Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 2

Suppose that a particle moves according to a Law of Motion

$f(t) = 0.01t^4 - 0.04t^3$, where $t$ is measured in
seconds and $s$ in feet.

a.) Determine the velocity at time $t$.


$
\begin{equation}
\begin{aligned}

\text{velocity } = s'(t) =& 0.01 \frac{d}{dt} (t^4) - 0.04 \frac{d}{dt} (t^3)
\\
\\
=& 0.01 (4t^3) - 0.04(3t^2)
\\
\\
=& 0.04t^3 - 0.12t^2

\end{aligned}
\end{equation}
$


b.) What is the velocity after $3 s$?


$
\begin{equation}
\begin{aligned}

\text{The velocity after $3 s$ is } v(3) =& 0.04(3)^3 - 0.12 (3)^2
\\
\\
v(3) =& 0 ft/s

\end{aligned}
\end{equation}
$


c.) When is the particle at rest?

The particle is at rest when $v(t) = 0$


$
\begin{equation}
\begin{aligned}

0 =& 0.04t^3 - 0.12t^2
\\
\\
0.12t^2 =& 0.04t^3
\\
\\
t =& 3 \text{ seconds}

\end{aligned}
\end{equation}
$


d.) When is the particle moving in the positive direction?

The particle is moving in the positive direction when $v(t) > 0$


$
\begin{equation}
\begin{aligned}

& 0.04t^3 - 0.12t^2 > 0
\\
\\
& 0.04t^2 (t - 3) > 0
\\
\\

\end{aligned}
\end{equation}
$


Assume $0.04t^2 (t - 3) = 0$ we have $t = 0$ and $t = 3$.

If we divide the interval into three parts we have, $t < 0$, $0 \leq t < 3$ and $t > 3$. However, we don't need to test $t < 0$ because we know that time can never be less than 0.

(i) $0 \leq t < 3$

Let's assume $t = 2: 0.04(2)^3 - 0.12(2)^2 = - 0.16 < 0$

(ii) $t > 3 $

Let's assume $t = 4: 0.04(4)^3 - 0.12(4)^2 = 0.64 > 0$

Therefore, we can conclude that the particle is speeding up at the
interval $t > 3$.

e.) Find the total distance traveled during the first $8 s$.

Since we know that the particle starts at and changes direction at
$t = 3$, we take the distance it traveled on the intervals $(0,3)$ and $(3,8)$


$
\begin{equation}
\begin{aligned}

\text{Total distance } =& |f(3) - f(0)| + |f(8) - f(3)| ; f(t) = 0.01t^4 - 0.04t^3
\\
\\
=& |-0.27 - 0| + |20.48 - (-0.27)|
\\
\\
=& |-0.27 | + |20.75|
\\
\\
=& 0.27 + 20.75
\\
\\
=& 21.02 ft

\end{aligned}
\end{equation}
$



f.) Illustrate the motion of the particle.







g.) Find the acceleration at time $t$ and after $3 s$.


$
\begin{equation}
\begin{aligned}

\text{acceleration } =& v'(t) = \frac{dv}{dt}
\\
\\
=& 0.04 \frac{d}{dt} (t^3) - 0.12 \frac{d}{dt} (t^2)
\\
\\
=& 0.04(3t^2) - 0.12 (2t)
\\
\\
=& 0.12t^2 - 0.24t
\\
\\
\text{acceleration at } t =& 3,
\\
\\
a(3) =& 0.12(3)^2 - 0.24(3)
\\
\\
a(3) =& 0.36 ft/s^2

\end{aligned}
\end{equation}
$


h.) Graph the position, velocity and acceleration functions $0 \leq t
\leq 8$.







i.) When is the particle speeding up? When is it slowing down?

Based from the graph, the particle is speeding up when the velocity
and acceleration have the same sign (either positive or negative) that is
at intervals $0 \leq t < 2$ and $t \geq 3$. On the other hand, the particle is
slowing down when the velocity and acceleration have opposite sign, that
is at intervals $2 \leq t < 3$.