Question

Hi yea thank you thank goodness for your help today please

Answer 1

Given:

`h(t)=3300-54t-300e^(-0.18t)`

Find-:

(1)

Velocity at the instant when t = 6 sec.

(2)

The time when velocity is -45 meter per sec

Explanation-:

Velocity is define as a:

`\begin{gathered} v(t)=(dh(t))/(dt) \\ \\ v(t)=h^(\prime)(t) \end{gathered}`

The value of the h'(t) is:

`\begin{gathered} h(t)=3300-54t-300e^(-0.18t) \\ \\ h^(\prime)(t)=-54-300(-0.18)e^(-0.18t) \end{gathered}`

Value is:

`\begin{gathered} h^(\prime)(t)=54e^(-0.18t)-54 \\ \\ v(t)=h^(\prime)(t) \\ \\ v(t)=54e^(-0.18t)-54 \end{gathered}`

At t=6 velocity is:

`\begin{gathered} v(t)=54e^(-0.18t)-54 \\ \\ v(6)=54e^(-0.18*6)-54 \\ \\ v(6)=54e^(-1.08)-54 \\ \\ v(6)=54*0.333-54 \\ \\ v(6)=18.333-54 \\ \\ v(6)=-35.66 \end{gathered}`

At t=6 velocity is -35.66 meters per sec.

(b)

Velocity is -45 meter per second then time is:

`\begin{gathered} v(t)=54e^(-0.18t)-54 \\ \\ v(t)=-45\text{ then time is:} \\ \\ -45=54e^(-0.18t)-54 \\ \\ 54-45=54e^(-0.18t) \\ \\ 9=54e^(-0.18t) \\ \\ (9)/(54)=e^(-0.18t) \\ \\ 0.166=e^(-0.18t) \end{gathered}`

Taking log both side then,

`\begin{gathered} \ln0.166=\ln e^(-0.18t) \\ \\ \ln0.166=-0.18t\ln e \\ \\ -1.79175=-0.18t \\ \\ t=(-1.79175)/(-0.18) \\ \\ t=9.95 \end{gathered}`

So, the time is 9.95 second