Derivative

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: \[\begin{aligned} A: \, s=2t^2+12t+1,\qquad&C:\, v=10t+4,\\ B:\, s=\frac13t^3+\frac{t^2}{2}+2,\qquad&D:\, v=\frac12t^2+1.\end{aligned}\] 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 \(s(t)\): \(\,a(t)=\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t} \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)=\frac{\mathrm{d}^2s}{\mathrm{d}t^2}\).

The motion of two bodies is given by equations \[s_1=\frac12t^2+6t+1\mbox{,}\quad s_2=\frac13t^3+t^2+4,\] where the positions \(s_1\) and \(s_2\) are given in meters and the time \(t\) in seconds. Determine at what time both bodies will move at the same speed. \[\] Hint: Instantaneous velocity can be expressed as the derivative of a position function \(s(t)\) with respect to time: \(v(t)=\frac{\mathrm{d} s}{\mathrm{d} t}\).