C

Suppose we throw an object vertically upwards at the initial speed \(v_0=60\,\mathrm{m}/\mathrm{s}\). Determine the time needed for the object to reach the maximum height and determine the corresponding maximum height as well. \[\] Hint: The vertical upwards motion of a body is the movement composed of uniformly rectilinear motion (vertically upwards) and free fall. The dependence of the instantaneous height of a body on the time is given by the relation \(h=v_0t-\frac12gt^2\), where \(v_0\) is the magnitude of the initial velocity and \(g\) is the gravitational acceleration. In this problem, calculate with the rounded value of \(g=10\,\frac{\mathrm{m}}{\mathrm{s}^2}\). We measure time \(t\) in second and height \(h\) in meters.

An electrical source is characterized by the electromotive force \(U_e=60\,\mathrm{V}\) and the internal resistance \(R_i=2\,\Omega\). Determine the value of the electric current for which the appliance power will be at its maximum and determine the value of this maximum power as well. \[\] Hint: The dependence of the power of an appliance (\(P\), unit Watt (\(\mathrm{W}\))) on the magnitude of the folowing current (\(I\), unit Ampere (\(\mathrm{A}\))) is given by the relation \(P=U_eI-R_iI^2\). The source properties have a role of parameters: \(U_e\) is the electromotive force and \(R_i\) is the internal resistance of the source.

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