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Solution:
\(\begin{aligned} \text{Velocity function, }v \text{ is given by } \ v&=\int a \ dt\\ v&=\int(4-2t) \ dt\\ v&=4t-t^2+c\\ \end{aligned}\)
\(\begin{aligned} \text{When }t=0 \text{ and }v=12, \text{ then, } \ 12&= 4(0)-0^2 +c\\ c&=12\\ \end{aligned}\)
\(\text{Hence, at time }t, \ v=12+4t-t^2\)
\(\begin{aligned} \text{Displacement function, } s \text{ is given by, }\ s&=\int v \ dt\\ s&=\int (12+4t-t^2) \ dt\\ s&=12t+2t^2-\dfrac{1}{3}t^3+c\\ \end{aligned}\)
When \(t=0 \text{ and }s=0\),
\(\begin{aligned} \text{then, } \ 0&=12(0)+2(0)^2-\dfrac{1}{3}(0)^3+c\\ c&=0\\ \end{aligned}\)
Hence, at time \(t,\ s=12t+2t^2-\dfrac{1}{3}t^3\)
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