Question

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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:

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

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:

$\begin{gathered} v(t)=54e^{-0.18t}-54 \\ \\ v(6)=54e^{-0.18\times6}-54 \\ \\ v(6)=54e^{-1.08}-54 \\ \\ v(6)=54\times0.333-54 \\ \\ v(6)=18.333-54 \\ \\ v(6)=-35.66 \end{gathered}$

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

(b)

Velocity is -45 meter per second then time is:

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

Taking log both side then,

$\begin{gathered} \ln0.166=\ln e^{-0.18t} \\ \\ \ln0.166=-0.18t\ln e \\ \\ -1.79175=-0.18t \\ \\ t=\frac{-1.79175}{-0.18} \\ \\ t=9.95 \end{gathered}$

So, the time is 9.95 second