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
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.
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!
Magnitude is a positive quantityOriginally 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.
Originally posted by: swaq
d
Originally posted by: Wnh5001
Magnitude is a positive quantityOriginally 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.
Originally posted by: swaq
Ask a hard question, that's like high school physics
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?.
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.
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.
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?.
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.