## 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

1. zzr0ck3r Group Title

take the first integral v(t) = At+(1/2)Bt^2 + C then plug in v(0) to solve for C

2. 3psilon Group Title

Then take the next integral and plug X(0)

3. zzr0ck3r Group Title

yerp

4. 0mega Group Title

i take the integral of the first integral? and then plug in x(0)?

5. zzr0ck3r Group Title

yep

6. zzr0ck3r Group Title

note you need to solve for c first

7. 0mega Group Title

how did you get (1/2)bt^2 ?

8. 3psilon Group Title

anti derivative

9. zzr0ck3r Group Title

the integral of Bt

10. zzr0ck3r Group Title

A*0+(1/2)B*0^2 + C = vo c = v0

11. 0mega Group Title

ohhhhhh. okay okay. i get it now @zzr0ck3r

12. zzr0ck3r Group Title

great... note: Newton was a p i m p

13. 3psilon Group Title

$V(0) = V_{i}$ 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