physics concept quesiton

[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

 

[lrad50]

C FTW!

(no idea)

 

[Cooler]

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

-9.81 the pull of graity in the negitive direction.

 

[Wnh5001]

hmm im not sure if thats the exact question. i think it went like this
" a particle thrown with a certain velocity reaches it maximum t=1.5 secs what is the magnitude of its acceleration." but not that it stuff matters anways.

 

[Wnh5001]

Originally posted by: Cooler

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

-9.81 the pull of graity in the negitive direction.

INCORRECT!

 

[Cooler]

Originally posted by: Wnh5001

Originally posted by: Cooler

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

-9.81 the pull of graity in the negitive direction.

INCORRECT!

Is the particle on earth? all falling objects(on earth) have that acceleration in the Y vector.

 

[Wnh5001]

Originally posted by: Cooler

Originally posted by: Wnh5001

Originally posted by: Cooler

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

-9.81 the pull of graity in the negitive direction.

INCORRECT!

Is the particle on earth? all falling objects(on earth) have that acceleration in the Y vector.

Magnitude is a positive quantity

 

[swaq]

d

[edit] well actually the absolute value of that, so b [/edit]

 

[Kevin1211]

b..

 

[Wnh5001]

Originally posted by: swaq
d

It was on my physics exam and i didnt get the results back yet+ the correct solutions so i dunno the answer sooo... i was hoping some physics enthusiast could answer it =|

 

[Cooler]

Originally posted by: Wnh5001

Originally posted by: Cooler

Originally posted by: Wnh5001

Originally posted by: Cooler

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

-9.81 the pull of graity in the negitive direction.

INCORRECT!

Is the particle on earth? all falling objects(on earth) have that acceleration in the Y vector.

Magnitude is a positive quantity

Well in that case it would have to be 9.81 if you want to be technical about it but it will go +180 of the original angle.

Magnitude = (Ax^2+AY^2)^(1/2)
which would be in this 9.81
But its acceleration is negitive.

 

[swaq]

Ask a hard question, that's like high school physics

 

[eelw]

Originally posted by: swaq
Ask a hard question, that's like high school physics

True, true. But if you notice, only one person who has posted in this topic (which appears to be more a guess than actually know why the answer is right) so far is right.

 

[Wnh5001]

the way i interpreted the question was, what is the acceleration of the particle, not the acceleration acting on the particle, anyone have a solution?.

 

[JDrake]

b

 

[Cooler]

9.81 gravity i have said it two times

 

[Atomicus]

Originally posted by: Wnh5001
the way i interpreted the question was, what is the acceleration of the particle, not the acceleration acting on the particle, anyone have a solution?.

The acceleration of the particle IS the acceleration acting on the particle. It is accelerating at 9.81 m/s^2 downwards since it is in freefall.

 

[czech09]

B

 

[czech09]

Originally posted by: Cooler
9.81 gravity i have said it two times

Exactly.

 

[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.

 

[Wnh5001]

Originally posted by: ledjani
Well I believe the answer is 0. The derivative of the velocity is acceleration. Thus whenever the graph of the acceleration 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.

😵 really? so were either at 0 or 9.81 HMM

 

[Heisenberg]

Magnitude is always a positive quantity. If you're talking about a particle on Earth after it has been released (i.e. no other forces acting on it), then its net acceleration would be 9.81 m/s^2 downward.

Edit: Another thing people often get confused about is the acceleration an object experiences and its net acceleration. Every object on Earth is being accelerated downward at g - from you sitting on the ground to a baseball in the air. The only difference is that the ground is exerting an acceleration upward (the normal force) so the net acceleration you experience is zero. While the baseball is in the air though, there is no other force besides gravity, so its net acceleration is equal to g.

 

[yhelothar]

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.

The acceleration is constant at 9.8m/s until it hits a surface, where the normal force cancels out the force of the gravity.

 

[yhelothar]

Originally posted by: Wnh5001
the way i interpreted the question was, what is the acceleration of the particle, not the acceleration acting on the particle, anyone have a solution?.

I think what you meant when you said, what is the acceleration of the particle, you meant what is the initial velocity of the particle.
The acceleration acting on the particle is called force, which is what causes acceleration.

 

[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.