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: \(v(t)=\frac{\mathrm{d} s}{\mathrm{d} t}\).

A flywheel rotates such that it sweeps out an angle at the rate of \[ \varphi = 4t^2, \] where an angle \(\varphi\) is measured in radians and time \(t\) is measured in seconds. At what time is instantaneous angular velocity of the flywheel equal to \(36\,\frac{\mathrm{rad}}{s}\)? (Hint: Instantaneous angular velocity can be expressed as the derivative of the function \(\varphi(t)\) with respect to time: \(\omega(t)=\frac{\mathrm{d}\varphi}{\mathrm{d}t}\).)

Given the position-versus-time graph (in black) of an object in motion and the tangent line to the graph at the time point of \(10\) seconds (in red), find the instantaneous velocity of this object at \(10\) seconds. (Hint: Instantaneous velocity can be expressed as the derivative of position function with respect to time: \(v(t)=\frac{\mathrm{d}s}{\mathrm{d}t}\).)