The position

The position of a body at any time T is given by the displacement function S=t³-2t²-4t-8. Find its acceleration at each instant time when the velocity is zero.

Correct result:

a =  8

Solution:

$$\begin{gathered} s=t^3-2t^2-4t-8 \\ v=ds\mathrm{/}dt=3t^2-4t-4 \\ a=dv\mathrm{/}dt=6t-4 \\ v=0 \\ 3\ast t^2-4\ast t-4=0 \\ 3\cdot t^2-4\cdot t-4=0 \\ 3t^2-4t-4=0 \\ a=3;b=-4;c=-4 \\ D=b^2-4ac=4^2-4\cdot 3\cdot (-4)=64 \\ D>0 \\ {t}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{4\pm \sqrt{64}}{6}} \\ {t}_{1,2}={\displaystyle \frac{4\pm 8}{6}} \\ {t}_{1,2}=0.66666667\pm 1.33333333333 \\ {t}_1=2 \\ {t}_2=-0.666666666667 \\ \text{ Factored form of the equation: } \\ 3(t-2)(t+0.666666666667)=0 \\ t>0 \\ a=6\cdot t_1-4=6\cdot 2-4=8 \end{gathered}$$

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