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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.

  1. Find the actual velocity $v=v(t)$ of the LM.
  2. Find the actual acceleration $A=a(t)$ of the LM.
  3. 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 $v(t)$.
  4. 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.

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  • $\begingroup$ This would be a linear problem of two equations and two unknowns. You can solve this by any algebraic method to your liking. Try substitution: solve for $J_A$ in one of the equations (you will have $S_A$ on the other side), then substitute it into the other equation for $J_A$. The result, you will have isolated $S_A$. Do the same for $J_A$. Does that help? $\endgroup$
    – MasterYoda
    May 14 '18 at 17:37
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$r(t) = R_T+V_Tt+\frac12A_Tt^2+\frac16J_At^3+\frac1{24}S_At^4$ $\quad\rightarrow (1)$

$v(t) = V_T+A_Tt+\frac12J_At^2+\frac16S_At^3$

$a(t) = A_T +J_At+\frac12S_At^2$

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 ;

$J_A =\frac{24}{t^3}\bigg(r(t) -R_T-\frac34V_Tt-\frac14A_Tt^2-{\frac t4v(t)}\bigg)$

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