I am sorry. What you said about increasing velocity after the driving force is removed is just plain wrong. F=ma. No F, no A.
Steve
Easy physics question
- Thread starter sahtt
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Steve
You have made an incorrect assumption. You said that once the pitcher releases the ball, there is no more acceleration. That is simply untrue. The fact is that once the pitcher releases the ball, he is no longer imparting acceleration to the ball and so the acceleration will decrease. The ball does not immediately go into negative acceleration. Let's say he was accelerating the ball at 3 ft/s^2. As soon as he releases it, acceleration does not go to 0 ft/s^2, which is your assumption.
Velocity is a function of initial velocity, time, and acceleration.
vf = vo*t + 0.5*a*t^2
Unless acceleration drops immediately below zero (it doesn't, unless you hit a wall), you will have a momentary increase in velocity as acceleration starts to decrease from positive and down through to negative.
I had a much longer explanation but I hit the "back" button by mistake and besides, that equation is much simpler.
Argue with the equation, not me. Unless you can prove that acceleration immediately goes from "some positive value" to zero (or less), you're arguing with Newton, not me.
Bullet ballistics are a completely different animal as the speeds are so high that drag/friction are huge factors and they are still covered under my initial post, where I said that high speed means high drag, which means that velocity would not increase since drag would completely defeat residual positive acceleration almost immediately.
Regarding the ball in space or a vacuum, Newton's First Law says that in a vacuum, an object tends to maintain its state of motion unless acted on by an unbalanced force. So, it covers constant velocity (since constant velocity means all forces are balanced) but if you impart an acceleration to it, you have an unbalanced force. The moment you remove that force, it will SETTLE to a state of equilibrium. That doesn't mean it will stop accelerating positively immediately, but rather that at some time, will stop accelerating and maintain a state of equilibrium (constant velocity).
Force causes acceleration, not motion per se, and I think you're forgetting that. Remove the force and the object will have to come to an equilibrium. At slower speeds, drag and friction are not high enough to quickly stop the car from continuing to gain speed for a short period of time.
Look at the velocity charts of these cars on page 2:
http://www.roadandtrack.com/assets/download/0608_comp_chart.pdf
If you look at the initial portion of the graph, you'll see how hard the cars are accelerating.
I plotted the acceleration of the C63 AMG in Excel and got an equation for the acceleration of the car from 10 mph up to 120 mph. That equation is 9668.1*(v^(-1.8115)) where v is velocity.
You should recognize that as a decaying function, with acceleration nearing 1 mph per second by the time you hit 100 mph.
Up to 50 mph, you're still accelerating at a rate of nearly 10 mph per second. If you chop throttle, you still have to get rid of that 10 mph per second acceleration before you start to LOSE speed. You don't drop 10 mph per second immediately.
Here's a graph of the acceleration of the car plotted against velocity.
Argue with the data, not some abstract idea you've come up with. You make assumptions without an understanding of the physics equations behind it all. Prove it to me with math, not your "understanding".
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You guys are missing the fact that when you disengage the clutch, the car still has acceleration imparted to it. As I pointed out in the REAL WORLD DATA above, the car in question is accelerating at 10 mph per second at 50 mph. You have to SHED that acceleration before the car will slow down. Disengaging a clutch doesn't shed the acceleration immediately, it merely removes the ability for the car to SUSTAIN acceleration. The clutch doesn't turn acceleration on and off, it merely turns of the ability to keep accelerating positively.
The acceleration of the car at the point in time when the clutch is disengaged has to be dissipated (ie, reduced to zero) before the car will stop increasing its velocity.
Like I said, the math (and data) don't lie.
The acceleration of the car at the point in time when the clutch is disengaged has to be dissipated (ie, reduced to zero) before the car will stop increasing its velocity.
Like I said, the math (and data) don't lie.
I'm no Ph.D and I don't know the answer, but I see a potential fallacy in the analogies with the baseball pitcher and the gun. The baseball and bullet are not necessarily still accelerating when the projectiles are released as they may already be at a constant velocity at that point.
This assumes I'm defining acceleration as a positive rate of change in velocity, and velocity is simply the speed of the object.
The motorcycle example says the bike was still accelerating when the clutch lever is pulled. It would seem to me that if the velocity was increasing at the moment the clutch lever is pulled, momentum could allow the velocity to increase still even after the driving force is removed. Now if the velocity had been constant at the time of pulling the clutch, then no doubt the velocity will decrease immediately.
I understand your POV as well and that was my original interpretation before writing things down. I was thinking if you are accelerating at 10m/s and pull in the clutch, in order for you to instantly lose velocity [assuming not in vacuum], the car would have to instantly go from a positive 10m/s to a negative rate of acceleration which I'd assume would be quite violent for the rider in this case. I know if I accelerate very hard and pull in the clutch, it doesn't 'feel' as if the change is that drastic.
But the issue is at whether the acceleration decreases, yet remains positive for some time after pulling in the clutch. Mathematically I cannot support that claim. And when you referred to the 'hit the wall' analogy, I think you meant velocity instead of acceleration. I'll check out your data.
But the issue is at whether the acceleration decreases, yet remains positive for some time after pulling in the clutch. Mathematically I cannot support that claim. And when you referred to the 'hit the wall' analogy, I think you meant velocity instead of acceleration. I'll check out your data.
A is acceleration, not velocity, right? Two different concepts.
Hitting the wall would induce an immediate negative acceleration since you'd go from some speed to virtually none. I meant what I said.
As I said, with high velocity you tend to get high drag, which very often does cause a near instantaneous negative acceleration. However, since positive acceleration in most vehicles is also fairly low at such high speeds, the immediate change (delta) between the two is not that big, which means it doesn't feel like much at all. Except in a Champ Car with lots of aero drag (1.2g, I believe was the deceleration just from lifting off the throttle).
I've done dozens (hundreds) of 0-70+ mph runs in my S2000 at full throttle and I roll from 57 through to 60 mph while I'm making the 2-3 shift. The car cannot mechanically go 60 mph in 2nd gear (redline and gearing won't allow it) so it's definitely getting to 60 while I'm making the shift.
Exactly. Decreasing acceleration (or removal of positive sustained acceleration) does not equate to an immediate reduction in velocity unless other forces are quite high (as I mentioned, drag or friction being the big ones).
If it was possible to instantaneously disengage the clutch (which it isn't people) of you car when using the engine to accelerate, your car's acceleration would instantaneously change direction because of aero and tire drag. It doesn't matter how small those drag values are, if they are there your velocity will not remain constant.
A change in acceleration is called 'jerk'. In the grand scheme of things here we are talking about some very small accelerations. You do feel them, and the changes in them, and no they don't put you through the windshield because the forces generated here are small.
Any 'real world' data people are citing here with their cars either holding a constant velocity after disengaging the clutch or spontaneously speeding up are either errors in your speedo, not a fine enough resolution in the speedo, or from you riding the clutch a bit without thinking you are.
This is physics. This is thermodynamics. You cannot increase the energy of a system (car speeding up, or holding a constant velocity somewhere outside of a perfect vacuum) without adding energy to it. If you can't agree with that then you missed some VERY important lessons in your thermo classes.
This is great. Way to go Sahtt! I didn't think you could top the plane on a treadmill controversy but you are off to a great start. :tongue: :biggrin:
Just like with the plane on a treadmill problem, I really love hearing all the different variety of wild theories and postulates. Unfortunately most of them are based on gut feelings and instinct, but not grounded on technical foundations. Most of these questions are actually very simple problems in freshman level engineering courses (mostly dynamics).
As with most engineering/dynamics problems, this problem can best be explained by simplifying the problem into the core issue and then extrapolating a “real world” explanation. Let’s simplify the problem by eliminating all minor intangibles and say we are in a space (or a complete vacuum which space is not, but close enough). Therefore there are no frictional forces, drag, anything. If I have an object, say it is a ball of some given mass, then unless I apply some force on it, it will not move. It will just sit there. If I apply a force, say a booster rocket on the back of it, then the ball will begin to move. No matter how small of a force I apply, because there are no forces holding it back, the ball will not only increase in velocity but in acceleration too. In fact, all things held constant, the ball will approach infinite acceleration and infinite velocity. Now say at some point, I turn the rocket off. What happens next? Since there are no forces on the object, the acceleration immediately goes to zero and the velocity stays constant. The ball will stay at constant velocity forever, unless some force acts upon that object.
Now, let’s think about why it is impossible for the ball to have any acceleration or change in velocity after I remove any acting forces on the object. First of all it is important to understand that the only way acceleration is affected (changed) is if we import a force on that object. No force = no acceleration change. As we mentioned before, if there are no forces acting on the ball, it will not move. Or if the ball is already moving, then nothing will affect its velocity or acceleration without some external force. So if in our example, the only thing providing a force was the booster rocket, then the only thing providing acceleration is the actual force of the rocket. If I immediately turn off the rocket, what force is continuing to affect the acceleration of the ball? How would that ball gain speed? What force is pushing the ball to move faster? Likewise, how would that acceleration decrease? How does the ball continue to have acceleration (aka a force)? What would cause that acceleration to decrease? There are no other forces acting on that object, so why would acceleration decrease? It takes a force to generate a change in acceleration, but since there are absolutely no forces acting at all on the object, there can be no change in acceleration. The point being is once an object has no forces applied to it, it becomes a completely unchanging event. Meaning, there are no dynamics (changes to anything) once forces are eliminated.
Again, it is important to understand that acceleration can only be changed through a force. If there are no forces, there is no acceleration change. If I turn off the forces, there are no acceleration changes. If forces equal 0, then acceleration must 0.
In our real world example, engaging the clutch would essentially be the same and eliminating any forces acting on that object. However, there actually is a considerable amount of resistance forces due to friction, wind resistance etc. That is a force in the negative direction. So the reality is that the acceleration actually becomes negative quite quickly and that is why you should immediately start slowing down.
Just like with the plane on a treadmill problem, I really love hearing all the different variety of wild theories and postulates. Unfortunately most of them are based on gut feelings and instinct, but not grounded on technical foundations. Most of these questions are actually very simple problems in freshman level engineering courses (mostly dynamics).
As with most engineering/dynamics problems, this problem can best be explained by simplifying the problem into the core issue and then extrapolating a “real world” explanation. Let’s simplify the problem by eliminating all minor intangibles and say we are in a space (or a complete vacuum which space is not, but close enough). Therefore there are no frictional forces, drag, anything. If I have an object, say it is a ball of some given mass, then unless I apply some force on it, it will not move. It will just sit there. If I apply a force, say a booster rocket on the back of it, then the ball will begin to move. No matter how small of a force I apply, because there are no forces holding it back, the ball will not only increase in velocity but in acceleration too. In fact, all things held constant, the ball will approach infinite acceleration and infinite velocity. Now say at some point, I turn the rocket off. What happens next? Since there are no forces on the object, the acceleration immediately goes to zero and the velocity stays constant. The ball will stay at constant velocity forever, unless some force acts upon that object.
Now, let’s think about why it is impossible for the ball to have any acceleration or change in velocity after I remove any acting forces on the object. First of all it is important to understand that the only way acceleration is affected (changed) is if we import a force on that object. No force = no acceleration change. As we mentioned before, if there are no forces acting on the ball, it will not move. Or if the ball is already moving, then nothing will affect its velocity or acceleration without some external force. So if in our example, the only thing providing a force was the booster rocket, then the only thing providing acceleration is the actual force of the rocket. If I immediately turn off the rocket, what force is continuing to affect the acceleration of the ball? How would that ball gain speed? What force is pushing the ball to move faster? Likewise, how would that acceleration decrease? How does the ball continue to have acceleration (aka a force)? What would cause that acceleration to decrease? There are no other forces acting on that object, so why would acceleration decrease? It takes a force to generate a change in acceleration, but since there are absolutely no forces acting at all on the object, there can be no change in acceleration. The point being is once an object has no forces applied to it, it becomes a completely unchanging event. Meaning, there are no dynamics (changes to anything) once forces are eliminated.
Again, it is important to understand that acceleration can only be changed through a force. If there are no forces, there is no acceleration change. If I turn off the forces, there are no acceleration changes. If forces equal 0, then acceleration must 0.
In our real world example, engaging the clutch would essentially be the same and eliminating any forces acting on that object. However, there actually is a considerable amount of resistance forces due to friction, wind resistance etc. That is a force in the negative direction. So the reality is that the acceleration actually becomes negative quite quickly and that is why you should immediately start slowing down.
Remember, acceleration is the rate of velocity change, so it would have to be 10m/s/s or 10 m/s^2 (not 10 m/s). If you let off the clutch, the car wouldn't go from 10 m/s/s to -10 m/s/s, it would just go to 0 m/s/s and the velocity would stay the same m/s as you were going when you let off the clutch.
The key to realize is in this case is that you feel velocity change more. If you were accelerating hard to 100 mph, then let off the accelerator, your acceleration would go to 0, but your velocity would stay at 100 mph. You wouldn't really feel it as much as if you were accelerating to 100 mph and at 100 mph you hit a brick wall. Then your acceleration would again go straight to 0, but more importantly your velocity would go to zero too. You would feel that. Remember, hitting a brick wall does not mean your acceleration goes negative, it just means it goes to zero. However, the force associated with making your velocity go to zero over a very short period of time, would be quite substantial (impulse). Ouch! :tongue:
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He's right. I'm wrong.
Except that hitting a brick wall, your acceleration doesn't go to zero, it goes way below zero.
(vf - vo) / t = a very large negative number if t is very small (more or less instantaneous stop within 0.1 seconds or so)
I will burn my engineering degree in a symbolic ceremony tomorrow. Major brain fart today...
Except that hitting a brick wall, your acceleration doesn't go to zero, it goes way below zero.
(vf - vo) / t = a very large negative number if t is very small (more or less instantaneous stop within 0.1 seconds or so)
I will burn my engineering degree in a symbolic ceremony tomorrow. Major brain fart today...
Have to agree with slurpee and would add that sampling time would be dependent on counting pulses available. Too fast of a sampling time(low pulse count) would loose accuracy and give a jumpy indicated display while a long sampling time would give great accuracy but would be unable to keep up with a quick change in speed. As in any design its all a compromise.
Easy enough to prove that your speedometers are not keeping up and giving you the illusion that you may still be accelerating. Brake hard from say 60mph then let off and see if your speedometer is keeping up?
Have to keep you busy. I studied economics/finance in college and through my careers but science has always been extremely interesting to me. I took a bunch of astronomy classes that covered black holes, all the forces involved in supernovas of several varieties, the rest will sound like I'm just trying to show off.
I guess I have a bit of an "empty spot" for sciences since I didn't get to learn about them as much as I would have liked. I was able to take most of the math that corresponds with it so I'm usually able to at least wrap my head around it once someone who actually knows what they are talking about explains it.
I knew the 10m/s deal, I was saying accelerating at 10m/s [10m/s, then 20m/s, then accelerate at 30m/s, etc.] as 'slang' but your right, have to throw the '^2' in there technically.
He's right. I'm wrong.
Except that hitting a brick wall, your acceleration doesn't go to zero, it goes way below zero.
(vf - vo) / t = a very large negative number if t is very small (more or less instantaneous stop within 0.1 seconds or so)
I will burn my engineering degree in a symbolic ceremony tomorrow. Major brain fart today...
Nener, nener,neener
jk... love debates...
Soo, whats next? Creationism or darwinism? haha
He's right. I'm wrong.
Except that hitting a brick wall, your acceleration doesn't go to zero, it goes way below zero.
(vf - vo) / t = a very large negative number if t is very small (more or less instantaneous stop within 0.1 seconds or so)
I will burn my engineering degree in a symbolic ceremony tomorrow. Major brain fart today...
Sorry to be Debbie Downer again, but I think I may have caught you again. Remember, this is no longer a dynamics equation but a collision equation. Dynamics describes the motion of objects, collisions describe the interaction of object in collision (energy). You are still applying Newton’s law of motion, when you should be looking at it in terms of collisions and conservation of energy.
Take two masses of equal size and mass heading toward each other at the exact same velocity (like two billiard balls heading toward each other at the exact same speed). If it is a perfect collision what happens? They will hit each other with the exact same force and stop and not move. If you look at the free body diagram for one of the objects, it is moving with momentum p=mv. When it stops, it is not because it has a force decelerating the ball (like a booster rocket pushing in the opposite direction). The ball stops because there is an external force that pushes on it. That force is of the other ball heading in the opposite direction and it is THAT mass and THAT velocity that stops the ball. That is an important distinction because in one case the object itself generates a force to stop (negative acceleration), while in the other case, an external force causes the object to stop. If I were to somehow use the motorcycle brakes to stop instantaneously then I would agree, the acceleration forces would be VERY negative. But hitting a brick wall is different in that it exerts and equal and opposite reaction force that nullifies the force and momentum.
However, you were right in that I did make a very egregious mistake in one of my statements above. When I spoke of the example of the object in space with a booster rocket, I said that given a constant force the velocity and acceleration will approach infinity. That is wrong! The velocity would approach infinity but the acceleration would remain a constant number. Doh! I guess that was my major brain fart for the day! :biggrin:
:biggrin: :biggrin: Yeah, the sad thing is I've been putting in 60-70 hour weeks and this is the last thing I should be doing. But I just can't help myself! :tongue:
And if it makes you feel any better, I'd say just based on what I've seen so far on the forums, you have a better understanding of engineering sytems and physics than half the engineers I've seen. Maybe you missed your true calling in life, unless you are already kicking ass in the business world. :tongue:
At the risk of making a bigger fool of myself, I have to disagree. What causes the vehicle to stop (external or internal force) is beside the point and not at all a point that I was arguing.
The vehicle is in motion and hits a brick wall. Velocity drops from some positive number to zero. That means that you get negative acceleration (per the equation I listed earlier). You cannot experience a change in velocity without also experiencing an acceleration. You're saying that there is no acceleration when it hits the wall. There is no NET acceleration between the car and the wall (as a system) but the car itself does experience a (negative) acceleration.
Seems to me you're looking at the wall and car (or two billiard balls) as a system and saying that since there is no net force, there is no acceleration. I'm just looking at the car. From the standpoint of the car, the car will experience negative acceleration because it drops in speed. What's causing that deceleration is not my point (and is not what I argued).
That said, are we far enough off topic yet?
Not so fast my friend. :smile: This is the key and I think you were able to prove my point. I am just looking at the car and that is why I say the acceleration goes to zero. Think of the force of the car as F1 = M1 x A1. When the car or motorcycle hits the wall it meets an equal and opposite force (F2 = M2 x A2) and like you said, the net force is zero which is why the car and wall both don’t move. But the key is that A1 doesn’t go negative, it goes to zero because the motion of the car stops (delta t equals zero, velocity goes to zero, acceleration goes to zero). The fact that it was an external force F2 = M2 x A2 is important. That is because it (A2) does not describe the acceleration of M1, it is a separate object. More importantly, there technically isn’t even an A2. The wall is stationary, it isn’t moving so therefore it should have no acceleration right? Therefore there should be no force right? No, remember, there are many types of forces other than motion forces. Gravity, friction, normal forces are all forces that don’t may not have an acceleration component to them, yet exert a force. The wall exerts a NORMAL or reactionary force back on the car (Newton's Third Law), which happens to be equal to F1, which happens to be equal to M1 x A1, but in the negative direction. We are all accelerating bodies. All day long we exert a force due to gravity (F=mg). But we aren’t moving because the earth pushes back up on us with a NORMAL force. That force happens to be equal to F=mg in the opposite direction and that is why I don’t move. But it doesn’t have to equal just that. It will push back as hard as I push on it, so if I jump out of a plane and then strap a rocket to my back and slam into the earth, with the force of gravity (F=mg) plus and additional thrust amount (F-rocket) then the earth will push back up on me with F + F-rocket. The net result is that I hit the ground and stay stationary. At no time do I ever experience a negative acceleration. In fact I still have a positive acceleration of gravity (less rocket) being applied on my dead carcass.
So in the case of the car, suppose I push it with a positive constant force. We can agree the car sees a positive acceleration, which is why it starts to move, right? Ok, say it slams into a wall, but say I'm still applying the positive force, what happens? Well the car will stop, but will still see a positive force (acceleration) being applied to it, but will remain stationary because the wall is pushing back with a reactionary normal force (equal to what ever that pushing force is in the opposite direction) which is why the car will no longer move despite the fact I am still applying a positive force (and positive acceleration) on it.
Or, try this, slowly walk towards a wall. You are moving with a certan force right? Carefully, walk into a wall but do not break your stride. You are still walking with the same force but now you are not moving, why? Because the wall is applying an equal force in the opposite direction so that the net result is zero and so you don't move anywhere. At no time during your stride did you acceleration or force ever change, it was constant. It did not go negative, the only thing that changed was that now, a wall introduced an opposing force that equaled your positive force to negate it so that your net result is that you move nowhere. And it doesn't have to be a wall, it could be wind or even a treadmill. :biggrin: :tongue: Same results, you still have positive acceleration at all times, yet you stand still because an opposing force negates it.
So have I convinced you yet? :biggrin:
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Nope.
You had a delta in velocity (say, from 100 mph to 0 mph), ergo you had an acceleration. That's the basis of my argument.
Acceleration GOES to zero, yes, I agree. The car, once it hits the wall and has been reduced to zero movement, has zero acceleration. When it hits the wall, though, it has an immediate (assuming an unmovable wall) reduction in velocity, which means that it has an immediate and massive negative acceleration, after which the acceleration obviously GOES to zero (since the car is not moving, ergo there is no delta v, therefore no acceleration). However, for the moment of impact, the car goes from moving to not moving, ergo, it has a massive deceleration.
When a car hits a wall, it (and the driver) are subjected to a certain amount of g's (as is regularly reported for heavy crashes). Those are negative accelerations.
Delta t is near zero but delta V is not zero (rather, it's fairly large). That's where you made the mistake, I believe. Delta v GOES to zero, after a very brief time, but you're ignoring the very brief time where you go from 100 mph to, say, 0 mph in 0.1 seconds.
In short, in a heavy collision with a virtually unmovable object, delta v is big, delta t is small, and therefore a is very large.
Do the equation for the motion of the car:
a = (vf - vi) / t
vf = 0
vi = 100
t = 0.1
a = - 1000 mph / s
Simple, right? There's a delta v, ergo there's an acceleration.
I'm not arguing what acceleration GOES to (zero), only that to get there, it must first go very, very negative (and big).
Yes, "A" is acceleration, not velocity. In this case, there is no force so there is no acceleration. Please read my statement again as far as "increase" of velocity.
Steve
Please, to make this discussion meaningful, you need to reread your original post and have some basic highschool physics. Again what you said about the car is still accelerating AFTER the engine power is cut off by stepping in the clutch is just totally wrong (assuming the car is not going downhill). There is simply no room for argument.
Steve
I'm sorry but that is entirely incorrect. Yes, they are subjected to a certain amount of g's, but they are not negative accelerations. They are measured forces of the wall pushing back on the accelerometer. I'm not sure I can do a better job of explaining it. Hopefully you can get reconciliation on this from a physics teacher or engineering professor. You completely forgetting about non-motion based forces. Not all forces are due to F=ma; as I pointed out with NORMAL forces. A man pushing on an immovable wall is exerting a lot of force, yet neither the man nor the wall is moving. How can that be? If the only force that exists is F=ma and “a” (acceleration) is zero, then force must be zero too, right? And if you say that the wall is pushing back with an equal force, how is it possible that a wall has an acceleration?
Furthermore, if I threw a marble at a wall, how fast does it decelerate? I would stay pretty instantaneously. In fact the definition of a true undeformable object and an immovable object would mean that the marble would have stop instantly. If that were the case, then the delta T would be 0. If the delta T were 0, the acceleration would be infinite (any number divided by zero). Some given mass of the marble times an infinite acceleration would be an infinite force. That’s impossible.
See below picture.
What I’m saying:
1) Object moves with some force (F)
2) Object hits wall. Wall exerts an equal and opposite force equal to F.
3) All forces cancel out. Object and wall remain stationary.
Now you are either trying to say:
1) Object moves with some force (F)
2) Object hits wall. Both wall and ball exert force in negative direction? If that was the case then the ball should fly backwards.
Or you are trying to say:
1) Object moves with some force (F)
2) Object hits wall. Wall exerts an equal and opposite force equal to F.
3) Object then takes on negative acceleration resulting in negative force. If that is the case, again, object should move backwards.
Or:
1) Object moves with some force (F)
2) Object hits wall. Object takes on negative acceleration and force stopping its motion. But if that were the case, what force did the wall impart on the object? What caused the object to take on a negative acceleration?
Now, there is one more possible scenario, and that would be that the wall imparts a force on the object. So basically you have an object with a force and a negative force that is applied to it that is equal to it. Now in that case that imparted force is equal to F in the opposite direction. Since F equals “m x a” I can see how one would think that –F = m x –a, but that is incorrect. That force is simply a reactionary force that happens to equal “m x a” but it is not an acceleration. Again, think of people standing on the earth analogy. We are constantly accelerating towards the ground but the earth is pushing back up on us.
You said:
Delta t is not zero in the real world, so acceleration is very much finite (and manageable). In a perfect world, yes, acceleration would be infinite (as would the force)...but that goes for a lot of things in "perfect" situations. That doesn't mean that in the real world, the same situation results in something far less than infinity.
I don't see how you can say that there is a finite delta v in a finite delta t, yet have an acceleration equal to zero. That doesn't add up.
We were assuming the car hits the wall at 100 mph (no acceleration - constant velocity). Therefore, there is no net unbalanced force on the car (otherwise it would be accelerating) before it hits the wall. To stop the car, you must have an unbalanced force (from the car's perspective) to cause it to stop (ie, a deceleration). Let me say that again - you MUST have an unbalanced force on the car to cause the car to stop. If there is an unbalanced force, there is also an acceleration (deceleration).
Once the car is stopped, it is not accelerating because the forces are balanced and it is at equilibrium (obviously). So acceleration might GO to zero but it goes through a really big number first.
Again, you have a delta v and a finite delta t. There's no way you cannot have an acceleration (deceleration).
Use the momentum/impulse equality:
F*t = m*(vf-vi)
You'll find that there is a force on the car and therefore an acceleration.
Plug the numbers in:
t = 0.1 s (random small number I made up to represent how long it takes for the car to come to a rest)
m = 1500 kg (3300 lbs)
vf = 0
vi = 44.44 m/s (100 mph)
You'll get a force of -666,666.7 N on the car. That means the acceleration is -444 m/s^2, or roughly 45 g's.
Delta t is not zero in the real world, so acceleration is very much finite (and manageable). In a perfect world, yes, acceleration would be infinite (as would the force)...but that goes for a lot of things in "perfect" situations. That doesn't mean that in the real world, the same situation results in something far less than infinity.
I don't see how you can say that there is a finite delta v in a finite delta t, yet have an acceleration equal to zero. That doesn't add up.
We were assuming the car hits the wall at 100 mph (no acceleration - constant velocity). Therefore, there is no net unbalanced force on the car (otherwise it would be accelerating) before it hits the wall. To stop the car, you must have an unbalanced force (from the car's perspective) to cause it to stop (ie, a deceleration). Let me say that again - you MUST have an unbalanced force on the car to cause the car to stop. If there is an unbalanced force, there is also an acceleration (deceleration).
Once the car is stopped, it is not accelerating because the forces are balanced and it is at equilibrium (obviously). So acceleration might GO to zero but it goes through a really big number first.
Again, you have a delta v and a finite delta t. There's no way you cannot have an acceleration (deceleration).
Use the momentum/impulse equality:
F*t = m*(vf-vi)
You'll find that there is a force on the car and therefore an acceleration.
Plug the numbers in:
t = 0.1 s (random small number I made up to represent how long it takes for the car to come to a rest)
m = 1500 kg (3300 lbs)
vf = 0
vi = 44.44 m/s (100 mph)
You'll get a force of -666,666.7 N on the car. That means the acceleration is -444 m/s^2, or roughly 45 g's.
The moment the clutch is depressed Velocity decreases. Just look at a dyno graph. You can see it in the shift. Your speedometer is lagging in the readout.
John
John
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