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

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

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

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

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

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

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

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

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