Finding Acceleration Using x=1/2 at2+v0t+x0 Equation | Homework Help

  • Thread starter Thread starter GodAllen
  • Start date Start date
Click For Summary
The equation x = 1/2 at² + v₀t + x₀ is being used to find acceleration, with initial conditions set to v₀t = 0 and x₀ = 0. The user calculated an acceleration of 29.72 m/s² after dropping an eye dropper from a height of 20 cm over 0.116 seconds, which is significantly higher than the standard gravitational acceleration of 9.8 m/s². Other participants pointed out potential errors in timing and questioned the accuracy of the drop height, suggesting that the time recorded might correspond to a shorter drop. They emphasized that human error in timing could introduce significant inaccuracies. Ultimately, the consensus is that the acceleration should approximate 9.8 m/s² under ideal conditions.
GodAllen
Messages
3
Reaction score
0

Homework Statement


Is this equation any good for finding a

x=1/2 at2+v0t +x0
Given that v0t=0 and x0=0

I am trying to find the acceleration give that I dropped an eye dropper at 20cm and the average time was 0.116sec

Homework Equations


I have isolated it to a, where it is A= 2x over t2


The Attempt at a Solution



I have calculated it to be 2972cm/s2, then i divided it by 100 to get 29.72m/s2. Isnt the standard 9.8m/s2. I have also timed other objects and it seems they are all different.
 
Physics news on Phys.org
Your math is OK, it's your time that was faulty.

Are you sure the starting height was 20 cm? That's 8 inches right? Your time corresponds to something that was dropped from 6 cm or 3 inches.

What did you use to time the event? If it involved any form of human starting or stopping of the timer, that throws a +/- of 0.2 seconds into anything.
 
you won't be able to find acceleration from the given data (I think). The acceleration will be 9.8 m/s/s.
 
1.72 is the av. velocity and initial velocity = 0 therefore final velocity = 3.45 m/s

acc = (3.45- 0)/0.116

but it is wrong, I guess
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
5K
Replies
7
Views
2K
Replies
2
Views
4K
Replies
14
Views
5K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 34 ·
2
Replies
34
Views
3K
Replies
2
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K