2010013703

Suppose $A$, $B$, $C$ and $D$ are bodies, which are set in motion at the same initial time $t$. We know how the position $s$ or speed $v$ of these bodies changes with time: $$\begin{array}{rl}A:s=2t^2+12t+1,& C:v=10t+4,\\ B:s=\frac{1}{3}{t}^3+\frac{t^2}{2}+2,& D:v=\frac{1}{2}{t}^2+1.\end{array}$$ Position $s$ is given in meters, time $t$ in seconds and speed $v$ in meters per second. Determine which body moves with the greatest acceleration at the time $t=1\mathrm{s}$. $$$$ Hint: Instantaneous velocity can be expressed as the derivative of a position function $s(t)$ with respect to time: $v(t)=\frac{\mathrm{d}s}{\mathrm{d}t}$, and the instantaneous acceleration can be expressed as the derivative of a function $v(t)$ with respect to time: $a(t)=\frac{\mathrm{d}v}{\mathrm{d}t}$. Since we can determine the velocity using the derivative of the position function $s(t)$, we as well can determine the acceleration using the second derivative of $s(t)$: $a(t)=\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)=\frac{{\mathrm{d}}^{2}s}{\mathrm{d}{t}^2}$.