\begin{equation*} \begin{split} & a = -8  \\\\ & v = \int -8 \: dt  \\\\ & v = -8t + c \rightarrow {\color {blue} c = v_o} \\\\ & v = v_o - 8t \end{split} \end{equation*}

Pada \(t = 1, v = 2\)

\begin{equation*} \begin{split} & v = v_o - 8t \\\\ & 2 = v_o - 8 \:.\: 1 \\\\ & v_o = 10 \end{split} \end{equation*}

Persamaan kecepatan \(\bbox[5px, border: 2px solid magenta] {v = 10 - 8t}\)

\begin{equation*} \begin{split} & v = 10 - 8t \\\\ & s = \int (10 - 8t) \: dt \\\\ & s = 10t - 4t^2 + d \rightarrow {\color {blue}d = s_o} \\\\ & s = s_o + 10t - 4t^2 \end{split} \end{equation*}

Pada \(t = 1, s = 8\)

\begin{equation*} \begin{split} & s = s_o + 10t - 4t^2 \\\\ & 8 = s_o + 10 \:.\: 1 - 4 \:.\: 1^2 \\\\ & s_o = 2 \end{split} \end{equation*}

Persamaan posisi:

\(\bbox[5px, border: 2px solid magenta] {s = 2 + 10t - 4t^2}\)

\begin{equation*} \begin{split} & a = 3v - 6  \\\\ & \frac {dv}{dt} = 3v - 6  \\\\ & dt = \frac {dv}{3v - 6} \\\\ & \int dt = \int \frac {dv}{3v - 6} \\\\ & t = \int \frac {1}{3v - 6} \: \frac {d(3v - 6)}{3} \\\\ & 3t = \int \frac {1}{3v - 6} \: d(3v - 6) \\\\ & 3t = \ln (3v - 6) + c \\\\ & \ln (3v - 6) = 3t - c \\\\ & 3v - 6 = e^{3t - c} \\\\ & 3v - 6 = e^{3t} \:.\: e^{- c} \\\\ & 3v - 6 = k \:.\: e^{3t} \quad {\color {blue} \text{(Konstanta } e^{- c} \text{ diganti k)}}  \\\\ & 3v = k \:.\: e^{3t} + 6 \\\\ & v = \frac 13 \:.\: k \:.\: e^{3t} + 2 \end{split} \end{equation*}

Substitusi \(t = 1, v = 4e^3 + 2\)

\begin{equation*} \begin{split} & v = \frac 13 \:.\: k \:.\: e^{3t} + 2 \\\\ & 4 e^3 + \cancel{2} = \frac 13 \:.\: k \:.\: e^{3} + \cancel{2} \\\\ & 4 \cancel{e^3} =\frac 13 \:.\: k \:.\: \cancel{e^3} \\\\ & 12 = k \end{split} \end{equation*}

\begin{equation*} \begin{split} & v = \frac 13 \:.\: k \:.\: e^{3t} + 2 \\\\ & v = \frac 13 \:.\: 12 \:.\: e^{3t} + 2 \\\\ & \bbox[5px, border: 2px solid magenta] {v = 4 \:.\: e^{3t} + 2} \end{split} \end{equation*}

\begin{equation*} \begin{split} & a = 2s + 1  \\\\ & \frac {dv}{dt} = 2s + 1  \\\\ & \frac {dv}{ds} \frac {ds}{dt} = 2s + 1 \quad {\color {blue} \left ( \frac {ds}{dt} = v \right) } \\\\ & \frac {dv}{ds} \:.\: v = 2s + 1  \\\\ & v \: dv = (2s + 1) \: ds  \\\\ & \int v \: dv = \int (2s + 1) \: ds  \\\\ & \frac 12 v^2 = s^2 + s + c \end{split} \end{equation*}

Subsitusi \(s = 1, v = −6\)

\begin{equation*} \begin{split} & \frac 12 v^2 = s^2 + s + c \\\\ & \frac 12 (-6)^2 = 1^2 + 1 + c \\\\ & 3 = 2 + c \\\\ & c = 1 \end{split} \end{equation*}

\begin{equation*} \begin{split} & \frac 12 v^2 = s^2 + s + 1 \\\\ & v^2 = 2s^2 + 2s + 2 \\\\ & v = \pm \sqrt{2s^2 + 2s + 2} \end{split} \end{equation*}

Karena pada \(s = 1\), nilai \(v\) bernilai \(-6\), maka persamaan \(v\) menggunakan nilai negatif

\(\bbox[5px, border: 2px solid magenta] {v = -\sqrt{2s^2 + 2s + 2}} \)