0mega
a(t) = A + Bt
x(0) =xi
v(0) = vi
derive the position as a function of time equation
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zzr0ck3r
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take the first integral
v(t) = At+(1/2)Bt^2 + C then plug in v(0) to solve for C
3psilon
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Then take the next integral and plug X(0)
zzr0ck3r
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yerp
0mega
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i take the integral of the first integral? and then plug in x(0)?
zzr0ck3r
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yep
zzr0ck3r
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note you need to solve for c first
0mega
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how did you get (1/2)bt^2 ?
3psilon
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anti derivative
zzr0ck3r
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the integral of Bt
zzr0ck3r
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A*0+(1/2)B*0^2 + C = vo
c = v0
0mega
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ohhhhhh. okay okay. i get it now @zzr0ck3r
zzr0ck3r
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great...
note: Newton was a p i m p
3psilon
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\[V(0) = V_{i} \] according to the boundary condition
So \[V(0) =\frac{ B0^{2} }{ 2 } + a(0) + C\]
\[ V(0) = C\]
\[ V(0) = vi\]
So \[ Vi = C\]
Plug \[ V(i)\]
into the orginal \[V(t)\] function as C