Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 7

The equation $s = t^3 - 4.5t^2 - 7t, t \geq 0$ represents the position function of a particle.

a.) At what time does the particle reach a velocity $5 m/s$?


$
\begin{equation}
\begin{aligned}

\text{velocity } =& \displaystyle s'(t) = \frac{ds}{dt}
\\
\\
=& \frac{d}{dt} (t^3) - 4.5 \frac{d}{dt} (t^2) - 7 \frac{d}{dt} (t)
\\
\\
=& 3t^2 - 4.5(2t) - 7(1)
\\
\\
=& 3t^2 - 9t - 7

\end{aligned}
\end{equation}
$


When velocity = $5 m/s$, solving for $t$ we have


$
\begin{equation}
\begin{aligned}

5 =& 3t^2 - 9t - 7
\\
\\
0 =& 3t^2 - 9t - 12

\end{aligned}
\end{equation}
$


Using Quadratic Formula

$t = 4s$ and $t = -1s$

The required time is $t = 4s$ since the position function $s$ is defined only for positive values of $t$.


b.) When is the acceleration ? What is the significance of this value of $t$?


$
\begin{equation}
\begin{aligned}

\text{acceleration } =& v'(t) = \frac{dv}{dt}
\\
\\
=& 3 \frac{d}{dt} (t^2) - 9 \frac{d}{dt} (t) - \frac{d}{dt} (7)
\\
\\
=& 3(2t) - 9(1) - 0
\\
\\
=& 6t - 9
\\
\\


\end{aligned}
\end{equation}
$


When $a(t) = 0$, we have


$
\begin{equation}
\begin{aligned}

0 =& 6t - 9
\\
\\
t =& \frac{3}{2} \text{ seconds}

\end{aligned}
\end{equation}
$


This means that, at time $\displaystyle t = \frac{3}{2}$ seconds, the velocity is not changing at this instant since the acceleration is zero.