physics concept quesiton

[Schfifty Five]

Originally posted by: ledjani
Even though all the feedback, I am going to stick with my answer being 0. When the original poster gets his exam back, I would appreciate it, if he could tell us the answer so we can put an end to this.

Why would you stick with your answer of 0?

The force of gravity is constantly pulling the object back down to earth...even from the very moment it went up away from earth.

For example, let's say the particle's velocity at the beginning of it leaving earth's ground is 50 m/s, and gravity is pulling down 9.8m/s^2, then after a few seconds, the velocity will be 0. When velocity is 0, that is when the particle reached the highest point of it's journey away from earth. After that, it falls back down to earth gaining velocity all the way until it hits earth again.

 

[DanFungus]

jesus christ. Everyone who says anything other than 9.8/-9.8 (because of the magnitude part of the question) needs to read this

 

[Cattlegod]

Originally posted by: DanFungus
jesus christ. Everyone who says anything other than 9.8/-9.8 (because of the magnitude part of the question) needs to read this

check out the awesome animation

 

[eits]

Originally posted by: blueshoe

Originally posted by: bonkers325

Originally posted by: Mo0o

Originally posted by: bonkers325
would have to be D, since it is the only negative quantity available. you are given no other infomation, so applying formulas and the such would be useless. if the acceleration were not negative, the particle would never stop.

it asks for magnitude, which is a positive value

if the acceleration is anything but negative, then the particle would never reach a -true- maximum height.

if acceleration > 0, then the particle will always be increasing in speed - this means that it will never reach maximum height.

if acceleration < 0, then the particle will reach a velocity of 0 m/s at some point in time, where it will begin to fall downward. that point in time (t=2s) is the apex which is the maximum height.

You are correct, the acceleration is negative. The magnitude of the acceleration is positive.

you're right. i was wrong. good point.

 

[clickynext]

The question is too vague to answer. Where is the particle? Are we taking about the net acceleration of the particle, and with respect to what? If the particle is on earth, are there any other forces acting on it, and which direction is regarded as positive in this particular case?

 

[Yossarian]

Originally posted by: clickynext
The question is too vague to answer.

No it isn't.

Where is the particle?

since the answers are in multiples of g, it's safe to assume it's the earth and not the fvcking phantom zone.

Are we taking about the net acceleration of the particle, and with respect to what?

Yes. The earth.

If the particle is on earth, are there any other forces acting on it, and which direction is regarded as positive in this particular case?

No. It doesn't matter, since the question asks the magnitude.

This is a simple question with a simple answer. Either you understand basic physics or you don't.

 

[Eeezee]

This is an intro physics test. The question is VERY easy. You are not supposed to overcomplicate it like this.

Everyone that answered 0 got the question wrong. Around this time of the semester the professor is likely going over very basic concepts like kinematics and gravitational acceleration. That's probably the only kind of natural force that these students even know about, and they haven't even touched on Newton's gravitational law for any 2 particles. I would bet a lot of money on the professor having covered gravity on Earth and not much else.

The particle started with an upward velocity and eventually came to v = 0. It is safe to assume that there is an acceleration in this case. It is safe to assume that the particle is on Earth and the acceleration is due to gravity (9.8 m/s^2 down).

 

[Wnh5001]

yah, this is mechanics, we went over motion in 1d, 2d, vectors, newtons 3 laws, thats about it =|. were doin more forces now, that relate to friction, drag force and terminal speed.

 

[clickynext]

Originally posted by: Yossarian

Originally posted by: clickynext
The question is too vague to answer.

No it isn't.

Where is the particle?

since the answers are in multiples of g, it's safe to assume it's the earth and not the fvcking phantom zone.

Are we taking about the net acceleration of the particle, and with respect to what?

Yes. The earth.

If the particle is on earth, are there any other forces acting on it, and which direction is regarded as positive in this particular case?

No. It doesn't matter, since the question asks the magnitude.

This is a simple question with a simple answer. Either you understand basic physics or you don't.

Maybe your experience was different, but in my first physics class our teacher loved to use questions where you would feel compelled to make assumptions (which was always inappropriate, because we were told to assume that enough information is given to properly answer the question without any doubt), and the correct answer would sometimes be "not enough information". If you need to make assumptions about things which are not specified in the question, then it is a bad question. And what's wrong with gravity being a multiple of g on some other mass in space? I've seen plenty of questions that tell you to assume gravity = 0.25g on the moon's surface, 2g on the planet tralfamadore, etc.

 

[yhelothar]

Originally posted by: Titan

Originally posted by: Mo0o

Originally posted by: Titan
Eeezee, I know all this. You don't have to give me remedial physics. I have condeded that 9.8 looks ok on paper. But I honestly ask you what use is that information to an engineer, or anybody for that matter? I look for the deeper meaning, which has to do with forces. It tells us there is a force we need to know about.

Or to answer your other statement, how do you measure acceleration? Predict it with a formula, or use an instrument?

I'm looking at the big picture, and remembering the fundamentals of how science works.

I say the question is ambiguous. But people are, so you need to know them.

Given all this "deep" physics talk, it's surprising you gave such a terrible initial answer of the acceleration somehow being zero because velocity is zero

In your judgment it's "terrible." In mine it's practical. I'm not judging your answer. We looked at it from very different contexts. I picture someone with high-speed camera measuring an object determining its acceleration. You picture the equation. We disagree on what the question as really asking. For a physics class where it usually boils down to numbers, i'll admidt you probably got the answer "right." Note the original post, which is what I read, did not specify this was for a class. It could have been Mythbusters. But out in the "real" world where science is determined by measurements, I am thinking of people measuring n object, and that could be what the word acceleration describes.

As many have pointed out, this is not a math question, but a concepts question. You do not understand the concept of this problem. When an object is in the air, the only force acting upon it is gravity, in the downwards direction. When there is a force applied to an object, the object will accelerate. PERIOD. Acceleration is directly related to force. If you had graphs of acceleration and force, they would look identical in shape. So you cannot have force without acceleration.
It's like you have a bowling ball rolling in a direction, and you tap it on the opposite direction with a consistent amount of force. The ball will begin to accelerate in the other direction. When it first starts accelerating, the velocity of the ball will begin to decrease, then pass 0, and begin moving in the opposite direction. The acceleration is constant on the ball. You don't say that when the velocity of the ball reached zero, the force stopped, and the acceleration is at 0. The force and acceleration is still constant.

 

[imported_Imp]

Think some of you have already gotten the general idea, but here's I go.

Acceleration is always -9.81 on an object. If the object was thrown, it stopped accelerating the instant it left the hand in the upward direction. ~Same can be said for other propellants for the most part, except for guns and other air stuff...too complicated for me.~ Assuming we're neglecting air resistance, no other forces act on said particle but gravity. Bla bla bla, most of you are probably thinking about v=0 when you say acceleration is zero, I give you Sum Fy=mg and v2 = v1 + at. I barely passed Dynamics in Uni, so someone else sum up the dynamic forces, I'll stick to my statics. Anyways, using the second equation, v2=0 at max height, and the only way that happens is if the net acceleration on the object is negative, which would also imply that it would be possible for an upward force <+9.81 to be acting on the object and I just lost it.

 

[ledjani]

I am sticking with my answer at 0 because when it reaches a maximum velocity it isn't accelerating or decelaring yet. Thus it is 0. The question asks for the acceleration, not for the force. I think people are confusing it with the forces on the object rather than the acceleration the ball is going.

 

[Mo0o]

Originally posted by: ledjani
I am sticking with my answer at 0 because when it reaches a maximum velocity it isn't accelerating or decelaring yet. Thus it is 0. The question asks for the acceleration, not for the force. I think people are confusing it with the forces on the object rather than the acceleration the ball is going.

It doesn't reach a maximum or minimum velocity. The velocity graph of the particle isn't parabolic or curved even. If you look at the velocity graph of the particle in flight, it's a linear equation. When you take the derivative of that equation you get a nonzero number.

 

[jagec]

Originally posted by: ledjani
I am sticking with my answer at 0 because when it reaches a maximum velocity it isn't accelerating or decelaring yet. Thus it is 0. The question asks for the acceleration, not for the force. I think people are confusing it with the forces on the object rather than the acceleration the ball is going.

Force IS acceleration for any object with mass. Unless the "particle" he is talking about is a photon, it's the same deal, yo. Remember, F=MA

However, since units are not specified, the answer could be anything except zero.

 

[Yossarian]

Originally posted by: ledjani
I am sticking with my answer at 0 because when it reaches a maximum velocity it isn't accelerating or decelaring yet. Thus it is 0. The question asks for the acceleration, not for the force. I think people are confusing it with the forces on the object rather than the acceleration the ball is going.

if there is a net force acting on something, there is acceleration regardless of the velocity.

 

[IAteYourMother]

ouch, magnitude. 9.81

 

[IAteYourMother]

Originally posted by: ledjani
I am sticking with my answer at 0 because when it reaches a maximum velocity it isn't accelerating or decelaring yet. Thus it is 0. The question asks for the acceleration, not for the force. I think people are confusing it with the forces on the object rather than the acceleration the ball is going.

have you taken physics?

 

[silverpig]

Originally posted by: Mo0o

Originally posted by: ledjani
I am sticking with my answer at 0 because when it reaches a maximum velocity it isn't accelerating or decelaring yet. Thus it is 0. The question asks for the acceleration, not for the force. I think people are confusing it with the forces on the object rather than the acceleration the ball is going.

It doesn't reach a maximum or minimum velocity. The velocity graph of the particle isn't parabolic or curved even. If you look at the velocity graph of the particle in flight, it's a linear equation. When you take the derivative of that equation you get a nonzero number.

Uh, yeah it does. It starts off at a max_vel, goes down linearly to 0 and then continues down to -max_vel.

Height wrt time = parabola
Velocity wrt time = line
Acceleration wrt time = horizontal line at 9.81

 

[Eeezee]

Originally posted by: clickynext

Originally posted by: Yossarian

Originally posted by: clickynext
The question is too vague to answer.

No it isn't.

Where is the particle?

since the answers are in multiples of g, it's safe to assume it's the earth and not the fvcking phantom zone.

Are we taking about the net acceleration of the particle, and with respect to what?

Yes. The earth.

If the particle is on earth, are there any other forces acting on it, and which direction is regarded as positive in this particular case?

No. It doesn't matter, since the question asks the magnitude.

This is a simple question with a simple answer. Either you understand basic physics or you don't.

Maybe your experience was different, but in my first physics class our teacher loved to use questions where you would feel compelled to make assumptions (which was always inappropriate, because we were told to assume that enough information is given to properly answer the question without any doubt), and the correct answer would sometimes be "not enough information". If you need to make assumptions about things which are not specified in the question, then it is a bad question. And what's wrong with gravity being a multiple of g on some other mass in space? I've seen plenty of questions that tell you to assume gravity = 0.25g on the moon's surface, 2g on the planet tralfamadore, etc.

No, it's a bad question when you need to specify "neglect relativistic effects, neglect the Lorentz Force Law, neglect air friction, neglect electromagnetic momentum" etc.

It's a fair assumption to say that the particle is on Earth. Obviously this professor is not the same guy that you had, otherwise there would be a response that is "There is not enough information"

Regardless, the acceleration is not 0

 

[Eeezee]

Originally posted by: ledjani
I am sticking with my answer at 0 because when it reaches a maximum velocity it isn't accelerating or decelaring yet. Thus it is 0. The question asks for the acceleration, not for the force. I think people are confusing it with the forces on the object rather than the acceleration the ball is going.

The particle is not losing mass, therefore Force is directly proportional to acceleration. Hell, you may as well say that it's a 1kg particle and equate the two.

 

[coomar]

wind resistance wouldn't matter at the apex, the acceleration is 9.8 m/s^2 downwards, as its falling wind resistance might play a role

the question is a little vague but it implies that there is an acceleration, reading the question you see that it mentions that the particle arrives at the apex at 2 seconds implying that it wasn't there in the neighborhood of 2 seconds so it clearly has an acceleration

magnitudes are always positive since they are the distance from the origin so the answer would be b

 

[Eeezee]

Wind resistance wouldn't matter at the apex, you are correct. I'm just saying that it's not that difficult of a question

 

[mobobuff]

The plane would definitely fly.


Oops...

 

[Mo0o]

Originally posted by: silverpig

Originally posted by: Mo0o

Originally posted by: ledjani
I am sticking with my answer at 0 because when it reaches a maximum velocity it isn't accelerating or decelaring yet. Thus it is 0. The question asks for the acceleration, not for the force. I think people are confusing it with the forces on the object rather than the acceleration the ball is going.

It doesn't reach a maximum or minimum velocity. The velocity graph of the particle isn't parabolic or curved even. If you look at the velocity graph of the particle in flight, it's a linear equation. When you take the derivative of that equation you get a nonzero number.

Uh, yeah it does. It starts off at a max_vel, goes down linearly to 0 and then continues down to -max_vel.

Height wrt time = parabola
Velocity wrt time = line
Acceleration wrt time = horizontal line at 9.81

Ah yes yes, you woud have a maximum velocity and assuming a flat surface the final is -max_vel. You're right.

 

[Eeezee]

In all honesty, this has been a hilarious thread