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0mega
Group Title
a(t) = A + Bt
x(0) =xi
v(0) = vi
derive the position as a function of time equation
 one year ago
 one year ago
0mega Group Title
a(t) = A + Bt x(0) =xi v(0) = vi derive the position as a function of time equation
 one year ago
 one year ago

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zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.1
take the first integral v(t) = At+(1/2)Bt^2 + C then plug in v(0) to solve for C
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.1
Then take the next integral and plug X(0)
 one year ago

0mega Group TitleBest ResponseYou've already chosen the best response.0
i take the integral of the first integral? and then plug in x(0)?
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.1
note you need to solve for c first
 one year ago

0mega Group TitleBest ResponseYou've already chosen the best response.0
how did you get (1/2)bt^2 ?
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.1
anti derivative
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.1
the integral of Bt
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.1
A*0+(1/2)B*0^2 + C = vo c = v0
 one year ago

0mega Group TitleBest ResponseYou've already chosen the best response.0
ohhhhhh. okay okay. i get it now @zzr0ck3r
 one year ago

zzr0ck3r Group TitleBest ResponseYou've already chosen the best response.1
great... note: Newton was a p i m p
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.1
\[V(0) = V_{i} \] 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
 one year ago
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