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Deriving kinematic relationships

Floorball Maniac

Posted 18 June 2005 - 10:19 AM

Does anyone have any notes on deriving the following:

w = w0 + at
theta.gif = w0t + 1/2at power2.gif
w power2.gif = w0power2.gif + 2a theta.gif

Thank you!

Jonny-Sensei

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

Dave

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

werlop

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
theta.gif = w0t +  1/2at  power2.gif
w power2.gif = w0power2.gif + 2a theta.gif

Thank you!

View Post



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

Dave

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

broughy

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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