Given the function;
[tex]s(t)=100\cos (0.75t).e^{-0.2t}+100[/tex]Part A:
We can find the change in the vertical position of the bungee jumper between t=0 and t=15 by subtracting the values derived after substituting the time values into the function.
When t=0
[tex]\begin{gathered} s(t)=100\cos (0.75\times0)\text{.e}^{-0.2\times0}+100 \\ s(t)=(100\cos 0\times1)+100 \\ s(t)=100+100 \\ s(t)=200 \end{gathered}[/tex]When t =15
[tex]\begin{gathered} s(t)=100\cos (0.75\times15)\text{.e}^{-0.2\times15}+100 \\ s(t)=100\cos (11.25)\text{.}e^{-3}+100 \\ s(t)=101.125 \end{gathered}[/tex]Therefore, the change in vertical position is
[tex]\begin{gathered} s(15)-s(0)=101.125-200 \\ =-98.875 \end{gathered}[/tex]Answer:
[tex]-98.875\cong-100[/tex]If we read the value of the graph, the answer also corresponds to -100.
Part B
The Jumper's average velocity would be given by
[tex]\frac{\triangle s}{\triangle t}=\frac{s(t_2)-s(t_1)}{t_2-t_1}_{}[/tex]We would read off the graph by tracing the "t" values to their corresponding "s" values
For [0,15]
[tex]\begin{gathered} _{} \\ \frac{\triangle s}{\triangle t}=\frac{100-200}{15-0} \\ \frac{\triangle s}{\triangle t}=-\frac{100}{15} \\ =-6.6667ms^{-1} \end{gathered}[/tex]Answer:
[tex]-6.6667ms^{-1}[/tex]For [0,2]
[tex]\begin{gathered} \frac{\triangle s}{\triangle t}=\frac{105-200}{2-0} \\ =-\frac{100}{2} \\ =-47.5ms^{-1} \end{gathered}[/tex]Answer:
[tex]-47.5ms^{-1}[/tex]For [1,6]
[tex]\begin{gathered} \frac{\triangle s}{\triangle t}=\frac{90-155}{6-1} \\ ==\frac{-65}{5} \\ =-13ms^{-1} \end{gathered}[/tex]Answer:
[tex]-13ms^{-1}[/tex]For [8,10]
[tex]\begin{gathered} \frac{\triangle s}{\triangle t}=\frac{105-120}{10-8} \\ =-\frac{15}{2} \\ =-7.5ms^{-1} \end{gathered}[/tex]Answer:
[tex]-7.5ms^{-1}[/tex]Part C
The bungee jumper achieves the highest average velocity at -6.6667m/s. Therefore the time interval would be;
Answer:
[tex]t=\lbrack0,15\rbrack[/tex]
