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Deriving kinematic relationships - HSN forum

# Deriving kinematic relationships

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### #1Floorball Maniac

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Posted 18 June 2005 - 10:19 AM

Does anyone have any notes on deriving the following:

w = w0 + at
= w0t + 1/2at
w = w0 + 2a

Thank you!

### #2Jonny-Sensei

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Posted 18 June 2005 - 02:25 PM

a = dv/dt = d/dt x v = ddS/dtdt (because v=dS/dt)
therefore a = d²S/dt²

a = v/t
at° = vt[^-1]
∫at° = ∫vt[^-1]
at + c = v
(at t=0 then v=u) c = u
so v = u + at

v = u + at
∫v = ∫u + ∫at
∫St[^-1] = ∫ut° + ∫at
St° = ut[^1] + at²/2 + c
S = ut + ½at² + c
(at t=0, S=0) c = 0
so S = ut + ½at²

I dont really understand the next one.

v = u + at
v² = u² + 2atu + a²t²
v² = u² + 2a(ut + ½at²)
so v² = u² + 2aS
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### #3Dave

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Posted 18 June 2005 - 03:20 PM

for the last one you square everything to get

v² = u² + 2atu + a²t²

ut + ½at² = S

so the final line is

v² = u² + 2aS

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### #4werlop

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Posted 18 June 2005 - 04:18 PM

QUOTE(Floorball Maniac @ Jun 18 2005, 11:19 AM)
Does anyone have any notes on deriving the following:

w = w0 + at
= w0t +  1/2at
w = w0 + 2a

Thank you!

You don't have to derive the those equations (for circular motion), you only need to be able to use and state them.

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### #5Dave

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Posted 18 June 2005 - 08:16 PM

yeah i thought that to but wasnt sure so said nothing

you are asked to go from displacemtn to velocity to acceleration and vice versa but thast just basic caculus

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### #6broughy

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Posted 18 June 2005 - 11:52 PM

basically, you need to be able to derive the equations of motion for uniform acceleration, but not the equations of motion for circular motion, which is what those are
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