Velocity & Acceleration of Elevator

AI Thread Summary
As the elevator approaches the top of a building and slows to a stop, its acceleration is directed downward due to the deceleration. The elevator does not maintain a constant velocity while slowing down; instead, its velocity decreases to zero at the moment it stops. At that point, the velocity is zero, meaning it lacks a defined direction. The discussion clarifies that a constant velocity would imply zero acceleration, which contradicts the scenario of the elevator coming to a halt. Therefore, while the elevator is slowing, its velocity is not constant, and it ultimately reaches a state of zero velocity before stopping.
mimo6267
Messages
2
Reaction score
0
When the elevator approaches the top of a building and slows to a stop, what direction is the elevator's velocity & acceleration?
 
Physics news on Phys.org
kindly explain what you believe is the answer and why? or what you have done so far. You cannot expect us to answer your questions directly.
 
I found that since the elevator is slowing to the stop, the direction of the elevator's acceleration would be downward. Since the elevator is always traveling at a constant velocity, does that mean when it slows to a stop, the velocity would still be constant, which means the direction would be horizontal? or since, the acceleration is downward that direction of velocity would also be downward? Thank you!
 
At the moment that the elevator is stopped, it doesn't really have a "direction." (E.g., Which way does an arrow with no length point?)

At any other moment the velocity certainly has a direction (which you seem able to describe)!
 
mimo6267 said:
I found that since the elevator is slowing to the stop, the direction of the elevator's acceleration would be downward. Since the elevator is always traveling at a constant velocity, does that mean when it slows to a stop, the velocity would still be constant, which means the direction would be horizontal? or since, the acceleration is downward that direction of velocity would also be downward? Thank you!

You're absolutely right about the acceleration. However, the elevator is NOT traveling at constant velocity!

For an object with constant velocity, its acceleration must be zero, which is clearly not the case here. Now at its top most point the elevator slows down and comes to a complete HALT before moving downwards again. At that point therefore, the velocity of the elevator is ZERO!

Also, the zero vector doesn't really have a defined direction.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top