANT you disengage the clutch, the motor is no longer pushing you (force). Another example - a train is pulling a car. It is accelerating, and while it is accelerating you pull the pin out. Does the car go faster than the train that was PULLING it? No, even though the train was accelerating, when the pin was pulled (equivalent to engaging the clutch in your example) the force accelerating the car stops, and it immediately stops accelerating positively, and begins to slow down.

 

[MrPickins]

Originally posted by: RGUN
You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?

You seem to be confusing velocity with acceleration.

I highly doubt you are a mechanical engineer, and if by some chance you are, I want a list of things you have designed so I know what I should avoid.

 

[SVT Cobra]

Originally posted by: RGUN
You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?

No, you see, you don't suddenly stop, you decelerate. Yes your velocity continues but the rate at which your velocity is will never increase (acceleration). People not being able to image this in their own head are being a pain.

This is of course that mass times velocity(with respect to position) is less than the friction in the drive train.

 

[RGUN]

Originally posted by: Mwilding
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
Im sorry to disappoint but I indeed am a Mechanical Engineer.

I dont know how else to explain this to you.... Just, look at all the forces in the system, without an impulse you CANNOT just go from accelerating to not, you have to have a linear (or parabolic) transition. There are no impulses!! The removal of the engines force will mean that it is no longer able to sustain its acceleration oppoising drag and the acceleration will begin to decline until it is zero.

</end quote></div>This demonstrates the difference between book smarts and street smarts. The slope of the curve you are talking about is, from a practical perspective, vertical. While technically correct, you are actually wrong.

Except for the fact that by you saying that Im wrong, meaning that it is not mearly a vertical line and that it HAS to have some time component proves me right....

 

[waggy]

Originally posted by: RGUN
You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?

i want to be 100% sure on what you are saying.

so saying you are accelerting to 100mph. at 100 acceleration stops. since accelration stops you stop at zero mPH?

 

[SVT Cobra]

Originally posted by: RGUN
<div class="FTQUOTE"><begin quote>Originally posted by: linen
For me thinking about it this way clears up the argument: acceleration is the second derivative of position, velocity is the first derivative of position, therefore acceleration is the first derivative of velocity....i.e. the rate of change of velocity [(m/s)/s]. When the clutch is engaged, and assuming (rightly) there are forces attempting to slow the bike down (i.e. friction in wheels and air resistance, etc.), the velocity continues to go up for a time, but the rate of change of velocity (i.e. acceleration) goes down, and therefore the bike is decelerating.</end quote></div>


Correct, except its not referred to as deceleration until velocity decreases, it is still acceleration but just in a diminishing amount.

no the second you let off the gas it is deceleration because of friction etc, as your velocity is not going to remain constant except for that slight second when you let off the gas, and your velocity will never increase after, one a level plane.


Also we need to clarrify, acceleration can be positive and negative, start specifying people.

 

[Zenmervolt]

Originally posted by: RGUN
I suspect you havent taken physics, Im a B. Eng with a specialization in solid mechanics.... This is my area of knowledge

When you get a job, let us know where so that we are never, ever using a product/machine/building for which you were on the design team.

ZV

 

[SVT Cobra]

Quote this is you agree the RGUN's is full of shit about being an engineer.

 

[RGUN]

Originally posted by: SVT Cobra
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?</end quote></div>

No, you see, you don't suddenly stop, you decelerate. Yes your velocity continues but the rate at which your velocity is will never increase (acceleration). People not being able to image this in their own head are being a pain.

This is of course that mass times velocity(with respect to position) is less than the friction in the drive train.

first of all, you dont decelerate, deceleration at least to me implies a reduction in velocity. Your velocity continues to increase (true, not at the same rate, which is why I have said all along you will continue to accelerate but at a decreasing rate) Accelerating at 10m/s^s then accelerating at 2m/s^2 is not decelerating, its just accelerating at a smaller rate.

 

[waggy]

Originally posted by: SVT Cobra
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
<div class="FTQUOTE"><begin quote>Originally posted by: linen
For me thinking about it this way clears up the argument: acceleration is the second derivative of position, velocity is the first derivative of position, therefore acceleration is the first derivative of velocity....i.e. the rate of change of velocity [(m/s)/s]. When the clutch is engaged, and assuming (rightly) there are forces attempting to slow the bike down (i.e. friction in wheels and air resistance, etc.), the velocity continues to go up for a time, but the rate of change of velocity (i.e. acceleration) goes down, and therefore the bike is decelerating.</end quote></div>


Correct, except its not referred to as deceleration until velocity decreases, it is still acceleration but just in a diminishing amount.
</end quote></div>

no the second you let off the gas it is deceleration because of friction etc, as your velocity is not going to remain constant except for that slight second when you let off the gas, and your velocity will never increase after, one a level plane.


Also we need to clarrify, acceleration can be positive and negative, start specifying people.

while true acceleration can be positive or negative. but thats just nit picking.

 

[SVT Cobra]

Originally posted by: RGUN
<div class="FTQUOTE"><begin quote>Originally posted by: SVT Cobra
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?</end quote></div>

No, you see, you don't suddenly stop, you decelerate. Yes your velocity continues but the rate at which your velocity is will never increase (acceleration). People not being able to image this in their own head are being a pain.

This is of course that mass times velocity(with respect to position) is less than the friction in the drive train.</end quote></div>

first of all, you dont decelerate, deceleration at least to me implies a reduction in velocity. Your velocity continues to increase (true, not at the same rate, which is why I have said all along you will continue to accelerate but at a decreasing rate) Accelerating at 10m/s^s then accelerating at 2m/s^2 is not decelerating, its just accelerating at a smaller rate.

Wow you are dense. No the point you let off the gas is t=0, from there it would be -8m/s/s in acceleration in your logic.

And, Zen don't worry this guy won't get an engineer job he's full of crap.

 

[linen]

<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
<div class="FTQUOTE"><begin quote>Originally posted by: linen
For me thinking about it this way clears up the argument: acceleration is the second derivative of position, velocity is the first derivative of position, therefore acceleration is the first derivative of velocity....i.e. the rate of change of velocity [(m/s)/s]. When the clutch is engaged, and assuming (rightly) there are forces attempting to slow the bike down (i.e. friction in wheels and air resistance, etc.), the velocity continues to go up for a time, but the rate of change of velocity (i.e. acceleration) goes down, and therefore the bike is decelerating.</end quote></div>


Correct, except its not referred to as deceleration until velocity decreases, it is still acceleration but just in a diminishing amount.
</end quote></div>


Yes...i agree with your statement. I mispoke, deceleration will not occur until after peak in velocity occurs.

 

[SVT Cobra]

Originally posted by: linen
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
<div class="FTQUOTE"><begin quote>Originally posted by: linen
For me thinking about it this way clears up the argument: acceleration is the second derivative of position, velocity is the first derivative of position, therefore acceleration is the first derivative of velocity....i.e. the rate of change of velocity [(m/s)/s]. When the clutch is engaged, and assuming (rightly) there are forces attempting to slow the bike down (i.e. friction in wheels and air resistance, etc.), the velocity continues to go up for a time, but the rate of change of velocity (i.e. acceleration) goes down, and therefore the bike is decelerating.</end quote></div>


Correct, except its not referred to as deceleration until velocity decreases, it is still acceleration but just in a diminishing amount.
</end quote></div>


Yes...i agree with your statement. I mispoke, deceleration will not until after peak in velocity occurs.

Wow you have multiple posts in this thread, yet you have 1 post?

 

[waggy]

Originally posted by: waggy
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?</end quote></div>

i want to be 100% sure on what you are saying.

so saying you are accelerting to 100mph. at 100 acceleration stops. since accelration stops you stop at zero mPH?

hey rgun can you please clerify this? is this what you are saying?

 

[linen]

Originally posted by: SVT Cobra
<div class="FTQUOTE"><begin quote>Originally posted by: linen
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
<div class="FTQUOTE"><begin quote>Originally posted by: linen
For me thinking about it this way clears up the argument: acceleration is the second derivative of position, velocity is the first derivative of position, therefore acceleration is the first derivative of velocity....i.e. the rate of change of velocity [(m/s)/s]. When the clutch is engaged, and assuming (rightly) there are forces attempting to slow the bike down (i.e. friction in wheels and air resistance, etc.), the velocity continues to go up for a time, but the rate of change of velocity (i.e. acceleration) goes down, and therefore the bike is decelerating.</end quote></div>


Correct, except its not referred to as deceleration until velocity decreases, it is still acceleration but just in a diminishing amount.
</end quote></div>


Yes...i agree with your statement. I mispoke, deceleration will not until after peak in velocity occurs.</end quote></div>

Wow you have multiple posts in this thread, yet you have 1 post?

Just getting started here, though I have perused topics here for years.

 

[RGUN]

Originally posted by: SVT Cobra
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
<div class="FTQUOTE"><begin quote>Originally posted by: SVT Cobra
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?</end quote></div>

No, you see, you don't suddenly stop, you decelerate. Yes your velocity continues but the rate at which your velocity is will never increase (acceleration). People not being able to image this in their own head are being a pain.

This is of course that mass times velocity(with respect to position) is less than the friction in the drive train.</end quote></div>

first of all, you dont decelerate, deceleration at least to me implies a reduction in velocity. Your velocity continues to increase (true, not at the same rate, which is why I have said all along you will continue to accelerate but at a decreasing rate) Accelerating at 10m/s^s then accelerating at 2m/s^2 is not decelerating, its just accelerating at a smaller rate.

</end quote></div>

Wow you are dense. No the point you let off the gas is t=0, from there it would be -8m/s/s in acceleration in your logic.

And, Zen don't worry this guy won't get an engineer job he's full of crap.

Excuse me? how will it be -8? At 2 m/s^2 your velocity will still increase.

 

[nakedfrog]

I bet RGUN is the OP's friend.

 

[Zenmervolt]

Originally posted by: RGUN
You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?

Acceleration is the result of force. Force is not the result of acceleration.

The motorcycle will stop gaining speed the instant the engine's force is taken away. It absolutely will not continue to speed up once the clutch is disengaged.

When the engine's force stops, acceleration forwards stops too.

A=F/M

Once F=0, then no matter what M is equal to, the equation F/M=0, and therefore A=0.

Your brick wall bit shows a complete lack of understanding since in your example, yes, the forward acceleration would stop, but the brick wall would impart a force against the vehicle causing severe backwards accleration, and thereby causing a large accident.

ZV

 

[SVT Cobra]

For those saying you still accelerate after letting off the gas/putting it in neutral (pushing in the clutch) you aren't even getting close with the convoluted reasons you are stating.

Ignoring going down a hill because that adds variables, if the momentum you have is greater than your drag and friction then yes you can still accelerate with no force, except since no forces are any longer acting on the vehicle and this is not a perfect world and drag etc increases with speed and mass, it would be impossible, because then theoretically you could have a perpetual motion machine.

 

[RGUN]

Originally posted by: waggy
<div class="FTQUOTE"><begin quote>Originally posted by: waggy
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?</end quote></div>

i want to be 100% sure on what you are saying.

so saying you are accelerting to 100mph. at 100 acceleration stops. since accelration stops you stop at zero mPH?

</end quote></div>


hey rgun can you please clerify this? is this what you are saying?

Im sorry, I dont understand the part Ive bolded

 

[waggy]

Originally posted by: RGUN
<div class="FTQUOTE"><begin quote>Originally posted by: waggy
<div class="FTQUOTE"><begin quote>Originally posted by: waggy
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?</end quote></div>

i want to be 100% sure on what you are saying.

so saying you are accelerting to 100mph. at 100 acceleration stops. since accelration stops you stop at zero mPH?

</end quote></div>


hey rgun can you please clerify this? is this what you are saying?


</end quote></div>


Im sorry, I dont understand the part Ive bolded

are you saying since the acceleration (force) of the motorcycle is zero the MPH the motorcycle is going goes to zero in the same instant? wich is not possible anyway.

 

[RGUN]

Originally posted by: waggy
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
<div class="FTQUOTE"><begin quote>Originally posted by: waggy
<div class="FTQUOTE"><begin quote>Originally posted by: waggy
<div class="FTQUOTE"><begin quote>Originally posted by: RGUN
You guys are really being a pain.... so at what exact instant are you telling me that you go from accelerating at whatever rate to not accelerating at all... do you realize how much force that would take and that during that time you would probably be subjected to many times the force of gravity?</end quote></div>

i want to be 100% sure on what you are saying.

so saying you are accelerting to 100mph. at 100 acceleration stops. since accelration stops you stop at zero mPH?

</end quote></div>


hey rgun can you please clerify this? is this what you are saying?


</end quote></div>


Im sorry, I dont understand the part Ive bolded
</end quote></div>

are you saying since the acceleration (force) of the motorcycle is zero the MPH the motorcycle is going goes to zero in the same instant? wich is not possible anyway.

No, you will start to decelerate until the point at which you have 0 velocity.