Question

Please do the homework without leaving me with nothing... The question is on the photo

Answer 1

Given the function;

`s(t)=100\cos (0.75t).e^(-0.2t)+100`

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

`\begin{gathered} s(t)=100\cos (0.75*0)\text{.e}^(-0.2*0)+100 \\ s(t)=(100\cos 0*1)+100 \\ s(t)=100+100 \\ s(t)=200 \end{gathered}`

When t =15

`\begin{gathered} s(t)=100\cos (0.75*15)\text{.e}^(-0.2*15)+100 \\ s(t)=100\cos (11.25)\text{.}e^(-3)+100 \\ s(t)=101.125 \end{gathered}`

Therefore, the change in vertical position is

`\begin{gathered} s(15)-s(0)=101.125-200 \\ =-98.875 \end{gathered}`

Answer:

`-98.875\cong-100`

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

`(\triangle s)/(\triangle t)=(s(t_2)-s(t_1))/(t_2-t_1)_{}`

We would read off the graph by tracing the "t" values to their corresponding "s" values

For [0,15]

`\begin{gathered} _{} \\ (\triangle s)/(\triangle t)=(100-200)/(15-0) \\ (\triangle s)/(\triangle t)=-(100)/(15) \\ =-6.6667ms^(-1) \end{gathered}`

Answer:

`-6.6667ms^(-1)`

For [0,2]

`\begin{gathered} (\triangle s)/(\triangle t)=(105-200)/(2-0) \\ =-(100)/(2) \\ =-47.5ms^(-1) \end{gathered}`

Answer:

`-47.5ms^(-1)`

For [1,6]

`\begin{gathered} (\triangle s)/(\triangle t)=(90-155)/(6-1) \\ ==(-65)/(5) \\ =-13ms^(-1) \end{gathered}`

Answer:

`-13ms^(-1)`

For [8,10]

`\begin{gathered} (\triangle s)/(\triangle t)=(105-120)/(10-8) \\ =-(15)/(2) \\ =-7.5ms^(-1) \end{gathered}`

Answer:

`-7.5ms^(-1)`

Part C

The bungee jumper achieves the highest average velocity at -6.6667m/s. Therefore the time interval would be;

Answer:

`t=\lbrack0,15\rbrack`