When $we$ differentiate and produce a slope function, first derivative, $or$ instantaneous rate $of$ change, $it$

means something concrete $to$ the original function. $In$ the context $of$ modeling $of$ motion, the first

derivative $is$ the velocity.

The first derivative $of$ position $is$ velocity.

$s^{\text{'}}(t)=\frac{ds}{dt}=v(t)=$ velocity

$We$ often, though not always, start our equation $of$ motion $at$ time $t=0,$ and restrict our domain $to$

$t0.$

$In$ this problem, $we$ have the position function, and $we$ want $to$ find the time $at$ which a certain

velocity $is$ attained. $So$ we're not going $to$ evaluate the velocity function, but instead solve $an$

equation using the velocity.

Here's the problem statement:

The position function $of$ a particle $is$ given $bys(t)=t^3-4t^2-8t,t0$

where $sis$ measured $in$ meters and $tin$ seconds.

When does the particle reach a velocity $of5\frac{m}{s}?$

$To$ start, $it$ would $be$ nice $to$ have a name for this time. It's a special time, $so$ let's call $it{t}^{*}.$ We're

asking you $to$ solve $an$ equation, $v(t^{*})=5.$

$()$ Determine $v(t)by$ differentiating, and solve for $t^{*}.If$ you find a negative time, disregard $it$ due $to$

the domain restriction.

$t^{*}=$