Velocity in Vacuum: Feather Dropped, 0.30 secs

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SUMMARY

A feather dropped in a vacuum chamber accelerates at 9.8 m/s² due to gravity. After 0.30 seconds, its velocity reaches 2.94 m/s, calculated using the formula v = g * t. The distance fallen in that time is 0.441 meters, derived from the kinematic equation d = 1/2 * a * t². This discussion clarifies the application of basic physics principles in a vacuum environment.

PREREQUISITES
  • Understanding of basic physics concepts, specifically acceleration and velocity
  • Familiarity with kinematic equations, particularly d = 1/2 * a * t²
  • Knowledge of gravitational acceleration, specifically 9.8 m/s²
  • Ability to perform calculations involving time and distance in physics
NEXT STEPS
  • Study the derivation and application of kinematic equations in physics
  • Learn about the effects of air resistance versus vacuum conditions on falling objects
  • Explore advanced topics in classical mechanics, such as projectile motion
  • Investigate the implications of gravity in different environments, including space
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Students studying physics, educators teaching mechanics, and anyone interested in understanding motion in a vacuum environment.

Carnivean
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Ok:

Note: Acceleration due to gravity: g = 9.8 m/s

A feather is dropped inside a vacuum chamber.

A) What is the Velocity after 0.30 sec?

B) How far has it fallen after 0.30 sec?

I'm stuck on this.. it seems simple.. but maybe too simple. I think there is more too it. Any help?
 
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According the the policy one is supposed to show ones efforts before asking for assistance.
 
Umm.. I said I was stuck. I do not know where to begin at? Help me?
 
Acceleration is the rate of change of velocity with respect to time. Now, if an object is accelerating at 9.8 m/s/s (note the units) how fast will it be going after 1 second? 0.3 seconds?
 
after 1 sec it will go 9.8 right? and after .3 sec it will go 2.94.. i think.
 
Carnivean said:
after 1 sec it will go 9.8 right? and after .3 sec it will go 2.94.. i think.
Sounds good to me. Now, for the second question what kinematic equations do you know?
 
for distance I have d = 1/2at (the 't' is squared , but I don't know how to type a small '2' next to it.. if you could show me that too please) this is an equation for the problems where acceleration info is already is included.

where initial velocity is included: d= InitialVelocity x (t) + 1/2a x (t squared)
 
Carnivean said:
for distance I have d = 1/2at (the 't' is squared , but I don't know how to type a small '2' next to it.. if you could show me that too please) this is an equation for the problems where acceleration info is already is included.

where initial velocity is included: d= InitialVelocity x (t) + 1/2a x (t squared)
Your first equation looks applicable here :smile:. To type superscript text simply enclose the text in [#sup#] [#/sup#] tags (without the #).
 
ok I got .441 meters for the distance which I think sounds right. Almost a half a meter. But could you give me an example of how to do the superscript thing again? Is it [/then the number]? or [#the number#] ? Just show me an example of how you would type it.
 
  • #10
Carnivean said:
ok I got .441 meters for the distance which I think sounds right. Almost a half a meter. But could you give me an example of how to do the superscript thing again? Is it [/then the number]? or [#the number#] ? Just show me an example of how you would type it.

For example if you wanted to type x2, you would type:
Code:
x[#sup]2[/sup#]
Or for subscript;
Code:
x[#sub]2[/sub#]
Again, without the # marks.
 
  • #11
ok thanks soo much for your help.. you don't know how much it is appreciated. I will be back many a times probably until about january.

let me try - x 2
 

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