physics concept quesiton

[yhelothar]

Originally posted by: JujuFish

Originally posted by: ledjani
Well I believe the answer is 0. The derivative of the velocity is acceleration. Thus whenever the graph of the velocity reaches its max, the derivative is 0. Thus the acceleration is 0. Because it is at rest acceleration wise, it has reached the top, and it isn't going up or down anymore.

If the graph of velocity has a local maximum, then acceleration is not constant. If that is the case, then yes, acceleration would be zero at that local maximum.

Nope the acceleration is still constant as the force acting on it is constant.

 

[Wnh5001]

Damn Gravity, u cost me 5 points =(

 

[Cattlegod]

when all else fails use the following pattern :

a
b
b
a

c
a
d
a
b
b
a

 

[coomar]

Originally posted by: Wnh5001
a particle reaches it max height t= 2 seconds, what is the magnitude of its acceleration?

|hint|- the t=2 seconds doesnt mean anything =S, choose wisely,

a)19.6
b)9.81
c)0
d)-9.81

at the highest point

velocity is zero, acceleration is 9.8 (last decimal place depends on where in the world you are) downwards, it would be more amusing to calculate the jerk

 

[JujuFish]

Originally posted by: virtualgames0

Originally posted by: JujuFish

Originally posted by: ledjani
Well I believe the answer is 0. The derivative of the velocity is acceleration. Thus whenever the graph of the velocity reaches its max, the derivative is 0. Thus the acceleration is 0. Because it is at rest acceleration wise, it has reached the top, and it isn't going up or down anymore.

If the graph of velocity has a local maximum, then acceleration is not constant. If that is the case, then yes, acceleration would be zero at that local maximum.

Nope the acceleration is still constant as the force acting on it is constant.

First, I was speaking in general. If a graph of velocity is curved, then acceleration is not constant. That is the type of graph referred to in ledjani's post. Second, you don't know what forces are acting on it, so you cannot make that claim regarding the OP. Third, notice the "if" in my original statement.

 

[yhelothar]

Originally posted by: JujuFish

Originally posted by: virtualgames0

Originally posted by: JujuFish

Originally posted by: ledjani
Well I believe the answer is 0. The derivative of the velocity is acceleration. Thus whenever the graph of the velocity reaches its max, the derivative is 0. Thus the acceleration is 0. Because it is at rest acceleration wise, it has reached the top, and it isn't going up or down anymore.

If the graph of velocity has a local maximum, then acceleration is not constant. If that is the case, then yes, acceleration would be zero at that local maximum.

Nope the acceleration is still constant as the force acting on it is constant.

First, I was speaking in general. If a graph of velocity is curved, then acceleration is not constant. That is the type of graph referred to in ledjani's post. Second, you don't know what forces are acting on it, so you cannot make that claim regarding the OP. Third, notice the "if" in my original statement.

Yeah you're right about the curved graph. I see what you were saying before.

 

[Trevelyan]

the answer is definately b

 

[Cawchy87]

The particle is no longer moving if it has reached its max height. The upward force is equal to the force pushing down on it (gravity). Thus 9.81 - 9.81 = 0.

That's the way I see it.

 

[yhelothar]

Originally posted by: Cawchy87
The particle is no longer moving if it has reached its max height. The upward force is equal to the force pushing down on it (gravity). Thus 9.81 - 9.81 = 0.

That's the way I see it.

That's how I thought before I took physics.
But if you really think about it, the downwards force is constant even when the object is still moving up. This is why the object deccelerates as it moves up. The downwards force due to gravity is always constant.

 

[Brainonska511]

Well, D would be the total vector quantity, but now after thinking about how you put magnitude, it would have to be B.

 

[Rock Hydra]

Originally posted by: ledjani
Well I believe the answer is 0. The derivative of the velocity is acceleration. Thus whenever the graph of the velocity reaches its max, the derivative is 0. Thus the acceleration is 0. Because it is at rest acceleration wise, it has reached the top, and it isn't going up or down anymore.

That's what I would have answered.

 

[jessicak]

b or d....depending on sign convention

 

[Goosemaster]

Originally posted by: Rock Hydra

Originally posted by: ledjani
Well I believe the answer is 0. The derivative of the velocity is acceleration. Thus whenever the graph of the velocity reaches its max, the derivative is 0. Thus the acceleration is 0. Because it is at rest acceleration wise, it has reached the top, and it isn't going up or down anymore.

That's what I would have answered.

except that I would have gone into detail about how the acceleration would decrease after a point and continue to decrease until it reached it's amxed height

||accelerationdeceleration|| = acceleration + deceleratin = = mag

 

[Brainonska511]

Originally posted by: Goosemaster

Originally posted by: Rock Hydra

Originally posted by: ledjani
Well I believe the answer is 0. The derivative of the velocity is acceleration. Thus whenever the graph of the velocity reaches its max, the derivative is 0. Thus the acceleration is 0. Because it is at rest acceleration wise, it has reached the top, and it isn't going up or down anymore.

That's what I would have answered.

except that I would have gone into detail about how the acceleration would decrease after a point and continue to decrease until it reached it's amxed height

||accelerationdeceleration|| = acceleration + deceleratin = = mag

It's not acceleration+deceleration in this case (and plus, you would just say acceleration in a - direction). There is only 1 acceleration which is in the downward direction due to gravity. Since you just want magnitude, you take acceleration due to gravity and leave out any signs.

 

[Fenixgoon]

Originally posted by: ledjani
Well I believe the answer is 0. The derivative of the velocity is acceleration. Thus whenever the graph of the velocity reaches its max, the derivative is 0. Thus the acceleration is 0. Because it is at rest acceleration wise, it has reached the top, and it isn't going up or down anymore.

WRONG. please retake physics/ calc

 

[Rainsford]

You know, for a technical message board, this thread is just sad...

Assuming a lot more than the OP stated as the question, the answer is 9.8m/s/s. When you throw something directly up into the air, it starts with a positive velocity. However, assuming it's a situation like throwing a ball up in the air and letting it fall back down, the only really important force acting on it is gravtiy. This force is constant during the entire flight of the ball, and afterwards as well of course. This force provides a downward acceleration of 9.8m/s/s, and absent any other force, that is the magnitude of acceleration.

 

[Rainsford]

Originally posted by: Cawchy87
The particle is no longer moving if it has reached its max height. The upward force is equal to the force pushing down on it (gravity). Thus 9.81 - 9.81 = 0.

That's the way I see it.

There is no upward force. Again, assuming a lot, the particle was given initial upward velocity, but the only force acting on it was gravity.

 

[Rainsford]

Originally posted by: jessicak
b or d....depending on sign convention

Magnitude is always positive

 

[Goosemaster]

Originally posted by: Rainsford
You know, for a technical message board, this thread is just sad...

Assuming a lot more than the OP stated as the question, the answer is 9.8m/s/s. When you throw something directly up into the air, it starts with a positive velocity. However, assuming it's a situation like throwing a ball up in the air and letting it fall back down, the only really important force acting on it is gravtiy. This force is constant during the entire flight of the ball, and afterwards as well of course. This force provides a downward acceleration of 9.8m/s/s, and absent any other force, that is the magnitude of acceleration.

obviously....


:laugh:

 

[slpaulson]

This is a pretty poorly worded question if you ask me. You have to assume a lot of information not given in the problem.

 

[Goosemaster]

Originally posted by: cRazYdood
This is a pretty poorly worded question if you ask me.

it states "its acceleration, " therefore it must require that all forces be taken into a cocount



if you were to draw a force diagram, you would ahve the particles aacceleration + weight (acceleration due to gravity)

the magnitude would be the overall acceleration.

For the particle, the acceleration it might have to reach it's max height will eventually equal its acceleration due to gravity at the maximum height, so the mag is o


<---sucks at physics

 

[Heisenberg]

/bangs head on desk at some replies



Newton must be spinning in his grave right now.

 

[Goosemaster]

Originally posted by: Heisenberg
/bangs head on desk at some replies



Newton must be spinning in his grave right now.

dude..by all means correct me...I sck at physics.


My first response would be to blurt out 9.8 but my reasoning, as you see it above, is what I would eventual utter to my furious and very angry physics prof.


by all means....enlghten us oh atomic master...

Plum pudding is delicious by the way...way better htan yoru stuff

 

[Eeezee]

Originally posted by: Wnh5001
a particle reaches it max height t= 2 seconds, what is the magnitude of its acceleration?

|hint|- the t=2 seconds doesnt mean anything =S, choose wisely,

a)19.6
b)9.81
c)0
d)-9.81

Wow, the world is so small. I graded a test with this exact question and these exact answers. Tell me, are you enrolled at the University of Arizona?

 

[Heisenberg]

Originally posted by: Goosemaster

Originally posted by: Heisenberg
/bangs head on desk at some replies



Newton must be spinning in his grave right now.

dude..by all means correct me...I sck at physics.


My first response would be to blurt out 9.8 but my reasoning, as you see it above, is what I would eventual utter to my furious and very angry physics prof.


by all means....enlghten us oh atomic master...

Plum pudding is delicious by the way...way better htan yoru stuff

I wasn't really meaning you specifically. I did post a reply earlier in the thread but apparently no one bothered to read it.