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

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Homework Help Overview

The discussion revolves around using the equation x = 1/2 at² + v₀t + x₀ to find acceleration in a scenario where an object (an eye dropper) is dropped from a height of 20 cm, with a recorded time of 0.116 seconds. Participants are examining the validity of the equation and the calculations involved.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to isolate acceleration (a) from the equation and calculates a value based on their measurements. Some participants question the accuracy of the time measurement and the initial height, suggesting that the time recorded may not correspond to the height given. Others explore the implications of using average velocity to find acceleration.

Discussion Status

Participants are actively discussing the calculations and measurements involved. Some guidance has been offered regarding potential errors in timing and assumptions about height. There is an ongoing exploration of the relationship between the measured values and the expected standard acceleration due to gravity.

Contextual Notes

There is a noted uncertainty regarding the initial height of the drop and the method of timing, which may affect the results. Participants are considering the implications of human error in timing and the validity of the initial conditions provided.

GodAllen
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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.
 
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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
 

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