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Lan wrote on 2010-11-15 05:04
If the initial velocity is equal to zero, then does Acceleration equal displacement divided by time squared? I'm doing a stupid lab that says to prove Newton's second law where in a chart they give you the unbalanced force, the displacement and you have to find the acceleration. I've tried the equation "Acceleration is equal to force divided by mass" But that gives me a totally different answer from the equation I derived. Which one should I use! HELP :T_T:
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EndlessDreams wrote on 2010-11-15 05:30
What is the final velocity? Acceleration is also the change of velocity over time.
If it isn't moving, the acceleration is zero.
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Lan wrote on 2010-11-15 05:33
The final velocity is just the displacement divided by the time. So I figured that I should just take the distance I got and divide it by time squared to get acceleration.
(The lab was using one of those air tables and seeing how far the puck slides down with each corresponding unbalanced force)
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Phunkie wrote on 2010-11-15 05:37
[Image: http://upload.wikimedia.org/math/4/7/f/47f9b222b5c53d00ec9583d3d8032877.png]
v_i is the body's initial velocity.
s_i is the body's initial position.
s is the displacement.
v is the final velocity.
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Lan wrote on 2010-11-15 05:40
Quote from Phunkie;216010:
[Image: http://upload.wikimedia.org/math/4/7/f/47f9b222b5c53d00ec9583d3d8032877.png]
v_i is the body's initial velocity.
s_i is the body's initial position.
s is the displacement.
v is the final velocity.
...yeah don't have calculus.
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Phunkie wrote on 2010-11-15 05:43
Quote from Lan;216016:
...yeah don't have calculus.
Don't need it. That's one of the equations of motion.
[Image: http://gyazo.com/2dcbc3ade4ff701c67de4d2ed0d2a2d9.png]
Your idea of acceleration being
a = d/(t^2)
would be correct, but it's missing a factor of 2.
a = 2*d/(t^2)
:D
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Lan wrote on 2010-11-15 05:49
Quote from Phunkie;216021:
Don't need it. That's one of the equations of motion.
[Image: http://gyazo.com/2dcbc3ade4ff701c67de4d2ed0d2a2d9.png]
Your idea of acceleration being
a = d/(t^2)
would be correct, but it's missing a factor of 2.
a = 2*d/(t^2)
:D
Er why do I need that two again?
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Phunkie wrote on 2010-11-15 05:53
That 2 comes out of the average velocity, which is the average of the final and initial velocities.
v_avg = (v_i + v_f)/2
a = v/t
v = d/t
v here is the average velocity
Average velocity is again v = (vi+vf)/2
but there's no initial velocity, as you said, so
vf = 2d/t
now plug vf in acceleration (a = vf/t)
----> a = 2d/(t^2)
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Lan wrote on 2010-11-15 05:55
I hate math sooo much. But thanks :) Saved me from failing.
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Phunkie wrote on 2010-11-15 05:56
Sure thing, buddy.
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Lan wrote on 2010-11-15 06:01
Hmm thought acceleration was delta v divided by delta t, and not v average :/ meh.
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Phunkie wrote on 2010-11-15 06:06
Quote from Lan;216035:
Hmm thought acceleration was delta v divided by delta t, and not v average :/ meh.
Acceleration still is change of velocity over change of time.
But you derive the equations of motion with the use of the average velocity.
Check out
http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html for more information. This site is gold for physics stuff.