Trouble with calculus problem I am working on a Calculus project and have come to an issue. 
Here's the problem: 

$$r(t)=R_{T}+V_T t+\frac12 A_T t^2+\frac16 J_A t^3+\frac{1}{24}S_A
 t^4.\qquad\qquad (2)$$ 
  We know the target parameters for position,
  velocity, and acceleration. We need to find the actual parameters for
  jerk and snap to know the proper force (acceleration) to apply.

  
*Find the actual velocity $v=v(t)$ of the LM.
  
*Find the actual acceleration $A=a(t)$ of the LM.
  
*Use equation (2) and the actual velocity found in Problem 6 to express $J_A$ and $S_A$ in terms of $R_T,\; V_T,\;A_T,\;r(t),$ and
  $v(t)$.
  
*Use the results of Problems 7 and 8 to express the actual acceleration $a=a(t)$ in terms of $R_T,\;V_T,\;A_T,\;r(t),$ and
  $v(t)$.
  

I get the first 2 questions, they are just asking for the first and second derivatives of the equation. However when I attempt to express $S_A$ and $J_A$ in terms of what's provided, I always end up with the other term in my equation. I know I'm probably missing something obvious here but I can't seem to figure it out. I am not looking to have it done for me, I just don't know where I'm going wrong.
 A: $r(t) = R_T+V_Tt+\frac12A_Tt^2+\frac16J_At^3+\frac1{24}S_At^4$ $\quad\rightarrow (1)$
$v(t) = V_T+A_Tt+\frac12J_At^2+\frac16S_At^3$
$a(t) = A_T +J_At+\frac12S_At^2$
solving $(1)$ for $J_A$ gives;
$J_A = \frac6{t^3}\bigg(r(t) -R_T-V_Tt-\frac12A_Tt^2-\frac1{24}{S_At^4}\bigg)$
$S_A = \frac6{t^3}\bigg(v(t)-V_T-A_Tt-\frac12J_At^2\bigg)$
$\implies S_A = \frac6{t^3}\bigg(v(t)-V_T-A_Tt-\frac12\frac6{t^3}\bigg(r(t) -R_T-V_Tt-\frac12A_Tt^2-\frac1{24}{S_At^4}
\bigg)t^2\bigg)$
$S_A =\frac6{t^3}\bigg(v(t)-V_T-A_Tt-\frac3{t}r(t) +\frac3tR_T+3V_T+\frac32A_Tt+\frac1{8}{S_At^3}\bigg) $
$S_A-\frac34S_A =\frac6{t^3}\bigg(v(t)-V_T-A_Tt-\frac3{t}r(t) +\frac3tR_T+3V_T+\frac32A_Tt\bigg) $
$S_A = \frac{24}{t^3}\bigg(v(t)-V_T-A_Tt-\frac3{t}r(t) +\frac3tR_T+3V_T+\frac32A_Tt\bigg)$
$\therefore S_A = \frac{24}{t^3}\bigg(v(t)+2V_T-\frac3{t}r(t) +\frac3tR_T+\frac12A_Tt\bigg)$
similarly $J_A =\frac6{t^3}\bigg(r(t) -R_T-V_Tt-\frac12A_Tt^2-\frac1{24}{\frac6{t^3}\bigg(v(t)-V_T-A_Tt-\frac12J_At^2\bigg)t^4}\bigg)  $
solving gives ;
$J_A =\frac{24}{t^3}\bigg(r(t) -R_T-\frac34V_Tt-\frac14A_Tt^2-{\frac t4v(t)}\bigg)$
