Posted 18 June 2005 - 10:19 AM
Does anyone have any notes on deriving the following:
w = w
0 + at

= w
0t + 1/2at

w

= w
0
+ 2a

Thank you!
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
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
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 = w
0 + at

= w
0t + 1/2at

w

= w
0
+ 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.
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
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